1
Affine Varieties
We will begin following Kempf’s Algebraic Varieties, and eventually will do
things more like in Hartshorne. We will also use various sources for commutative
algebra.
What is algebraic geometry? Classically, it is the study of the zero sets of
polynomials.
We will now fix some notation. k will be some fixed algebraically closed field,
any ring is commutative with identity, ring homomorphisms preserve identity,
and a k-algebra is a ring R which contains k (i.e., we have a ring homomorphism
ι : k → R).
P ⊆ R an ideal is prime iff R/P is an integral domain.
Algebraic Sets
We define affine n-space, An = k n = {(a1 , . . . , an ) : ai ∈ k}.
Any f = f (x1 , . . . , xn ) ∈ k[x1 , . . . , xn ] defines a function f : An → k :
(a1 , . . . , an ) 7→ f (a1 , . . . , an ).
Exercise If f, g ∈ k[x1 , . . . , xn ] define the same function then f = g as
polynomials.
Definition 1.1 (Algebraic Sets). Let S ⊆ k[x1 , . . . , xn ] be any subset. Then
V (S) = {a ∈ An : f (a) = 0 for all f ∈ S}.
A subset of An is called algebraic if it is of this form.
e.g., a point {(a1 , . . . , an )} = V (x1 − a1 , . . . , xn − an ).
Exercises
1. I = (S) is the ideal generated by S. Then V (S) = V (I).
2. I ⊆ J ⇒ V (J) ⊆ V (I).
P
3. V (∪α Iα ) = V ( Iα ) = ∩V (Iα ).
4. V (I ∩ J) = V (I · J) = V (I) ∪ V (J).
Definition 1.2 (Zariski Topology). We can define a topology on An by defining
the closed subsets to be the algebraic subsets. U ⊆ An is open iff An \ U = V (S)
for some S ⊆ k[x1 , . . . , xn ].
Exercises 3 and 4 imply that this is a topology.
The closed subsets of A1 are the finite subsets and A1 itself.
Definition 1.3 (Ideal of a Subset). If W ⊂ An is any subset, then I(W ) =
{f ∈ k[x1 , . . . , xn ] : f (a) = 0 for all a ∈ W }
Facts/Exercises
1. V ⊆ W ⇒ I(W ) ⊆ I(V )
2. I(∅) = (1) = k[x1 , . . . , xn ]
3. I(An ) = (0).
1
Definition 1.4 (Affine Coordinate Ring). W ⊂ An is algebraic. Then A(W ) =
k[W ] = k[x1 , . . . , xn ]/I(W )
We can think of this as the ring of all polynomial functions f : W → k.
Definition 1.5 (Radical Ideal).
Let R be a ring and I ⊆ R be an ideal, then
√
the radical of I is the ideal I = {f√∈ R : f i ∈ I for some i ∈ N}
We call I a radical ideal if I = I.
Exercise
√
If I is an ideal, then I is a radical ideal.
Proposition 1.1. W ⊆ An any subset, then I(W ) is a radical ideal.
p
Proof. We have that
p I(W ) ⊆ I(Wi ).
Suppose f ∈ I(W ). Then f ∈ I for some i. That is, for all a ∈ W ,
f i (a) = 0. Thus, f (a)m = 0 = f (a). And so, f (a) ∈ I.
Exercises
1. S ⊆ k[x1 , . . . , xn ], then S ⊆ I(V (S)).
2. W ⊆ An then W ⊆ V (I(W )).
3. W ⊆ An is an algebraic subset, then W = V (I(W )).
√
√
4. I ⊆ k[x1 , . . . , xn ] is any ideal, then V (I) = V ( I) and I ⊆ I(V (I))
Theorem 1.2 (Nullstellensatz).
√ Let k be an algebraically closed field, and I ⊆
k[x1 , . . . , xn ] is an ideal, then I = I(V (I)).
√
Corollary 1.3. k[V (I)] = k[x1 , . . . , xn ]/ I.
To prove the Nullstellensatz, we will need the following:
Theorem 1.4 (Nöther’s Normalization Theorem). If R is any finitely generated k-algebra (k can be any field), then there exist y1 , . . . , ym ∈ R such that
y1 , . . . , ym are algebraically independent over k and R is an integral extension
of the subring k[y1 , . . . , ym ].
Proof is in Eisenbud and other Commutative Algebra texts.
Theorem 1.5 (Weak Nullstellensatz). Let k be an algebraically closed field,
and I ( k[x1 , . . . , xn ] any proper ideal, then V (I) 6= ∅.
Proof. We may assume without loss of generality that I is actually a maximal
ideal. Then R = k[x1 , . . . , xn ]/I is a field. R is also a finitely generated kalgebra, and so by Normalization, ∃y1 , . . . , ym ∈ R such that y1 , . . . , ym are
algebraically independent over k and that R is integral over k[y1 , . . . , ym ].
Claim: m = 0. Otherwise, y1−1 ∈ R is integral over k[y1 , . . . , ym ], and so then
−p
y1 +y11−p f1 +. . .+y1−1 fp−1 +fp = 0 for fi ∈ k[y1 , . . . , ym ]. Multiplying through
2
by y1p gives 1 = −(y1 f1 + . . . + y1p−1 fp−1 + y1p fp ) ∈ (y1 ), which contradicts the
algebraic independence.
Thus, the field R is algebraic over k. As k is algebraically closed, R = k.
k ⊆ k[x1 , . . . , xn ] → R = k
Let ai = the image in k of xi . Then xi −ai ∈ I. Thus, the ideal generated by
(x1 −a1 , . . . , xn −an ) ⊆ I ( k[x1 , . . . , xn ], and so they I = (x1 −a1 , . . . , xn −an ),
as it is a maximal ideal.
V (I) = V (x1 − a1 , . . . , xn − an ) = {(a1 , . . . , an )} =
6 ∅
Note: Any maximal ideal of k[x1 , . . . , xn ] is of the form (x1 −a1 , . . . , xn −an )
with ai ∈ k.
This is NOT true over R, look at the ideal (x2 + 1) ⊆ R[x]. It is, in fact,
maximal.
Now, we can prove the Nullstellensatz.
√
Proof. Let I ⊆ k[x1 , . . . , xn√
] be any ideal. We will prove that I(V (I)) = I.
It was an exercise that I ⊆ I(V (I)).
√
Let f ∈ I(V (I)). We must show that f ∈ I.
Looking at An+1 , we have the variables, x1 , . . . , xn , y. Set J = (I, 1 − yf ) ⊆
k[x1 , . . . , xn , y].
Claim: V (J) = ∅ ⊂ An+1 . This is as, if p = (a1 , . . . , an , p) ∈ V (J), then
(a1 , . . . , an ) ∈ V (I), then (1 − yf )(p) = 1 − bf (a1 , . . . , an ). But f (a1 , . . . , an ) =
0, so (1 − yf )(p) = 1, and so p ∈
/ V (J).
By the Weak Nullstellensatz, J = k[x1 , . . . , xn , y]. Thus 1 = h1 g1 + . . . +
hm gm + q(1 − yf ) where g1 , . . . , gm ∈ I and h1 , . . . , hm , q ∈ k[x1 , . . . , xn , y].
Set y = f −1 , and multiply by some big power of f to get a polynomial
equation once more.
N
Then f N = h̃1 g1 + . . . + h̃m gm where the
√ h̃i = f hi (x1 , . . . , xn ).
N
And so, we have f ∈ I, and thus, f ∈ I, by definition.
exercise: V (I(W )) = W in the Zariski Topology.
Irreducible Algebraic Sets
Recall: V (y 2 − xy − x2 y + x3 ) = V (y − x) ∪ V (y − x2 ), and V (xz, yz) =
V (x, y) ∪ V (z).
Definition 1.6 (Reducible Subsets). A Zariski Closed subset W ⊆ An is called
reducible if W = W1 ∪ W2 where Wi ( W and Wi closed.
Otherwise, we say that W is irreducible.
Proposition 1.6. Let W ⊆ An be closed. Then W is irreducible iff I(W ) is a
prime ideal.
Proof. ⇒: Suppose I(W ) is not prime. Then ∃f1 , f2 ∈
/ I(W ) such that f1 f2 ∈
W . Set W1 = W ∩ V (f1 ) and W2 = W ∩ V (f2 ).
As fi ∈
/ I(W ), W 6⊆ V (fi ), and so Wi ( W . Now we must show that W =
W1 ∪ W2 . Let a ∈ W . Assume a ∈
/ W1 . Then f1 (a) 6= 0, but f1 (a)f2 (a) = 0, so
f2 (a) = 0, thus a ∈ W2 .
⇐: Exercise
3
This gives us the beginning of an algebra-geometry dictionary.
Algebra
Geometry
k[x1 , . . . , xn ]
An
radical ideals
closed subsets
prime ideals
irreducible closed subsets
maximal ideals
points
In fact, this is an order reversing correspondence. So I ⊆ J ⇐⇒ V (I) ⊇
V (J), but this requires I, J to be radical.
Definition 1.7 (Nötherian Ring). A ring R is called Nötherian if every ideal
I ⊆ R is finitely generated.
Exercise A ring R is Nötherian iff every ascending chain of ideals I1 ⊆ I2 ⊆
. . . stabilizes, that is, ∃N such that IN = IN +1 = . . ..
Theorem 1.7 (Hilbert’s Basis Theorem). If R is Nötherian, then R[x] is
Nötherian.
Corollary 1.8. k[x1 , . . . , xn ] is Nötherian.
Definition 1.8 (Nötherian Topological Space). A topological space X is Nötherian
if every descending chain of closed subsets stabilizes.
Corollary 1.9. An is Nötherian.
W1 ⊇ W2 ⊇ . . . closed in An , then I(W1 ) ⊆ I(W2 ) ⊆ . . . ideals in k[x1 , . . . , xn ],
and so must stabilize.
Theorem 1.10. Any closed subset of a Nötherian Topological Space X is a
union of finitely many irreducible closed subsets.
Proof. Assume the result is false. ∃W a closed subset of X which is not the
union of finitely many irreducible closed sets.
As X is Nötherian, we may assume that W is a minimal counterexample. W
is not irreducible, and so W = W1 ∪ W2 , where Wi ( W and Wi closed. The Wi
can’t be counterexamples, as W is a minimal one, but then W = W1 ∪ W2 and
each Wi is the union of finitely many irreducible closed sets. Thus, W cannot
be a counterexample.
Corollary 1.11. Every closed W ⊆ An is union of finitely many irreducible
closed subsets.
Example: V (xy) = V (x) ∪ V (y) ∪ V (x − 1, y).
Recall: X is a topological space, then if Y ⊆ X is any subset, it has the
subspace topology, that is, U ⊆ Y is open iff ∃U 0 ⊆ X open such that U =
U0 ∩ Y .
Note:
1. W ⊆ Y is closed iff W = W ∩ Y , where the closure is in X.
2. X is Nötherian implies that Y is Nötherian in the subspace topology.
4
Definition 1.9 (Zariski Topology on X ⊆ An ). If X ⊆ An , then the Zariski
Topology on X is the subspace topology.
Definition 1.10 (Components of X). If X is any Nötherian Topological Space,
then the maximal irreducible closed subsets of X are called the (irreducible)
components of X.
Exercises
1. X has finitely many components.
2. X = the union of its irreducible components.
3. X 6= union of any proper subset of its components.
4. A topological space is Nötherian if and only if every subset is quasicompact.
5. A Nötherian Hausdorff space is finite.
Recall: X ⊆ An closed. Then A(X) = k[x1 , . . . , xn ]/I(X).
Definition 1.11. If f ∈ A(X), set D(f ) = {a ∈ X : f (a) 6= 0}.
Proposition 1.12. The sets D(f ) form a basis for the Zariski Topology on X.
Proof. Let p ∈ U ⊆ X, U open. Show that p ∈ D(f ) ⊆ U for some f ∈ A(X).
Z = X \U a closed subset of X, and Z ( Z ∪{p} implies that I(Z) ) I(Z ∪{p}).
Take any f ∈ I(Z) \ I(Z ∪ {p}). Then f vanishes on Z but not at p, so
p ∈ D(f ).
Regular Functions
Let X ⊆ An be an algebraic subset, and U ⊆ X is a relatively open subset
of X.
Definition 1.12 (Regular Function). A function f : U → k is called regular if f
is locally rational. That is, ∃ open cover U = ∪α Uα and functions pα , qα ∈ A(X)
such that ∀a ∈ Uα , qα (a) 6= 0 and f (a) = pα (a)/qα (a).
We define k[U ] to be the set of regular functions from U to k.
Note:
1. k[U ] is a k-algebra.
2. A(X) ⊆ k[X].
Example
Let X = V (xy − zw) ⊆ A4 . f : U → k can be defined by f = x/w on D(w)
and f = z/y on D(y). Thus, f ∈ k[U ].
Exercise; 6 ∃p, q ∈ A(X) such that q(a) 6= 0 and f (a) = p(a)/q(a) for all
a ∈ U.
5
Lemma 1.13. Let q1 , . . . , qn ∈ A(X). Then D(q1 ) ∪ . . . ∪ D(qn ) = X iff
(q1 , . . . , qm ) = (1) = A(X).
P
Proof. ⇐: 1 =
hi qi , hi ∈ A(X), then the qi cannot all vanish at any point,
and so we are done.
⇒: Take Qi ∈ k[x1 , . . . , xn ] such that qi = Qi ∈ A(X). D(q1 )∪. . .∪D(qm ) =
X, so X ∩ V (Q1 ) ∩ . . . ∩ V (Qm ) = ∅ = V (I(X), Q1 , . . . , Qm ) = ∅, and so,
by the weak nullstellensatz, (I(X), Q1 , . . . , Qm ) = (1) ⊆ k[x1 , . . . , xn ], and so
(q1 , . . . , qm ) = (1) = A(X).
Theorem 1.14. Let X ⊆ An be an algebraic set. Then k[X] = A(X).
Proof. Let f ∈ k[X]. Then X = U1 ∪ . . . ∪ Um and there are pi , qi ∈ A(X) such
that qi 6= 0 and f = pi /qi on Ui .
We can refine the open cover such that each Ui = D(gi ) for some gi . Note:
f = pi /qi = pqiiggii on Ui = D(gi ) = D(gi qi ). We can replace pi with pi gi and qi
by qi gi .
2
2
Then we can assume that Ui is D(qi ) =
i ). Thus, X = D(q1 ) ∪ . . . ∪
PD(q
m
2
2
D(qm ). By the lemma, we know that 1 = i=1 hi qi , hi ∈ A(X). Note qi2 f =
qi pi on Ui , and
of Ui .
Pqmi = 0 outside
Pm
f = 1f = i=1 hi qi2 f = i=1 hi qi pi , so f ∈ A(X).
Definition 1.13 (Spaces With Functions). A space with functions (SWF) is
a topological space X together with an assignment to each open U ⊆ X of a
k−algebra k[U ] consisting of functions U → k. These are called regular functions. It must also satisfy the following:
1. If U = ∪Uα is an open cover and f : U → k any function, then f is
regular on U iff f |Uα is regular on Uα for all α.
2. If U ⊆ X is open, f ∈ k[U ], then D(f ) = {a ∈ U : f (a) 6= 0} is open and
1
f ∈ k[D(f )].
Note: OX (U ) = k[U ] is another common notation.
Examples:
1. Algebraic sets. These are called Affine Algebraic Varieties
2. M is a differentiable manifold, k = R, k[U ] = {C ∞ functions U → R}.
3. X is a SWF, U ⊆ X open subset, then U is a SWF. OU (V ) = OX (V ).
Definition 1.14 (Morphism of SWFs). Let X, Y be SWFs, then a morphism
ϕ : X → Y is a continuous map which pulls back regular functions to regular
functions. i.e., if V ⊆ Y is open, f ∈ OY (V ), then ϕ∗ (f ) ∈ OX (ϕ−1 (V )),
ϕ∗ (f ) = f ◦ ϕ.
Definition 1.15 (Isomorphism). ϕ : X → Y is an isomorphism if ϕ is a
morphism and ∃ a morphism ψ : Y → X such that ϕ ◦ ψ = idY and ψ ◦ ϕ = idX .
6
Exercises
1. The id function of a SWF is a morphism.
2. Compositions of morphisms are morphisms.
3. Let X be any SWF and Y ⊆ An closed, that is, an affine variety. Then
f = (f1 , . . . , fn ) : X → Y , is a morphism iff fi ∈ k[X] for all i.
Example: A1 \ {0} is (isomorphic to) an affine variety defined by V (1 − xy).
Localization
Let R be a ring and S ⊆ R multiplicatively closed subset. That is, s, t ∈
S ⇒ st ∈ S and 1 ∈ S.
We define S −1 R = {f /s : f ∈ R, s ∈ S}. We consider f /s to be the same
element as g/t iff there exists u ∈ S such that u(f t − sg) = 0. This is a ring by
f g
fg
f
g
f t+gs
s t = st and s + t =
st .
Exercise: Check these assertions.
Special Case: If f ∈ R, then Rf = S −1 R where S = {f n : n ∈ N}. In fact,
Rf ∼
= R[y]/(1 − f y).
Definition 1.16 (Reduced Ring).pR is a reduced ring iff f n = 0 implies f = 0
for all f ∈ R. Equivalently, (0) = (0).
Facts:
1. R reduced implies S −1 R is reduced
√
2. R/I reduced iff I = I
Proposition 1.15. Let X ⊆ An be a closed affine variety and f ∈ A(X). Then
D(f ) if an affine variety, with affine coordinate ring A(X)f .
Proof. Let I = I(X) ⊆ k[x1 , . . . , xn ] define J = (I, yf − 1) ⊆ k[x1 , . . . , xn , y].
Let φ : D(f ) → V (J) ⊆ An+1 by (a1 , . . . , an ) 7→ (a1 , . . . , an , f (a1 , . . . , an )−1 ).
Note: φ is an isomorphism.
What remains is to compute the coordinate ring. A(X) is reduced, and
so A(X)
√ f is reduced. A(X)f = k[x1 , . . . , xn , y]/J, so J is a radical ideal, so
J = J = I(V (J)). Therefore, k[D(f )] = k[V (J)] = k[x1 , . . . , xn , y]/J =
A(X)f .
Definition 1.17 (Prevariety). A prevariety is a space with functions X such
that X has a finite open cover X = U1 ∪ . . . ∪ Um where Ui is an affine variety.
Example: Any affine variety is a prevariety.
Exercise: Any prevariety is a Nötherian Topological Space
Example: An open subset of a prevariety is a prevariety. This follows from
the previous proposition and the fact that principle open sets are a basis for the
topology of any affine variety.
7
Proposition 1.16. X is a space with functions, Y ⊆ An an affine variety, we
have a 1-1 correspondence:
{morphisms X → Y } ⇐⇒ {k-algebra homomorphism A(Y ) → k[X]} by
ϕ ⇐⇒ ϕ∗
Proof. Note φ 7→ φ∗ is a well defined map. Write A(An ) = k[y1 , . . . , yn ], then
I(Y ) ⊆ k[y1 , . . . , yn ]. Then y¯i is the image of yi in A(Y ). Assume that α :
A(Y ) → k[X] is a k-algebra homomorphism. We define φ : X → An by φ(x) =
(φ1 (x), . . . , φn (x)).
If f ∈ I(Y ) then f (y¯1 , . . . , y¯n ) = 0, so f (φ1 , . . . , φn ) = α(f (y¯1 , . . . , y¯n ) ) = 0,
and so φ(X) ⊆ Y . Note, φ∗ (y¯i ) = yi ◦ φ = φi = α(y¯i ). Thus, φ∗ = α.
If φ : X → Y is a morphism, then φi = yi ◦ φ = φ∗ (y¯i )
Thus, φ is the morphism that we construct from φ∗ .
Corollary 1.17. Two affine varieties are isomorphic iff their affine coordinate
rings are isomorphic as k-algebras
Exercise: An \ {(0, . . . , 0)} is not affine for n ≥ 2.
Proposition 1.18. We have a one-to-one correspondence between affine varieties and reduced finitely generated k-algebras, up to isomorphism, by X 7→ k[X].
Proof. Last time, we proved that two affine varieties are isomorphic iff their
coordinate rings are isomorphic. Thus, X 7→ k[X] is injective.
Let R be a finitely generated reduced k-algebra generated by r1 , . . . , rn ∈
R. There is a k-alg homomorphism φk[x1 , . . . , xn ] → R by xi 7→ ri which is
surjective. Set I = ker φ and let X = V (I) ⊆ An .
I is radical, as R is reduced, so k[X] = k[x1 , . . . , xn ]/I ' R.
Note: Assume m ⊆ R is a maximal ideal, then φ : k[x1 , . . . , xn ] → R as
in proof, then M = φ−1 (m) is maximal, and M = (x1 − a1 , . . . , xn − an ).
R/m = k[x1 , . . . , xn ]/M = k.
Canonical Construction
Let R be a finitely generated reduced k-algebra. Then define Spec −m(R) =
{m ⊆ R max ideals}.
The topology will be that the closed sets V (I) = {m ⊇ I|I ⊆ R and ideal }.
Let f ∈ R. We define f : Spec −m(R) → k by f (m) =image of f in
R/m = k. So f is a function from Spec −m to k.
I.E., f (m) ∈ k ⊆ R is the unique element such that f − f (m) ∈ m.
Finally, if U ⊆ Spec −m is open, f : U → k is some function, then f is
regular if f is locally of the form f (m) = p(m)/q(m) where p, q ∈ R.
Exercise: Spec −m(R) ∼
= X, where X is the affine variety with coordinate
ring R, as spaces with functions.
Subspaces of SWFs
Let X be any space with functions, and Y ⊆ X any subset. Then give Y an
”inherited” SWF structure as follows:
We give Y the subspace topology, and if U ⊆ Y is open and f : U → k is a
function, then f is regular iff f can be locally extended to a regular function on
8
X. That is, for every point y ∈ U , there is an open subset U 0 ⊆ X containing
y and F ∈ OX (U 0 ) such that f (x) = F (x) for all x ∈ U ∩ U 0 .
Exercises
1. Y is a SWF
2. i : Y → X the inclusion map is a morphism.
3. Let Z be a SWF, φ : Z → Y function. Then φ is a morphism iff i ◦ φ is a
morphism.
4. The SWF structure on Y is uniquely determined by (2) and (3) together.
5. Let Z ⊆ Y ⊆ X. Then Z inherits the same structure from Y and X.
Example: X ⊂ An an algebraic set inherits structure from An . If Y ⊆ X is
closed, then Y inherits structure from X (or An ).
Proposition 1.19. A closed subset of a prevariety is a prevariety.
Proof. Let X be a prevariety, and Y ⊆ X a closed subset. X = U1 ∪ . . . ∪ Um
where Ui are open affine subsets of X.
Ui ∩ Y is a closed subset of Ui , which implies that Ui ∩ Y is affine, and so Y
has the open cover (U1 ∩ Y ) ∪ . . . ∪ (Un ∩ Y ), and so is a prevariety.
Projective Space
Theorem 1.20. Two distinct lines in the place intersect in exactly one point.
(except when parallel)
Theorem 1.21. A line meets a parabola in exactly two points. (except when
false)
We need projective space to remove the bad cases.
Definition 1.18 (Projective Space). Define an equivalence relation on An+1 \
{0} by
(a0 , . . . , an ) ∼ (λa0 , . . . , λan ) where λ ∈ k ∗ = k \ {0}.
So Pn = (An+1 \ {0})/ ∼ and π : An+1 \ {0} → Pn the projection.
There is a topology on Pn given by U ⊆ Pn is open iff π −1 (U ) ⊆ An+1 is
open.
The regular functions on Pn are f : U → k such that π ∗ (f ) = π ◦ f :
−1
π (U ) → k is regular.
Thus, Pn is a SWF called Projective Space. Note: Pn = {lines through the
origin in An+1 }, and this method of thinking is often very helpful.
We will use the notation (a0 : . . . : an ) for the image π(a0 , . . . , an ) ∈
Pn . If f ∈ k[x0 , . . . , xn ] is a homogeneous polynomial of total degree d, then
f (λa0 , . . . , λan ) = λd f (a0 , . . . , an ). Thus it is well-defined to ask if f (a0 : . . . :
an ) = 0 or not.
9
Definition 1.19. D+ (f ) = {(a0 : . . . : an ) ∈ Pn : f (a0 : . . . : an ) 6= 0}.
Theorem 1.22. Pn is a prevariety.
Proof. Let Ui = D+ (xi ) ⊆ Pn for 0 ≤ i ≤ n.
Claim: Ui ' An .
φ : An → Ui : (a0 , . . . , ai−1 , ai+1 , . . . , an ) 7→ (a0 : . . . : ai−1 : 1 : ai+1 : . . . : an )
n
ψ : Ui → A : (a0 : . . . : an ) 7→
aˆi
an
a0
,..., ,...,
ai
ai
ai
`
`
Note:Pn = D+ (x0 ) V+ (x0 ) = An Pn−1 , that is, An = {(1 : a1 : . . . an )}
the usual n-space and Pn−1 = {0 : a1 : . . . : an } points at ∞.
The points at ∞ correspond to lines through the origin in An , that is, we
can think of them as being directions.
Example: P2 has ”homogeneous coordinate ring” k[x, y, z]. We can think of
that as A2 = D+ (z) ⊂ P2 and we know that k[A2 ] = k[x/z, y/z]. We want to
intersect a parabola with a line.
The vertical line is V (x/z − 1) = V+ (x−z) and the parabola is V (y/z − (x/z)2 ) =
V+ (yz − x2 ).
And so, V+ (x − z) ∩ V+ (yz − x2 ) = {(1 : 1 : 1), (0 : 1 : 0)}, where (0 : 1 : 0)
is the point at infinity in the direction ”up”.
Exercise: k[Pn ] = k.
Exercise: X a SWF, φ : Pn → X a function, then φ is a morphism iff
φ ◦ π : An+1 \ {0} → X is a morphism.
Definition 1.20 (Projective Coordinate Ring of Pn ). We define k[x0 , x1 , . . . , xn ]
to be the coordinate ring of Pn . An ideal I ⊆ k[x0 , . . . , xn ] is homogeneous if it
is generated by homogeneous polynomials. Equivalently, f ∈ I iff each homogeneous component is in I.
Definition 1.21. If W ⊆ Pn is a subset, then I(W ) = I(π −1 (W )) ⊆ k[x0 , . . . , xn ].
Notice that I(W ) is homogeneous. Let f = f0 + . . . + fd ∈ I(W ), fi a form of
degree i, then (a0 : . . . : an ) ∈ W , so 0 = f (λa0 , . . . , λan ) = f0 (a0 , . . . , an ) +
. . . + λd fd (a0 , . . . , an ). As this is true for all λ, fi (a0 , . . . , an ) = 0 for all i, and
so fi ∈ I(W ).
Definition 1.22. If I ⊆ k[x0 , . . . , xn ] is a homogeneous ideal, then define
V+ (I) = {(a0 : . . . , an ) ∈ Pn : f (a0 , . . . , an ) = 0 for all f ∈ I}
Theorem 1.23 (Projective Nullstellensatz). If I ⊆ k[x0 , . . . , xn ] is a homogeneous ideal, then
√
1. V+ (I) = ∅ ⇒ (x0 , . . . , xn )N ⊆ I for some N > 0. That is, I = (1) or
(x0 , . . . , xn ).
10
2. V+ (I) 6= ∅ then I(V+ (I)) =
√
I.
Proof.
1. V√
+ (I) = ∅ ⇐⇒ V (I) = ∅ or V (I) = {0}. By the regular nullstellensatz, I = I(V (I)) = (1) or (x0 , . . . , xn ).
2. V+ (I) 6= ∅. Then π −1 √
(V+ (I)) = π −1 (V+ (I)) ∪ {0} = V (I) ⊆ An+1 . So
I(V+ (I)) = I(V (I)) = I.
This gives us a 1-1 correspondence between closed subsets of Pn and radical
homogeneous ideals in k[x0 , . . . , xn ] except for (x0 , . . . , xn ). This ideal is often
called the irrelevant ideal.
Definition 1.23 (Locally Closed subset). X is a topological space, W ⊆ X a
subset is locally closed if it is the intersection of an open set in X and a closed
set in X.
Note: A locally closed subset of a prevariety is a prevariety.
Terminology: a projective variety is any closed subset of Pn considered as
a space with functions. A Quasi-projective variety is a locally closed subset of
Pn . An affine variety is a closed subset of An . A quasi-affine variety is a locally
closed subset of An .
We notice that anything affine is also quasi-affine and anything quasi-affine
is quasi-projective. Something that is projective will also be quasi-projective.
Exercise: Pn is not quasi-affine for n ≥ 1. Later: If X is both projective and
quasi-affine, then X is finite.
Definition 1.24 (Projective Coordinate Ring). X ⊆ Pn is a closed projective
variety, then the projective coordinate ring of X = k[x0 , . . . , xn ]/I(X).
Warning: This definition depends on the embedding of X in Pn .
Example: φ : P1 → P2 by φ(a : b) = (a2 : ab : b2 ). This is a morphism. In
fact, it is an isomorphism of P1 and V+ (xz − y 2 ), but the coordinate ring of P1
is k[s, t] and the coordinate ring of V+ (xz − y 2 ) is k[x, y, z]/(xz − y 2 ). These
two rings are NOT isomorphic as k-algebras.
Definition 1.25 (Projective Closure of an affine variety). X ⊆ An is affine.
Then we know that An = D+ (x0 ) ⊆ Pn , and X ⊂ An ⊆ Pn makes X a quasiprojective variety, so we take X̄ = the closure of X in Pn .
I = I(X) ⊆ k[x1 , . . . , xn ] if f = f0 + . . . + fd ∈ k[x1 , . . . , xn ] where fi is a
form of degree i. Then we define f ∗ = xd0 f0 + xd−1
f1 + . . . + fd ∈ k[x0 , . . . , xn ].
0
And I ∗ is the ideal generated by {f ∗ : f ∈ I} in k[x0 , . . . , xn ].
Exercise: I(X̄) = I(X)∗ .
Example: I = (y −x2 , z −x2 ) ⊆ k[x, y, z]. Then X = V (I) ⊆ A3 = D+ (w) ⊂
Pn . I(X) = I ∗ = (yw − x2 , y − z) ⊇ (wy − x2 , wx − x2 )
So V+ (wy − x2 , wz − x2 ) = X̄ ∪ V+ (x, w).
We now recall that a graded ring is a ring R with decomposition R = ⊕d≥0 Rd
as an abelian group such that Rd · Re ⊆ Rd+e .
11
e.g., R = k[x0 , . . . , xn ]. f ∈ Rd ⇒ Rf is a Z-graded ring g ∈ Rp implies that
g/f m ∈ Rf is homogeneous of degree p − md.
Definition 1.26. R(f ) = {homogeneous elements of degree zero} = (Rf )0 =
{g/f m : g ∈ Rdm }.
Exercise: f ∈ k[x0 , . . . , xn ] homogeneous implies that k[x0 , . . . , xn ](f ) is a
finitely generated reduced k-algebra.
Theorem 1.24. f ∈
/ k ⇒ D+ (f ) ⊆ Pn is affine and in fact k[D+ (f )] =
k[x0 , . . . , xn ](f ) .
Proof. k[D+ (f )] = {h ∈ k[D(f )] : h(λx) = h(x), ∀λ ∈ k ∗ , x ∈ D(f )}.
If h ∈ k[D(f )] = k[x0 , . . . , xn ]f , h = g/f m , g ∈ k[x0 , . . . , xn ].
g
g
(λa0 , . . . , λan ) = m (a0 , . . . , an ) ⇐⇒ g homogeneous of degree md
fm
f
Therefore, k[D+ (f )] = k[x0 , . . . , xn ](f ) .
The identity map k[D+ (f )] → k[D+ (f )] gives a morphism φ : D+ (f ) →
Spec −m(k[D+ (f )]) by φ(x) = Mx where Mx = I({x}) ⊆ k[D+ (f )]
Observe that if x, y ∈ D+ (f ), x 6= y then ∃ homogeneous g ∈ k[x0 , . . . , xn ]
such that deg(g) = d and g(x) = 0 with g(y) 6= 0. So Mx 6= My . fg ∈ Mx , ∈
/ My .
xd
Thus, φ is injective. Set hi = fi ∈ k[D+ (f )] for 0 ≤ i ≤ n.
Ui = D(hi ) ⊆ D+ (f ), Vi = D(hi ) ⊆ Spec −m(k[D+ (f )]) = {m 63 hi }.
Now we must check that D+ (f ) = ∪ni=0 Ui and Spec −m(k[D+ (f )]) = ∪ni=0 Vi .
It is enough to prove that φ : Ui → Vi is an isomorphism for all i.
D+ (xi ) ⊆ Pn is affine. k[D+ (xi )] = k[x0 /xi , . . . , xn /xi ] = k[x0 , . . . , xn ](xi ) .
Thus, Ui = D(f /xdi ) ⊆ D+ (xi ) is affine, so k[Ui ] = (k[x0 , . . . , xn ](xi ) )f /xdi ,
which is k[x0 , . . . , xn ](x,f ) = k[D+ (f )]hi = k[Vi ]. Thus, Ui ' Vi .
Example: f = xz − y 2 ∈ k[x, y, z]. X = D+ (f ) ⊆ P2 , R = k[x, y, z](f ) . R
is generated by A = x2 /f , B = y 2 /f , C = z 2 /f , D = xy/f , E = yz/f , and
F = xz/f .
So X ' V (AB − D2 , AC − F 2 , BC − E 2 , F − B − 1) ⊆ A6 by (x : y : z) 7→
(A, B, C, D, E, F ).
Exercise:X ⊆ Pn a projective variety f ∈ R = k[x0 , . . . , xn ]/I(X) is homogeneous, then D+ (f ) ⊆ X is affine with affine coordinate ring k[D+ (f )] = R(f ) .
2
Algebraic Varieties
Products
Let X, Y be two sets. Then X × Y , the cartesian product, is the set {(x, y) :
x ∈ X, y ∈ Y }.
What is X ×Y , really? Well, it is a set with projection πX : X ×Y → X and
πY : X × Y → Y . This set with the projections satisfies a universal property in
the category of sets.
12
For any set Z with arbitrary functions f : Z → X and g : Z → y, there
exists a unique function φ : Z → X × Y such that f = πX ◦ φ and g = πY ◦ φ.
...... X
........... ...
...........
f........................π
X ...............
.
.
.
.
.
.
.
.
.
.
.
.
........
.....
..........
........... ∃!
.
Z ............................... ....... ......... X × ..Y
.....
...........
........... g
..... πY
...........
.....
...........
......
...........
............. ........
......
Y
Definition 2.1 (Product of SWFs). Let X, Y be spaces with functions. A
product of X and Y is a SWF called X × Y with morphism πX : X × Y → X
and πY : X × Y → Y which satisfies the above universal property except with
”morphisms” rather than ”functions”.
0
, πY0 ) are two products of X
Exercise: Assume that (P, πX , πY ) and (P 0 , πX
and Y . Then they are isomorphic by unique isomorphism. (See homework
problem)
Example: A1 × A1 = A2 . NOTE: A2 does not have the product topology!
General Construction: X, Y spaces with functions. Then X × Y = {(x, y) :
x ∈ X, y ∈ Y } is the point set. If U ⊂ X and V ⊂ Y are open, then U × V ⊂
−1
X × Y is open, as it is πX
(U ) ∩ πY−1 (V ).
Pn
Let g1 , . . . , gu ∈ OX (U ) and h1 , . . . , hn ∈ OY (V ) set f (u, v) = i=1 gi (u)hi (v).
Then f : U × V → k must be regular. Thus, DU ×V (f ) = {(u, v) ∈ U × V :
f (u, v) 6= 0} must be open in X × Y . And so, we define our topology by
S ⊆ X × Y is open iff it is a union of sets DU ×V (f ). The regular functions
F : S → k are the functions that can locally be written as f 0 (u, v)/f (u, v) on
some DU ×V (f ).
⊆ X open and V ⊆ Y open and f (u, v) =
PnThat is, ∀(x, y) ∈0 S ∃U P
m
0
0
0
0
g
(u)h
(v)
and
f
(u,
v)
=
i
i=1 i
j=1 gj (u)hj (u) with gi , gj ∈ OX (U ) and hi , hj ∈
0
OY (V ) such that (x, y) ∈ DU ×V (f ) ⊆ S and F (u, v) = f (u, v)/f (u, v) for all
(u, v) ∈ DU ×V (f ).
Exercises (for X × Y above)
1. X × Y is an SWF
2. πX : X × Y → X and πY : X × Y → Y are morphisms.
3. X × Y is the product of X and Y .
Remark: X, Y SWFs, and U ⊆ X, V ⊆ Y are arbitrary subsets, U, V have
inherited structure as SWFs. Then U × V has the product space with functions
structure and subspace SWF structure U × V ⊆ X × Y . These are in fact the
same, due to the universal properties.
For now, we call U × V the product.
We obtain the following diagram.
.
U ....................................................... X
πU .................
πX .................
....
....
.....
.....
φ
i
...............................................
....... ....... ..............
U × .V
X × ..Y
Z
.....
.....
..... πV
..... πY
.....
.....
.
......
........
V
......
........
........................................................
Y
13
φ is a morphism iff i ◦ φ is one, and so we see that the two structures are the
same.
Definition 2.2 (Separated SWF). A SWF X is separated if ∀ SWFs Y and
morphisms f, g : Y → X the set {y ∈ Y : f (y) = g(y)} ⊆ Y is closed.
Example: Let X = (A1 \ {0}) ∪ {O1 , O2 }. We can define φi : A1 → X by
a a 6= 0
taking a 7→
Oi a = 0
1
We define a topology by U ⊆ X is open iff φ−1
i (U ) ⊆ A is open for all i. A
−1
∗
function f : U → k is regular iff φi (f ) = f ◦ φi : φi (U ) → k is regular for all i.
X is a prevariety as X = φ1 (A1 ) ∪ φ2 (A1 ) and φi (A1 ) ' A1 . However, it is
not separated, as {a ∈ A1 : φ1 (a) = φ2 (a)} = A1 \ {0} is not closed in A1 .
Definition 2.3 (Algebraic Variety). An algebraic variety is a separated prevariety.
Exercise:
1. Any subspace of a separated SWF is separated.
2. A product of separated SWFs is separated.
Remark: If X is any SWF, then ∆ : X → X × X : x 7→ (x, x) is a morphism.
Now we set ∆X = ∆(X) ⊆ X × X. Then ∆ : X → ∆X is an isomorphism.
Lemma 2.1. X is separated iff ∆X ⊆ X × X is closed.
Proof. ⇒: πi : X × X → X be the projections. ∆X = {z ∈ X × X : π1 (z) =
π2 (z)} is closed.
⇐: Let Y be a SWF, f, g : Y → X maps. Define φ : Y → X × X by
φ(y) = (f (y), g(y)) is a morphism. Now {y ∈ Y : f (y) = g(y)} ⇒ φ−1 (∆X ) is
closed.
Exercise: A topological space X is Hausdorff iff ∆X ⊆ X × X is closed.
NB: Product topology!
Exercise: An × Am = An+m .
Proposition 2.2. All affine varieties are varieties.
Proof. Enough to show that An itself is separated.
∆An ⊆ An × An = A2n is closed, as k[A2n ] = [x1 , . . . , xn , y1 , . . . , yn ], so
∆An = V ({xi − yi }).
Products of Affine Varieties
Lemma 2.3. Let A, B be finitely generated reduced k-algebras. Then A ⊗k B
is a finitely generated reducted k-algebra and k algebraically closed.
Recall: ring structure on A ⊗ B by (a1 ⊗ b1 )(a2 ⊗ b2 ) = a1 a2 ⊗ b1 b2 .
14
Proof. If A is generated by a1 , . . . , an and B is generated by b1 , . . . , bm then
A ⊗ B is generated by a1 ⊗ 1, . . . , an ⊗ 1, 1 ⊗ b1 , . . . , 1 ⊗ bm . For example,
k[x1 , . . . , xn ] ⊗ k[y1 , . . . , ym ] = k[x1 , . . . , xn , y1 , . . . , ym ].
Let X, Y be affine varieties such that k[X] = A, k[Y ] = B. We define
φ : A ⊗k B →the set of all functions X × Y → k by f ⊗ g 7→ (x, y) 7→ f (x)g(x)
is a k-algebra homomorphism.
Pn
φ is injective (which implies that A⊗B is reduced): Suppose φ ( i=1 fi ⊗ gi ) =
0.
PnWLOG we can assume g1 , . . . , gn are linearly independent. Let x ∈ X. Then
i=1 fi (x)gi = 0 ∈ B, but {gi } is linearly independent so fi (x) = 0 for all i
and x, thus fi = 0 ∈ A.
Theorem 2.4.
1. If X, Y are affine then X × Y is affine and k[X × Y ] =
k[X] ⊗k k[Y ].
2. A product of prevarieties is a prevariety.
Proof. 1 ⇒ 2: X, Y prevarieties, X = ∪Ui , Y = ∪Vj with Ui , Vj affine. Then
X × Y = ∪i,j Ui × Vj affine.
Now, we must only prove 1. Set P = Spec-m(k[X]⊗k[Y ]), We have k-algebra
homomorphisms k[X] → k[X] ⊗ k[Y ] : f 7→ f ⊗ 1 and k[Y ] → k[X] ⊗ k[Y ] : g 7→
1 ⊗ g. These give morphism πX : P → X and πY : P → Y . Claim: P = X × Y .
Let Z be a SWF, p : Z → X and q : Z → Y .
Define k-alg homomorphism k[X] 7→ k[Y ] → k[Z] by f ⊗ g 7→ p∗ (f )q ∗ (g).
This gives us a morphism φ : Z → P demonstrating that P is the product.
Remark: If Y is affine and X ⊆ Y is closed, then k[X] ' k[Y ]/I(X).
On the other hand, assume X, Y are affine, ϕ : X → Y is a morphism and
ϕ∗ : k[Y ] → k[X] is surjective. Then set I = ker(ϕ∗ ) ⊆ k[Y ], we get
ϕ∗
............................
k[Y ] .................................................. k[X]
Y ............................ϕ
X
.
.......
.....
.....
...........
.....
.....
.....
.
'
.
.....
......
..
.....
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
......
..... '
.....
inj .....
k[Y ]/I
V (I)
Therefore ϕ is an embedding of X as a closed subset of Y .
Recall: ∆ : X → X × X : x 7→ (x, x) gives X ' ∆X = ∆(X) = {(x, x) : x ∈
X}.
Proposition 2.5. A prevariety X is separated iff ∀ open affine U, V ⊆ X, U ∩V
is affine and k[U × V ] = k[U ] ⊗ k[V ] → k[U ∩ V ] = k[∆U ∩V ] is surjective.
Proof. ⇒: U ∩ V ' ∆U ∩V = ∆X ∩ (U × V ) ⊆ U × V is closed, thus U ∩ V is
affine and k[U × V ] → k[U ∩ V ] is surjective.
⇐: If U, V, U ∩ V are affine and k[U × V ] → k[U ∩ V ] is surjective, then
∆ : U ∩V → U ×V is an inclusion of closed subsets. Thus, ∆X ∩(U ×V ) ⊆ U ×V
closed. So if X × X = ∪U × V is an open cover then ∆X ⊆ X × X is closed.
Exercises
1. X is a prevariety such that ∀x, y ∈ X there is an open affine U ⊆ X such
that x, y ∈ U . Then X is separated.
15
2. Pn has this property.
Corollary 2.6. Quasi-projective varieties are separated.
We want to show that the products of projective varieties are again projective.
Let X ⊆ Pn and Y ⊆ Pm are closed. Then X × Y ⊆ Pn × Pm is closed. It
is enough to show that Pn × Pm is projective, that is, Pn × Pm ⊂ PN is closed.
Segre Map: If N = (n + 1)(m + 1) − 1 = nm + n + m, then we can define
s : Pn × Pm → PN : (x0 : . . . : xn ) × (y0 : . . . : ym ) 7→ (x0 y0 : x0 y1 : . . . : x0 ym :
x1 y0 : . . . : xn ym ).
We call the projective coordinates on PN as zij for 0 ≤ i ≤ n, 0 ≤ j ≤ m.
Exercise: s : Pn × Pm → V+ ({zij zpq − ziq zpj }) ⊆ PN
2
Note: Pn × Pn ⊆ Pn +2n is closed.
Exercise: ∆Pn = V+ ({zij − zji }) ⊆ PN .
Complete Varieties: analogues of compact manifolds
Definition 2.4 (Complete). A variety X is complete if for any variety Y , the
projection πY : X × Y → Y is closed. (i.e.: Z ⊆ X × Y is closed implies that
πY (Z) ⊆ Y is closed)
Note: 1) closed subsets of complete varieties are complete.
2) Products of complete varieties are complete.
Examples:Points are complete.
A1 is not complete, as Z = V (xy − 1) ⊆ A1 × A1 = A2 = X × Y is such that
πY (Z) is not closed, as it is A1 \ {0}
Proposition 2.7. Let ϕ : X → Y be a morphism of varieties. If X is complete,
then ϕ(X) is closed in Y and is complete.
Proof. Γ(ϕ) = {(x, ϕ(x)) ∈ X × Y : x ∈ X} ⊆ X × Y = (ϕ × 1)−1 (∆Y )
As Y is separated, Γ(ϕ) ⊆ X × Y is closed. X is complete implies that
ϕ(X) = πY (Γ(ϕ)) ⊆ Y is closed.
Now, let Z ⊆ ϕ(X) × Y 0 be closed. Then
ϕ×1
0 ......................... ϕ(X) × Y 0
X ×Y
.
.....
...
.....
.....
...
.....
...
.....
...
.....
.....
...
..... Y 0
...
.....
...
.....
...
.....
.....
...
.....
..
.....
...... ..........
.
.......
π̃
πY 0
Y0
W = (ϕ × 1) (Z) ⊆ X × Y 0 is closed, πY 0 (Z) = πY 0 ((ϕ × 1)(W )) = π̃Y 0 (W )
is closed.
−1
Exercise: ϕ : X → Y is a continuous map of topological spaces then X is
irreducible implies that ϕ(X) is irreducible.
Proposition 2.8. If X is an irreducible complete variety then k[X] = k.
16
Proof. Let f ∈ k[X]. f : X → A1 a morphism. As X is complete, f (X) must
be is irreducible, closed and complete. Thus, it must be a point. Thus, f is
constant.
Proposition 2.9. A complete quasi-affine variety is finite.
Proof. X is such a variety, without loss of generality X is irreducible. X ⊆ An is
locally closed, then xi : X ⊆ An → k must be constant, and so x is a point.
Theorem 2.10. Pn is complete.
Note: I ⊆ S = k[x0 , . . . , xn ] a homogeneous ideal, then V+ (I) is not empty
iff Id ( Sd for all d ∈ N.
Proof. Let Y be a variety and Z ⊆ Pn × Y be closed. Show that πY (Z) ⊆ Y is
closed.
Y = ∪Yi and open affine cover. It is enough to show that πY (Z) ∩ Yi ⊆ Yi
is closed, that is, πYi (Z ∩ (Pn × Yi )) is closed. So we take πYi : Pn × Yi → Yi .
Thus, WLOG, we assume Y is affine. Then let C(Z) = (π × id)−1 (Z) ⊆
n+1
A
× Y . k[An+1 × Y ] = S ⊗ k[Y ] = k[Y ][x0 , . . . , xn ] = ⊕d≥0 Sd ⊗k k[Y ], so it
is a graded ring.
Note that (y, (a0 , . . . , an )) ∈ C(Z) implies that (y, (λa0 , . . . , λan )) ∈ C(Z)
for λ ∈ k.
Thus, I(C(Z)) ⊆ k[An+1 × Y ] is a homogeneous ideal. We write I(C(Z)) =
(f1 , . . . , fm ) with fi ∈ Sdi ⊗ k[Y ].
For y ∈ Y , fi (y) = fi (−, y) ∈ Sdi . We observe that y ∈ πY (Z) iff ∃x ∈ Pn
such that (x, y) ∈ Z. This happens iff V+ (f1 (y), . . . , fm (y)) 6= ∅ ⊆ Pn . This is
true iff (f1 (y), . . . , fm (y))d 6= Sd for all d ≥ 0.
Fix d ≥ 0, P
then y ∈ Y defines a linear map ΦY : ⊕m
i=1 Sd−di → Sd :
m
(g1 , . . . , gm ) 7→ i=1 fi (y)gi .
Note: Every entry of the matrix ΦY is a regular function of Y.
n+d
Now, (f1 (y), . . . , fm (y))d 6= Sd iff rank(ΦY ) < dim(Sd ) =
which
n
n+d
holds iff all minors in ΦY of size
vanish. Therefore, Wd = {y ∈ Y :
n
(d1 (y), . . . , fm (y))d 6= Sd } ⊆ Y is closed. Finally, πY (Z) = ∩d≥0 Wd , and so is
closed.
Challenge: Find a complete variety that is not projective.
Exercise:
1. Let X be a topological space and W ⊆ X is a subset. W = X iff W ∩U 6= ∅
for all nonempty open sets U ⊆ X.
2. f : X → Y continuous and W = X and f (X) = Y then f (W ) = Y
3. X is irreducible and ∅ =
6 U ⊆ X is open. Then U = X and U is irreducible.
17
Rational Map: X and Y are irreducible varieties. The ideal is that a morphism f : X → y is uniquely determined by restriction to any non-empty open
subset of X.
Consider pairs (U, f ) where U ⊆ X nonempty and open and f : U → Y is a
morphism. Relation: (U, f ) ∼ (V, g) iff f = g on U ∩ V because Y is separated
and X is irreducible, this is an equivalence relation. Checking this is an exercise.
Definition 2.5 (Rational Map). A rational map f : X 99K Y is an equivalence
class for ∼.
X irreducible implies that U ∩ W ⊇ U ∩ V ∩ W dense, and Y separated
implies that f = h on a closed subset of U ∩ W .
Remark: If f : X 99K Y then there is a maximal open U ⊆ X where f is
defined as a morphism. U = ∪(V,g)∼f V ⊂ X.
Example: f : A2 99K A2 : (x, y) 7→ (x/y, y/x2 ) defined as a morphism of
D(xy).
Exercise: f : A2 99K A2 ⊆ P2 . Find the max open where f : A2 99K P2 is
defined.
Definition 2.6 (Rational Function). A rational function on X is a rational
map f : X 99K A1 = k. f is given by a regular function f : U → k where
∅=
6 U ⊆ X open.
k(X) = {f : X 99K k} is the field of rational functions on X.
Note: If (U, f ), (V, g) ∈ k(X) then f + g, f − g, f g : U ∩ V → k define
rational functions on X. If f 6= 0 in k[U ] then ∅ 6= D(f ) ⊆ U is open, and
1/f : D(f ) → k is regular. Thus, 1/f = (D(f ), 1/f ) ∈ k(X).
Examples: k(An ) = k(x1 , . . . , xn ). k(Pn ) = k(x1 /x0 , . . . , xn /x0 ).
Proposition 2.11. Let X be an irreducible variety
1. If ∅ =
6 U ⊆ X open, then k(X) = k(U ).
2. If X is affine, then k(X) = k[X]0 =field of fractions of k[X].
Proof.
1. k(X) → k(U ) : (V, g) → (V ∩ U, g|V ∩U ) is isomorphism.
2. Define k[X]0 → k(X) : f /g 7→ (D(g), f /g).
Injective: As this is a homomorphism of fields, it is enough to say that
it is not identically zero, and it maps 1 to (X, 1), which is not the zero
function.
Surjective: If f : U → k is regular, ∅ 6= U ⊆ X open. Find 0 6= g ∈ k[X]
such that ∅ =
6 D(g) ⊆ U . Then f ∈ k[D(g)] = k[X]g ⊆ k[X]x0 .
Definition 2.7 (Dominant). (U, f ) : X 99K Y is dominant if f (U ) = Y .
18
Indep. of rep.: If ∅ =
6 V ⊆ U open, then f (V ) = Y by the homework.
Suppose (U, f ) : X 99K Y is dominant and (V, g) : Y 99K Z is any rational
map, then we can compose g ◦ f : X 99K Z, as f (U ) = Y so f (U ) ∩ V 6= ∅, so
f −1 (V ) 6= ∅ ⊆ U .
g ◦ f = (f −1 (V ), g ◦ f ).
Exercise: If f, g both dominant, then g ◦ f dominant.
Proposition 2.12. X, Y irreducible varieties, then there is a one to one correspondence between {φ : X → Y with φ dominant} and field extensions k(Y ) ⊆
k(X) by φ 7→ φ∗ = [h 7→ h ◦ φ]
Proof. WLOG X, Y are affine. The map φ 7→ φ∗ is:
Injective: So we let ψ : X 99K Y dominant and ψ ∗ = φ∗ . Take D(h) ⊆ X
such that φ and ψ are both defined on D(h).
φ∗ = ψ ∗
k(Y ) ............................................... k(X)
..
.........
..
...
...
...
...
...
...
...
...
..
.........
..
...
...
...
...
...
...
...
...
⊆
⊆
k[Y ] ............................................... k[X]h
This diagram commutes, and so φ = ψ on D(h).
Surjective: Let α : k(Y ) → k(X) be a k-algebra homomorphism. k[Y ] is
generated by f1 , . . . , fn . α(fi ) = gi /hi , gi , hi ∈ k[X]. Let h = h1 . . . hn ∈ k[X].
So α : k[Y ] → k[X]h is a k-alg hom. This gives us a morphism φ : D(h) → Y
and φ∗ = α.
Definition 2.8 (Birational). Let f : X 99K Y be a rational map. It is birational
if f is dominant and there is a dominant g : Y 99K X such that f ◦ g = idY and
g ◦ f = idX as rational maps.
Definition 2.9 (Birationally Equivalent). X and Y are birationally equivalent
(often X and Y are birational) written X ≈ Y if there exists f : X 99K Y a
birational map.
Examples: A2 ≈ P2 ≈ P1 × P1 .
If U, V ⊆ X open and X irred., then U ≈ Y .
Theorem 2.13. The following are equivalent
1. X ≈ Y
2. k(X) ' k(Y ) as k-algebras
3. ∃∅ =
6 U ⊂ X, V ⊂ Y open such that U ' V as varieties.
Proof. 3 ⇒ 2 is clear from the first prop on this topic.
2 ⇒ 1 is clear from the second prop on this topic.
1 ⇒ 3: Let (U, f ) : X 99K Y and (V, g) : Y 99K X be inverses. Set
U0 = f −1 (V ) ⊆ U . Then g ◦ f : U0 → V → X must be the inclusion of U0 into
19
X. Thus, g(f (U0 )) ⊆ U0 , so f (U0 ) ⊆ g −1 (U0 ). Set V0 = g −1 (U0 ) ⊆ V . Then
U0 ' V0 as varieties by f, g.
Definition 2.10 (Rational Variety). An irreducible variety is rational if it is
birational to An for some n.
Examples: Any curve C ⊆ P2 of degree 2 is rational.
Let C = V (y 2 −x3 −x2 ) ⊆ A2 is rational. Let φ : C 99K A1 by φ(x, y) = y/x.
The inverse should be ψ : A1 → C by ψ(t) = (1 − t2 , t − t3 ). So φ ◦ ψ(t) =
(t − t3 )/(1 − t2 ) = t = idA1 , and the opposite is also an identity (Exercise, show
this).
Challenge: Show that E = V (y 2 − x3 + x) ⊆ A2 is not rational.
Big Challenge: If C is any irreducible variety and there exists a dominant
rational map A1 99K C then C is rational.
Transcendence Degree
Let k ⊆ L a field extension. Then L is algebraic over k if for all f ∈ L, there
is a polynomial equation f n + a1 f n−1 + . . . + an = 0 for ai ∈ k.
S ⊆ L subseteq, then S is algebraically independent over k if for all s1 , . . . , sn ∈
S with si 6= sj for i 6= j, then k[x1 , . . . , xn ] → L by xi 7→ si is injective.
Definition 2.11 (Transcendence Basis). A transcendence basis for L over k is
a set B ⊆ L such that B is algebraically independent over k and L is algebraic
over k(B).
Theorem 2.14.
1. All transcendence bases have the same cardinality.
2. If S ⊆ Γ ⊆ L subsets such that S is alg indep over k and L is alg over
k(Γ), then there exists a transcendence basis B for L over k such that
S ⊆ B ⊆ Γ.
Proof. The idea is as ”any vector space has a basis.”
Exercise: Prove where L is a finitely generated extension of k.
Lang’s Algebra contains the proof.
Definition 2.12 (Transcendence Degree). The transcendence degree tr degk (L) =
tr deg(L) = the number of elements in any transcendence basis for L over k.
Definition 2.13 (Dimension). Let X be an irreducible variety. Then define
dim(X) = tr deg(k(X)).
Examples:
1. dim(An ) = tr degk k(x1 , . . . , xn ) = n
2. If X is irreducible and dim(X) = 0, then k ⊆ k(X) is algebraic extension,
then k(X) = k. Thus, X is a point.
20
Some terminology: a curve is a variety of dimension 1, a surface is a variety
of dimension 2, and an n-fold is a variety of dimension n.
Notation: If R is a finitely generated domain over k, then we write tr deg(R) =
tr deg(R0 ).
We will state the following without proof.
Theorem 2.15 (Principle Ideal Theorem). If R is a finitely generated domain
over k and 0 6= f ∈ R and P ⊆ R is a minimal prime, then P is a minimal
prime containing f . Then tr deg(R/P ) = tr deg(R) − 1.
Geometric Statement: If X is any irreducible variety and 0 6= f ∈ k[X] and
if Z ⊆ V (f ) is an irreducible component then dim Z = dim X − 1.
Proof. Take U ⊆ X open affine such that U ∩ Z 6= ∅. Then Z ∩ U = V (P ) ⊆ U ,
P ⊆ k[U ] prime ideal. Z is a component of V (f ) ⇒ P is minimum over
(f ) ⊆ k[U ].
Thus, dim(Z) = tr deg(k[U ]/P ) = tr deg k[U ] − 1 = dim X − 1.
Theorem 2.16. Let X be an irreducible variety, and let ∅ =
6 X0 ( X1 ( . . . (
Xn = X be a maximal chain of irreducible closed subsets. Then dim(X) = n.
Proof. WLOG, X is affine. Take 0 6= f ∈ I(Xn−1 ). Then Xn−1 ⊆ V (f ) is a
component. PIT says that dim Xn−1 = dim X − 1.
Induction implies that dim Xn−1 = n − 1, so dim X = n.
Definition 2.14. If X is any variety, set dim(X) =the supremum of all n such
that ∃ a chain ∅ =
6 X1 ( X1 ( . . . ( Xn ⊆ X where Xi ⊆ irreducible and closed
for all i.
Exercises
1. X = X1 ∪ . . . ∪ Xm and Xi ⊆ X closed, then dim X = max dim(Xi )
2. dim(X × Y ) = dim X + dim Y
Recall: If R is a ring, then dim(R) is the supremum of all n such that ∃
Pn ( Pn−1 ( . . . ( P0 ⊆ R where Pi is a prime ideal.
Note: If X is affine then dim X = dim k[X].
Theorem 2.17 (PIT For Several Equations). If X is an irreducible variety
and f1 , . . . , fr ∈ k[X] and Z ⊆ V (f1 , . . . , fr ) are components, then dim Z ≥
dim X − r.
Proof. Enough to show that if W ⊆ X is a closed subset and each component
of W has dim ≥ d, then each component of W ∩ V (f ) has dim ≥ d − 1 for all
f ∈ k[X].
Let Z ⊆ W be a component. If f |Z = 0 then V (f ) ∩ Z = Z. If f |Z 6= 0 then
every component of Z ∩ V (f ) has dim= dim(Z) − 1 ≥ d − 1.
Therefore W ∩ V (f ) = union of finitely many irreducible closed subsets of
dim ≥ d − 1.
21
Lemma 2.18 (Prime Avoidance). X is an affine variety, Z ⊆ X an irreducible
closed subset and X1 , . . . , Xm ⊆ X are also irreducible closed subsets, then if
Xi 6⊆ Z then ∃f ∈ I(Z) such that f ∈
/ I(Xi ).
Proof. Induction on m.
If m = 1, then X1 6⊆ Z ⇒ I(Z) 6⊆ I(X1 ) ⇒ ∃f ∈ I(Z) \ I(X1 ).
For m ≥ 2, take fi ∈ I(Z) such that fi ∈
/ I(Xj ) for j 6= i. If any fi ∈
/ I(Xj ),
then done. Take f = fi .
If fi ∈ I(Xi ) for all i, then f = f1 + f2 f3 . . . fm ∈ I(Z).
Definition 2.15 (Codimension). If X is any variety, Z ⊆ X closed and irreducible, let X1 , . . . , Xm be the components of X containing Z. Set codim(Z; X) =
dim(X1 ∪ . . . ∪ Xm ) − dim Z.
E.g. X is the union of a line and a plane, Z is a single point of X. Then
codim(Z; X) is 2 if it is a point in the plane, 1 otherwise.
Theorem 2.19 (Reverse PIT). X affine, Z ⊆ X irreducible closed and c =
codim(Z; X). Then ∃f1 , . . . , fc ∈ k[X] such that Z ⊆ V (f1 , . . . , fc ) irreducible
component.
Proof. If Z is a component of Z, then c = 0.
Otherwise, no components of X are contained in Z, so the lemma implies
that there exists f1 ∈ k[X] such that f1 ∈ I(Z) and f1 does not vanish on any
component of X. PIT implies that codim(Z; V (f )) < c.
Induction on c gives us that there are f2 , . . . , fc ∈ I(Z) such that Z is a
component of V (f1 , . . . , fc ).
Resultants
Let K be an arbitrary field, and f (T ) = an T n + . . . + a1 T + a0 and g(T ) =
bm T m + . . . + b1 T + b0 ∈ K[T ].
Q: Do f (T
 ) and g(T ) have a common
 factor?
an a2 a1 a0 0

an a2 a1 a0 



.
Set A =  bm b1 b0



bm b1 b0
bm b1 b0
Definition 2.16 (Resultant). We define Res(f, g) = det A ∈ K.
Let ~v = (cm−1 , c0 , dn−1 , d1 , d0 ) ∈ K n+m , then ~v · A = (rn+m−1 , . . . , r1 , r0 ) ∈
n−1
K
. Then (cm−1
+ . . . + d1 T + d0 )g(T ) =
m−1 + . . . + c1 T + c0 )f (T ) + (dn−1 T
m+n−1
rm+n−1 T
+ . . . + r1 T + r0 .
n+m
Proposition 2.20. Suppose an 6= 0, then Res(f, g) 6= 0 ⇐⇒ (f, g) = 1 ∈
K[T ].
Proof. Res(f, g) = 0 iff ∃~v ∈ K m+n such that ~v · A = 0 iff ∃p(T ), q(T ) of deg
≤ m − 1, n − 1 such that p(T )f (T ) = q(T )g(T ), iff (f, g) 6= 1.
22
Pn
= i=0 ai T i then we allow formal differentiation, that is, f 0 (T ) =
PnIf f (T )i−1
∈ K[T ].
i=1 iai T
Note that (f g)0 = f 0 g + f g 0 , and similar rules still hold.
Corollary 2.21. If f (T ) = an T n + . . . + a1 T + a0 then f (T ) has n different
roots in K iff Res(f, f 0 ) 6= 0.
Qr
Proof.
(T − αi )di , where αi 6= αj for i 6= j. Then f 0 (T ) =
i=1Q
Pr f (T ) = an di −1
dj
an i=1 di (T − αi )
j6=i (T − αj ) .
0
0
Res(f, f ) = 0 iff (f, f ) 6= 1 iff f 0 (α` ) = 0 for some ` iff d` ≥ 2 for some
`.
Definition 2.17 (Discriminant). The discriminant of f (T ) is Res(f, f 0 ).
Exercise: a 6= 0 and f (T ) = aT 2 + bT + c then discriminant= −a(b2 − 4ac).
Remark: If char(K) = 0 and if f (T ) ∈ k[T ] is an irreducible polynomial,
then (f (T ), f 0 (T )) = 1 so Res(f, f 0 ) 6= 0. In char(K) = p, then f 0 (T ) may be
zero, for example (T p + 1)0 = 0.
Remark: ϕ : X → Y is a morphism, then dim(ϕ(X)) ≤ dim(X). This is as
k(ϕ(X)) ⊆ k(X).
Theorem 2.22. φ : X → Y is a dominant morphism of irreducible varieties,
such that k(Y ) ⊆ k(X) is a finite extension of degree d. Suppose that char(k) = 0
or k(X)/k(Y ) is separable. Then ∃ dense open V ⊆ Y such that |φ−1 (y)| = d
for all y ∈ V .
Proof. Assume X, Y affine and k[X] = k[Y ][f ]. Let P (T ) = ad T d + . . . +
a1 T + a0 ∈ k(Y )[T ] be the minimum polynomial for f ∈ k(X) over k(Y ). ie
P (f ) = 0 ∈ k(X).
WLOG, ai ∈ k[Y ] for all i and we can replace Y with D(ad ) and X with
φ−1 (D(ad )) ⊆ X. We may assume that ad = 1. Now k[X] = k[Y ][T ]/(P (T )).
πY
This implies that X ' V (P ) ⊆ Y × A1 →
Y , and φ : X → Y goes through this
path.
Pd
If (y, t) ∈ Y × A1 then we set Py (T ) = i=0 ai (y)T i ∈ k[T ]. (y, t) ∈ X iff
Py (t) = 0. Let ∆ = Res(P, P 0 ) ∈ k[Y ]. P (T ) irreducible and char(k) = 0 imply
that ∆ 6= 0. Note that Res(Py , Py0 ) = ∆(y).
Thus, if y ∈ D(∆), then Py (t) = 0 has exactly d solutions.
Now the general case: X and Y are irreducible varieties and φ : X → Y dominant. Let V ⊆ Y and U ⊆ φ−1 (V ) ⊆ X be open affines. Then dim(φ(X \ U )) ≤
dim(X \ U ) < dim(X) = dim(Y ). Thus φ(X \ U ) ( Y , and so ∃h ∈ k[V ] such
that D(h) ∩ φ(X \ U ) = ∅.
We can replace X with D(φ∗ h) and Y with D(h). And so, WLOG, X, Y
affine.
φ dominant implies that k[Y ] ⊆ k[X] and k[X] generated by f1 , . . . , fn .
Then k[Y ] ⊆ k[Y ][f1 ] ⊆ . . . ⊆ k[Y ][f1 , . . . , fn ] = k[X] gives X = Xn → Xn−1 →
ψ
. . . → X1 → Y a sequence of dominant maps.
Induction on n: ∃ a dense open U ⊆ X1 such that all points of U are hit by
d1 = [k(X) : k(X1 )] pts of X.
23
As above: ψ(X1 \ U ) ( Y implies ∃h ∈ k[Y ] such that D(h)∩ψ(X1 \U ) = ∅,
and so ψ −1 (D(h)) = D(φ∗ h) ⊆ U . ψ : D(φ∗ h) → D(h) gives k[D(φ∗ h)] =
k[X]φ∗ h = k[X]h . Thus, k[Y ][f1 ]h = k[Y ]h [f1 ] = k[D(h)][f1 ].
Thus, the first case implies that ∃∅ =
6 V ⊆ D(h) ⊆ Y open such that
|ψ −1 (y)| = [k(X1 ) : k(Y )]. Since ψ −1 (D(h)) ⊆ U we have |φ−1 (y)| = [k(X1 ) :
k(Y )] · d1 = d.
Exercise: πY : X × Y → Y is an open map. That is, if U ⊆ X × Y is open
then πY (U ) is open in Y .
Corollary 2.23. φ : X → Y is a dominant morphism of irreducible varieties.
Then φ(X) contains a dense open subset of Y .
Proof. We can assume that X, Y are affine. Choose B = {f1 , . . . , fn } ⊆ k[X]
such that B is a transcendence basis for k(X)/k(Y ).
ψ
π
Y
Y and φ is the
Then k[Y ] ⊆ k[Y ][f1 , . . . , fn ] ⊆ k[X] gives X → Y × An →
composition.
The theorem says that there is a open subset U ⊆ Y × An such that U ⊆
ψ(X). As πY is an open mapping, πY (U ) is open.
Definition 2.18. Let X be a variety.
1. W ⊆ X is locally closed if W = open ∩ closed.
2. W ⊆ X is constructible if W =the union of finitely many locally closed
subsets.
Example: W = D(xy) ∪ {0} ⊆ A2 . Notice: φ : A2 → A2 : (x, y) 7→ (x2 y, xy),
then φ(A2 ) = W .
Exercise: φ : X → Y is an arbitrary morphism of varieties, then φ(X) is
constructible.
3
Nonsingular Varieties
Local Rings
Definition 3.1 (Local Ring at a point). If X is an irreducible variety and x ∈ X
then OX,x = {f ∈ k(X) : f (x) defined}. This is a local ring with maximal ideal
mx = {f ∈ OX,x : f (x) = 0}.
T
Note: If U ⊆ X is any open subset, then k[U ] = x∈U OX,x ⊆ k(X).
Let U ⊆ X open affine, x ∈ U then M = I({x}) ⊆ k[U ]. f ∈ OX,x then f is
defined on D(h) ⊆ U for some h ∈ k[U ] \ M , so f = g/hn where g ∈ k[U ]. And
so, OX,x = {g/h : g, h ∈ k[U ], h(x) 6= 0} = k[U ]M .
Remark: X not irreducible implies that OX,x = limU 3x OX (U ). If U ⊆ X
−→
open affine, x ∈ U we still have that OX,x = k[U ]I({x}) .
Note: If X is irreducible then dim(X) = dim OX,x for any x ∈ X. If we let
X be any variety, then dim(X) = maxx∈X dim OX,x .
24
Definition 3.2 (Regular Local Ring). If (R, m) is a local ring, then F = R/m
is a field called the residue field of R and m/m2 is an F -vector space.
R is a regular local ring if dimF (m/m2 ) = dim R.
Exercise: If R is a Nötherian Local Ring then dimF (m/m2 ) =min number
of generators for m ≥ dim(R). The equality is from Nakayama, the inequality
from PIT.
Definition 3.3. Let X be a variety and x ∈ X.
1. X is nonsingular at x if OX,x is a regular local ring.
2. Otherwise, X is singular at x.
3. X is nonsingular if all points x ∈ X are nonsingular.
Exercise: S is a commutative ring, and M ⊆ S is a maximal ideal, then
set R = SM a local ring. Then unique maximal ideal m = M SM . Show that
S/M ' R/m and M/M 2 = m/m2 .
Example: Let C = V (f ) ⊆ A2 a curve. Let P = (0, 0) ∈ C. So f = ax+by +
HOT . And k[C] = k[x, y]/(f ). Let M = I({P }) = (x̄, ȳ) = (x, y)/(f ) ⊆ k[C].
mP /m2P = M/M 2 = (x, y)/(x2 , xy, y 2 , f ) = (x, y)/(x2 , xy, y 2 , ax + by). So
P ∈ C is nonsing iff dimk (M/M 2 ) = dim(C) = 1. This is iff ax + by 6= 0 ∈
k[x, y], note that ax + by 6= 0 implies that V (f ) looks like V (ax + by) close to
the point.
Exercise: Find all singular points of V (y 2 − x3 − x2 ), V (y 2 − x3 ).
X ⊂ An a closed affine variety. Then I = I(X) = (f1 , . . . , ft ) ⊆ S =
k[x1 , . . . , xn ]. Idea: X is nonsing at P ∈ X iff X has a tangent space at P .
h
i
∂fi
Definition 3.4 (Jacobi Matrix). Let JP = ∂x
(P
)
be a t × n matrix, we call
j
this the Jacobi matrix.
Note: If ~v ∈ k n then Jp · ~v ∈ k t is the partial derivative of (f1 , . . . , ft ) at p
in the direction ~v .
ker(JP ) = {~v ∈ k n : Jp · ~v = ~0} is a candidate for a tangent space.
Lemma 3.1. Let P ∈ X ⊆ An , then rank(Jp ) + dimk (mP /m2P ) = n.
∂f
∂f
Proof. Set M = I({P }) ⊆ S. Define d : M → k n by d(f ) = ( ∂x
(P ), . . . , ∂x
(P )).
1
n
d is surjective as d(xi − pi ) = ei . Note: f, g ∈ S that d(f g) = f (P )d(g) +
d(f )g(P ). So d(M 2 ) = 0. Thus, d : M/M 2 → k n is an isomorphism if it is
injective. It is injective as the two vector spaces are of the same dimension and
it is surjective.
P
P
d(fi ) = ith row of JP and d( gi fi ) =
gi (P )d(fi ) so d(I) =row span of
n
2
2
JP in k . Thus d : I + M /M → row span of JP is an isomorphism. Thus
rank(Jp ) + dim(M/I + M 2 ) = dim(M/M 2 ) = n. Finally OX,P = (S/I)M and
mP /m2P = (M/I)/(M/I)2 = M/I + M 2 .
25
Theorem 3.2. P ∈ X ⊆ An , then rank(JP ) ≤ n − dim(OX,P ) and rank JP =
n − dim OX,P ⇐⇒ P nonsingular.
Proof. rank(Jp ) = n − dim(mP /m2P ) ≤ n − dim OX,P .
We have equality iff dimk (mP /m2P ) = dim OX,P .
Example: X = V (z 2 − x2 y 2 ) ⊆ A3 where char(k) 6= 2, 3. Then J =
[−2xy 2 , −2x2 y, 3z 2 ]. p ∈ X is a nonsingular point iff rank(Jp ) = 3 − x = 1,
so Xsing = V (z 3 − x2 y 2 , xy 2 , x2 y, z 2 ) = V (xy, z).
Exercise: X is a variety and p ∈ X. Then OX,P is a domain iff p is in only
one component.
Theorem 3.3. Any Nötherian regular local ring is a domain. (in fact, a UFD,
and even Macaulay)
Conclude: The points on an intersection of two components are singular.
Proposition 3.4. Xsing ⊆ X is a closed subset of X.
Proof.
Let X =S X1 ∪ . . . ∪ Xm be the components of X. Then Xsing =
Sm
(X
i )sing ∪
i=1
i6=j Xi ∩ Xj . The latter are closed, so without loss of generality, X irreducible and affine.
X ⊆ An closed, I(X) = (f1 , . . . , ft ) ⊆ k[x1 , . . . , xn ], P ∈ Xsing iff p ∈ X
and rank(Jp ) < n − dim OX,p = n − dim(X).
∂fi
]. Xsing =
Let m1 , . . . , mN be all of the minors of size n−dim(X) in J = [ ∂x
j
X ∩ V (m1 , . . . , mN ).
Fact: Xsing 6= X.
Lemma 3.5. If p ∈ X = V (g1 , . . . , gr ) ⊆ An and if rank Jp (g1 , . . . , fr ) = r
then OX,p is regular local of dimension n − r.
Proof. PIT implies that dim OX,p ≥ n − r.
I(X) = (f1 , . . . , ft ) ⊇ (g1 , . . . , gr ). Thus, row span Jp (f1 , . . . , ft ) ⊇ row span
Jp (g1 , . . . , gr ), so r = rank Jp (g1 , . . . , gr ) ≤ rank Jp (f1 , . . . , ft ) ≤ n−dim OX,p ≤
r.
Theorem 3.6 (Implicit Function Theorem). If f1
, . . . , fc are holomorphic functions in a classical nbhd of p ∈ Cn . Suppose det
∂fi
∂xj (p)
6= 0. Then ∃
1≤i,j≤c
n−c
holomorphic functions w1 , . . . , wc on classical open subset of C
and classical
open subset V ⊆ Cn such that p ∈ V and so that for all z ∈ V , f1 (z) = . . . =
fc (z) = 0 iff z − i = wi (zc+1 , . . . , zn ) for all 1 ≤ i ≤ c.
Theorem 3.7. X ⊆ Cn a complex affine variety. p ∈ X a nonsingular point.
Then a classical neighborhood of p in X is holomorphic to a classical open subset
of Cd where d = dim OX,p .
26
Proof. WLOG, X is irreducible. I(X) = (f1 , . . . , ft ) and rank Jp (f1 , . . . , ft ) =
∂fi
n − d = c. WLOG, det( ∂x
(p)) 6= 0. Set Y = V (f1 , . . . , fc ) ⊆ Cn , then
j
rank Jp (f1 , . . . , fc ) = c implies that p is a nonsingular point of Y . So OY,p is
regular local of dimension d.
Now, only one component of Y contains p, p ∈ X and X ⊆ Y . dim X
is the same as the dimension of the component of Y containing p, and as X
is irreducible, X is the component of Y containing p. Then there exists open
U ⊆ An such that X ∩ U = Y ∩ U = V (f1 , . . . , fc ) ∩ U . We apply the IFT, let
V ⊂ U , p ∈ V , and w1 , . . . , wc be as in the IFT.
Define π : Cn → Cd by π(z) = (zc+1 , . . . , zn ). Then π : X ∩V → π(X ∩V ) ⊆
d
C is an holomorphism, as we can get an inverse map π −1 : π(X ∩ V ) → X ∩ V
by (zc+1 , . . . , zn ) 7→ (w1 (z), . . . , wc (z), zc+1 , . . . , zn ).
Corollary 3.8. Every nonsingular complex variety variety is a complex manifold.
If X is an affine variety, recall that pts in X are in 1-1 correspondence with
max ideals in k[X].
So X an irreducible variety, p ∈ X corresponds to local rings OX,P = {f ∈
k(X) : f (P ) defined }.
Lemma 3.9. X an irred var, x, y ∈ X, if OX,x ⊆ OX,y then x = y.
Proof. Take open affine x ∈ U ⊆ X, y ∈ V ⊆ X. X is separated, so U ∩ V is
affine and k[X] ⊗ k[Y ] → k[U ∩ V ] is surjective, and k[U ] ⊆ OX,x ⊆ OX,y with
k[V ] ⊆ OX,y , thus, k[U ∩ V ] ⊆ OX,y .
k[V ] ⊆ k[U ∩ V ] ⊆ OX,y proves that y ∈ U ∩ V . Then k[U ∩ V ] ∩ my is a
max ideal in k[U ∩ V ] so it corresponds to a point in U ∩ V which maps to y
under U ∩ V → V the inclusion.
Thus, x, y ∈ U ⊆ X. U is affine. If x 6= y, then there is an f ∈ k[U ] such
that f (x) 6= 0, f (y) = 0. Then f1 ∈ OX,x and f1 ∈
/ OX,y .
Nonsingular Curves
Definition 3.5 (Curves). A curve is an irreducible variety of dimension one.
Example: C = A1 . k[C] = k[t] and k(C) = k(t). Let 0 6= f ∈ k(t). What is
the order of vanishing of f at 0?
f = p/q, p, q ∈ k[t], we can write p = tn p0 and q = tm q0 with p0 (0) 6= 0 and
q0 (0) 6= 0. f = (p0 /q0 )tn−m , so v(f ) = n − m is the order of vanishing.
Note that OC,0 = k[t](t) with m0 = (t) ⊆ OC,0 is a maximal ideal. Then
f = utn−m where u = p0 /q0 is a unit in OC,0 .
Definition 3.6 (Discrete Valuation Ring). A discrete valuation ring, or DVR,
is a Nötherian regular local ring of dimension 1.
Examples are OC,P where C is a curve and P a nonsingular point.
Let (R, m) be a DVR, m = (t), t is a uniformizing parameter, and K = R0
is the field of fractions of R.
27
Claim: Any f ∈ K ∗ = K \ {0} can be written f = utn where u ∈ R \ m a
unit in R and n ∈ Z. WLOG, f ∈ R \ {0}.
Assume that the claim is false, then choose a counterexample such that (f )
is maximal. f is not a unit implies that f ∈ m, so f = gt for some g ∈ R.
(f ) ( (g) as (f ) = (g) ⇒ g = hf ⇒ f = thf ⇒ th = 1, contradiction.
So g = utn , u ∈ R a unit implies that f = utn+1 .
Check that if f = utn , u a unit, then u, n are unique. n = min{p ∈ Z : f ∈
p
(t ) ⊆ K}.
Definition 3.7 (Valuation Map). v : K ∗ → Z is a valuation map v(f ) = n if
f = utn with u ∈ R \ m.
Note that R = {f ∈ K ∗ : v(f ) ≥ 0}∪{0} and m = {f ∈ K ∗ : v(f ) > 0}∪{0}.
Rules: v(f g) = v(f ) + v(g). v(f + g) ≥ min(v(f ), v(g)) if f, g, f + g ∈ K ∗ .
If f = utn , g = vtm and n ≤ m, then f + g = (u + vtm−n )tn = u0 tr tn .
Example: C a curve, p ∈ C nonsing, then R = OC,p is a DVR with K =
R0 = k(C). vp : K(C)∗ → Z is a valuation, f ∈ k(C)∗ vp (f ) =the order of
vanishing of f at p. vp (f ) > 0 iff f ∈ mp , vp (f ) = 0 iff f (p) 6= 0 and vp (f ) < 0
iff f is not defined at p.
Lemma 3.10. Let R be a DVR, R0 = K and S any ring such that R ⊆ S ⊆ K.
Then S = R or S = K.
Proof. If S 6= R, then take f ∈ S, f ∈
/ R. m = (t), f = utn , n < 0, S ⊇ R[f ] =
K.
Recall: A domain R is called integrally closed iff for all f ∈ R0 , f is integral
over R implies that f ∈ R.
Theorem 3.11. If R is any Nötherian Local Domain of dimension one, then
R is regular iff R is integrally closed.
Exercise: Prove ⇒ (easy, as DV R ⇒ P ID ⇒ U F D ⇒ int closed, so prove
the last one)
Proposition 3.12. Let A be a domain. A is integrally closed iff AP is integrally
closed for all maximal ideals P ⊆ A.
Proof. ⇒: Easy
⇐: A = ∩P ⊆A AP over maximal ideals P .
Note: A a f.g.
∩p∈X OX,p = ∩Ap .
domain over k, X = Spec −m(A), then A = k[X] =
Definition 3.8 (Dedekind Domain). A Dedekind Domain is an integrally closed
Nötherian domain of dimension 1.
Note: If X is an irred affine variety, X nonsingular curve iff k[X] is a
Dedekind domain.
28
OX,p a DVR for all p ∈ X iff k[X]p integrally closed for all max ideals p by
the theorem, each of these is integrally closed iff k[X] is, and that is the def of
a Dedekind domain.
Finiteness of Integral Closure
Let R be a finitely generated domain over k, K = R0 , and K ⊆ L a finite
field extension. Then R is defined to be the integral closure of R in L. That is,
{f ∈ L : f integral over R}. Fact R is an integrally closed domain with field of
fractions L.
Let f ∈ L. Then f n + a1 f n−1 + . . . + an = 0 with ai ∈ K. Take b ∈ R such
that bai ∈ R for all i. Then (bf )n + ba1 (bf )n−1 + . . . + bn an = 0. Thus bf ∈ R.
b ∈ R, so f ∈ (R)0 .
Theorem 3.13 (Finiteness of Integral Closure). R is a finitely generated Rmodule.
In particular: R is a finitely generated K algebra.
Definition 3.9 (DVR of K/k). k ⊆ K is a field extension, a DVR of K/k is a
subring R ⊆ K such that:
1. R is a DVR
2. R0 = K
3. k ⊆ R.
Let K be a function field of dimension 1 over k.
i.e., k ⊆ K is finitely generated as a field extension and it has transcendence
degree 1.
Q: ∃ a nonsingular curve C such that k(C) = K?
Key construction: Let f ∈ K \ k. Then k(f ) ⊆ K is a finite field extension.
It is finitely generated by assumption, and {f } must be a transcendence basis
for K/k, thus it is algebraic, and so it is a finite field extension.
Set B = k[f ] ⊆ K.
Finiteness of integral closure says that B is a finitely generated k-algebra,
and so B is a Dedekind domain with B0 = K.
Proposition 3.14. X = Spec −m(B) is a nonsingular curve.
1. Points in X are in 1-1 correspondence with DVRs R of K/k such that
f ∈ R.
2. Points in V (f ) correspond to the DVRs R of K/k such that f ∈ mR .
Proof. X → {DVRs R 3 f } is well-defined and injective.
Let R be a DVR of K/k with f ∈ R.
k[f ] ⊆ R ⇒ B ⊆ R.
Set M = B ∩ mR ⊆ B a prime ideal. Then BM ⊆ R 6= K ⇒ M 6= 0 and so
BM is a DVR of K/k.
And so, the lemma implies that BM = R.
Now we prove (b). M ∈ V (f ) ⇐⇒ f ∈ M ⇐⇒ f ∈ M BM = m.
29
Corollary 3.15. Every DVR R of K/k is the local ring of a nonsingular curve
at some point. In particular, R/mR = K.
Proof. Let f ∈ R \ k, B = k[f ] ⊆ K.
Then R = BM for some maximal ideal M ∈ Spec −m(B). So R = OX,p .
Corollary 3.16. Given f ∈ K ∗ , there are only finitely many DVRs R of K/k
such that f ∈ mR .
Also have finitely many R such that f ∈
/ R.
Proof. WLOG, f ∈
/ k. Set B = k[f ] ⊆ K. {R : f ∈ mR } is in correspondence
with V (f ) ⊆ Spec −m(B). PIT implies that dim V (f ) = 0. Thus, V (f ) is a
finite set.
Note: f ∈
/ R iff 1/f ∈ mR .
Definition 3.10. CK = {DVRs of K/k}
Elements of CK will be called ”points” P . DVR given by P is RP with
maximal ideal mP .
We define a topology on CK to have as closed sets the finite sets and all of
CK . We also let f ∈ K and P ∈ CK , assume f ∈ RP .
Definition 3.11. f (P ) is defined to be the image of f by RP → RP /mP = k.
i.e. f (P ) ∼
= f mod mP .
The regular functions on a nonempty open subset U ⊆ CK are then the set
k[U ] = ∩p∈U RP ⊆ K.
This makes CK a SWF.
Note: If U ⊆ CK open, f ∈ k[U ] then D(f ) = {P ∈ U : f (P ) 6= 0} = {P ∈
U :f ∈
/ mP } is open by the second corollary.
Example Let K = k(t). The DVRs of K/k are k[t](t−a) for a ∈ k and
k[1/t](1/t) .
Then CK and P1 are in 1-1 correspondence, with k[t](t−a) corresponding to
a ∈ A1 and the other point corresponding to the point at infinity.
Note: If f ∈ k[t](t−a) then f (t) ∼
= f (a) mod (t − a), f (k[t](t−a) ) = f (a).
Theorem 3.17. CK is a non-singular curve and k(CK ) = K.
Proof. Take any f ∈ K \ k. B = k[f ] ⊆ K. U = {p ∈ CK : f ∈ RP } ⊆ CK is
open.
We define φ : Spec −m(B) → U by M 7→ BM . The prop implies that this is
bijective.
φ is a homeomorphism, as the closed sets are the finite sets in both. It
remains to show that this is a morphism of spaces with functions.
For V ⊆ Spec −m(B) open, then k[V ] = ∩M ∈V BM = ∩P ∈φ(V ) RP =
k[φ(V )].
Thus, φ : Spec −m(B) → U is an isomorphism of spaces with functions.
Note: If P ∈ CK then f ∈ RP or 1/f ∈ RP . Thus, Ck = Spec −m(k[f ]) ∪
Spec −m(k[f −1 ]). This is an open affine cover, and so Ck is a prevariety.
30
Let P, Q ∈ CK . Enough to find an open affine U ⊆ CK such that P, Q ∈ U .
Take f ∈ RP \ mP and f ∈
/ k. (f = 1 + t, (t) = mP ).
If f ∈ RQ then P, Q are both in Spec −m(k[f ]).
Otherwise, 1/f ∈ RQ , so both are in Spec −m(k[1/f ]).
Thus CK is a variety. It is nonsingular and of dimension one by construction.
We must show that it is irreducible.
CK is, in fact, irreducible because its proper closed sets are finite and CK is
infinite.
Proposition 3.18. Let C be an irreducible curve and P ∈ C a nonsingular
point. Let Y be any projective variety and φ : C \ {P } → Y is any morphism of
varieties. Then ∃! extension φ : C → Y .
Note: No points in Y are ”missing”.
Proof. Y ⊆ Pn closed subset. It is enough to make φ : C → Pn .
WLOG φ(C \ {p}) 6⊆ V+ (xi ) for all i. Set U = D(x0 , x1 , . . . , xn ) ⊆ Pn .
φ(C \ {p}) ∩ U 6= ∅.
Set fij = xi /xj ◦ φ ∈ k(C). Defined on φ−1 (U ) 6= ∅.
vP : k(C)∗ → Z is the valuation given by OC,P .
Set ri = vP (fi0 ) for 0 ≤ i ≤ n. Choose j such that rj minimal.
0
As xxji = xxji /x
/x0 , fij = fi0 /fj0 , so vP (fij ) = ri − rj ≥ 0. Thus, fij ∈ OC,P for
all i. Note that if Q ∈ φ−1 (U ), then φ(Q) = (f0j (Q) : f1j (Q) : . . . : fjj (Q) = 1 :
. . . : fnj (Q)).
We can then extend φ to P by this expression. Then φ is a morphism on
φ−1 (U ) ∪ {p} and so φ : C → Pn is a morphism.
Lemma 3.19. R ⊆ K is a local ring, k ⊆ R, R is not a field. Then R is
contained in some discrete valuation ring of K/k.
Proof. Set B = R ⊆ K. Lying over implies that there exists some maximal
ideal M ⊆ B such that M ∩ R = mR .
Claim: BM is a DVR of K/k.
Let 0 6= f ∈ mR . S = k[f ] is a Dedekind domain, S ⊆ B. M̃ = M ∩ S is a
maximal ideal of S.
Thus SM̃ is a DVR of K/k. SM̃ ⊆ BM ( K and a lemma from before says
that we have equality of SM̃ and BM .
Theorem 3.20. CK is a projective curve.
Proof. Let f ∈ K \ k. U = Spec −m(k[f ]), V = Spec −m(k[f −1 ]), and CK =
U ∪ V an open affine cover.
U ⊆ AN closed. U ⊆ PN projective closure.
The proposition implies that the inclusion U → U extends to a morphism
ϕ1 : C K → U .
Similarly, we take V to be the projective closure of V . Then V → V extends
to ϕ2 : CK → V .
31
We now define ϕ : Ck → U × V by ϕ(P ) = (ϕ1 (P ), ϕ2 (P )). Set Y =
ϕ(CK ) ⊆ U × V .
Y is a projective variety.
Claim: ϕ : CK → Y is an isomorphism.
Note: ϕ(U ) ⊆ U × V is a closed subset. Let ψ = ϕ2 × id : U × V → V × V .
ϕ(U ) = {(u, v) ∈ U × V : ϕ2 (u) = v} = ψ −1 (∆V ) closed. Thus, ϕ(U ) =
Y ∩ (U × V ), which implies that ϕ : U → Y ∩ (U × V ) is bijective, isomorphism.
So πU : Y ∩ (U × V ) → U is the inverse.
Similarly, ϕ : V → Y ∩ (U × V ) is an isomorphism.
Note: k(Y ) = k(CK ) = K, and forall P ∈ CK , OY,ϕ(P ) = RP ⊆ K. Thus, ϕ
is injective.
For surjective, let y ∈ Y . Then k ⊆ OY,y ⊆ K is a local ring. By the lemma,
OY,y ⊆ RP for some P ∈ CK . OY,y ⊆ RP = OY,ϕ(P ) so y = ϕ(P ).
Corollary 3.21. Any curve is birational to some nonsingular projective curve.
Corollary 3.22. X is any nonsingular curve, then X ∼
= some open subset of
CK , K = k(X).
Proof. ϕ : X → CK by ϕ(x) = P where P ∈ CK such that RP = OX,x ⊆ K.
Injectivity is clear. Claim: ϕ(X) ⊆ CK is open. Take U ⊆ X open affine.
k[U ] is generated by f1 , . . . , fn . P ∈ ϕ(U ) iff k[U ] ⊆ RP , iff fi ∈ RP for all i.
Thus, ϕ(U ) = ∩ni=1 Spec −m(k[fi ]) open. Thus ϕ(X) ⊆ CK is open.
ϕ : X → ϕ(X) is a homeomorphism. To check that ϕ is an isomorphism,
then U ⊆ X open gives k[U ] ∩x∈U OX,x = ∩P ∈ϕ(U ) Rϕ(x) = k[ϕ(U )].
Exercise: Two nonsingular projective curves are isomorphic iff they have the
same function field.
Degree of Projective Varieties in Pn
Bezout: f1 , . . . , fn ∈ k[x0 , . . . , xn ] homogeneous of degrees d1 , . . . , dn . Then
V+ (f1 , . . . , fn ) has cardinality at most d1 . . . dn or is infinite. If it is finite and
counted with multiplicity, then it is equal to d1 . . . dn .
Classical Definition: X ⊆ Pn closed, then deg(X) = #(X ∩V ) where V ⊆ Pn
is a linear subspace with dim V + dim X = n.
e.g. f ∈ S = k[x0 , . . . , xn ] a square-free homogeneous polynomial. Then
#(V+ (f )∩ general line) = deg f .
Warning: V+ (xz − y 2 ), V+ (x) ⊆ P2 . These are isomorphic but have different
degrees. So degree is not a property of a projective variety, but rather one of
the embedding into projective space.
Example: f ∈ S is square-free, deg f = d. Then X = V+ (f ), I(X) = (f ),
R = S/(f ) is the projective
ring.
coordinate
m+n
Note: dimk (Sm ) =
, where Sm is all forms of degree m. This is
n
1
n! (m + n)(m + n − 1) . . . (m + 1), which is actually a polynomial in m of degree
1
n with lead coefficient n!
.
32
f
Consider 0 → S → S → R → 0 implies that 0 → S
m−d →Sm→ Rm → 0is
m+n
m+n−d
exact. Then dim Rm = dim Sm − dim Sm−d , which is
−
,
n
n
1
1
(m + n) . . . (m + 1) − n!
(m + n − d) . . . (m + 1 − d). Which is a
which is n!
polynomial of degree n − 1.
P
P
d
1
( i − (i − d)) = nd
This has lead coefficient n!
n! = (n−1)! .
Recall: A graded S-module is a module M with a decomposition M =
⊕d∈Z Md as abelian group such that Sm Md ⊂ Mm+d .
Definition 3.12. Ann(M ) = {f ∈ S : f M = 0} is a homogeneous ideal,
Supp(M ) = V+ (Ann(M )) ⊆ Pn .
Reason for Supp is x ∈ Pn , P = I({x}) ⊆ S is a homogeneous prime ideal.
MP 6= 0 iff P ⊇ Ann(M ), iff x ∈ Supp(M ).
Example: X ⊆ Pn closed, then Ann(S/I(X)) = I(X), so Supp(S/I(X)) =
V+ (I(X)) = X.
Note: Md is a k-vector space for all d ∈ Z because S0 Md ⊆ Md . If M is a
finitely generated graded S-module, then dimk Md < ∞ forall d.
Definition 3.13 (Hilbert Function). HM (d) = dimk Md is the Hilbert function
of M .
Theorem 3.23. If M is a finitely generated graded S-module then ∃!PM (z) ∈
Q[z] such that PM (d) = dimk (Md ) for all d sufficiently large. We call PM (z)
the Hilbert Polynomial.
Definition 3.14 (Numerical Polynomial). P (x) ∈ Q[z] is a numerical polynomial if P (d) ∈ Z for all d sufficiently large in Z.
z
1
= m!
z(z − 1) . . . (z − m + 1).
Example:
m
z
Note:
: m ∈ N is a basis over Q for Q[z].
m
z
z
Lemma 3.24. P (z) = c0 + c1
+ . . . + cr
∈ Q[z], ci ∈ Q. Then TFAE
1
r
1. P (z) is a numerical polynomial.
2. P (d) ∈ Z for all d ∈ Z.
3. ci ∈ Z.
Proof. 3 ⇒ 2 ⇒ 1 are
⇒ 3. easy,so we
need
1
z+1
z
z
We know that
−
=
. Thus P (z + 1) − P (z) =
m
m
m
−1
z
z
c1 + c2
+ . . . + cr
.
1
r−1
We perform induction on r: P (z) is numeric, then P (z + 1) − P (z) is also
numeric. Thus, c1 , . . . , cr ∈ Z, and so c0 must also be an integer.
33
Theorem 3.25 (Affine Dimension Theorem). X, Y ⊆ An closed and irreducible,
then Z ⊆ X ∩ Y component has dim Z ≥ dim X + dim Y − n.
Proof. X ∩ Y = (X × Y ) ∩ ∆An = (X × Y ) ∩ V (x1 − y1 , . . . , xn − yn ) ⊆ An × An ,
use PIT.
WARNING: Does not prevent X ∩ Y = ∅.
Theorem 3.26 (Projective Dimension Theorem). X, Y ⊆ Pn closed irreducible,
Z ⊆ X ∩Y a component, then dim Z ≥ dim X +dim Y −n. If dim X +dim Y −n
is nonnegative, then X ∩ Y is nonempty.
Proof. First statement follows from ADT.
Set s = dim X, t = dim Y . C(X) = π −1 (X) ⊆ An+1 , then dim C(X) = s+1,
dim C(Y ) = t + 1.
Every component of C(X) ∩ C(Y ) has dimension ≥ s + 1 + t + 1 − n − 1 =
s + t − n + 1 ≥ 1, and 0 ∈ C(X) ∩ C(Y ), so such a component exists.
Definition 3.15 (Twisted Module). Let ` ∈ Z, M (`) is the twisted module
given by M (`)d = M`+d . i.e., we are shifting the grading, but doing nothing
else.
Definition 3.16 (Homogeneous Homomorphism). A homomorphism ϕ : M →
N of graded S-modules is homogeneous if ϕ(Md ) ⊆ Md for all d.
Definition 3.17 (Homogeneous Submodule). A submodule N ⊆ M is homogeneous if N = ⊕d∈Z (N ∩ Md ).
This implies that N and M/N = ⊕Md /(N ∩ Md ) are graded and 0 → N →
M → M/N → 0 is a short exact sequence of homogeneous homomorphisms.
Note: Let m ∈ M be homogeneous of degree `, that is, m ∈ M` , then
I = Ann(m) ⊆ S is a homogeneous ideal. Set N = S · m ⊆ M , then 0 →
I → S → N → 0 is a short exact sequence, but it is not quite of homogeneous
maps, unless we change S, I to S(−`), I(−`). Thus, (S/I)(−`) is isomorphic to
N = S · m ⊆ M.
Exercise: M f.g. module over a Nötherian ring, then M is a Nötherian
module.
Lemma 3.27. M a f.g. graded module over a Nötherian graded ring S, then
∃ filtration 0 = M 0 ⊆ . . . ⊆ M r = M such that M i ⊆ M is a homogeneous
submodule and M i /M i−1 ' S/Pi (`i ) where Pi is a homogeneous prime ideal
and `i ∈ Z.
Proof. Let N ⊆ M be maximal homogeneous submodule such that the lemma
is true for N .
Claim: N = M . Else M 00 = M/N 6= 0, Take 0 6= m ∈ M 00 homogeneous
such that Ann(m) ⊆ S is as large as possible.
Claim: P = Ann(m) prime ideal. P 6= S. Let f, g ∈ S \ P . Enough to
show that f g ∈
/ P . Note: gm 6= 0 and Ann(gm) ) Ann(m), thus Ann(gm) =
Ann(m). So we must have f g ∈
/ Ann(m), else f ∈ Ann(gm).
34
Thus, Sm ' (S/P )(−`), ` = deg m. So M → M/N = M 00 ⊇ Sm, Ñ ⊆ M
is the inverse image of Sm, and N ( Ñ , and the lemma is true for Ñ .
Definition 3.18 (Eventually Polynomial). f : Z → Z is eventually polynomial
if ∃P (z) ∈ Q[z] such that f (n) = P (n) for all n >> 0.
Set ∆f (n) = f (n + 1) − f (n)
Lemma 3.28. f is eventually polynomial of degree r iff ∆f is eventually polynomial of degree r − 1.
Proof. ⇒: Obvious.
z
⇐: Assume ∆f (n) = Q(n) for all n >> 0 where Q(z) = c1 + c2
+...+
1
z
z
z
cr
. Set P (z) = c1
+ . . . + cr
.
r−1
1
r
Then ∆P = Q, so ∆(f − P )(n) = 0 for all n >> 0. Thus f (n) − P (n) = c0
a constant for n >> 0.
Set S = k[x0 , . . . , xn ], M a f.g. graded S-module.
Hilbert Function: HM (d) = dimk (Md ), and Supp(M ) = V+ (Ann(M )) ⊆ Pn .
Note: If 0 → M 0 → M → M 00 → 0 is a short exact sequence of graded
S-modules, then Supp(M ) = Supp(M 0 ) ∪ Supp(M 00 ).
⊇: Ann(M ) ⊆ Ann(M 0 ) ∩ Ann(M 00 ).
⊆: Let x ∈ Pn . If x ∈
/ Supp(M 0 ) ∪ Supp(M 00 ), then ∃f ∈ Ann(M 0 ) such
that f (x) 6= 0 and ∃g ∈ Ann(M 00 ) such that g(x) 6= 0, then f g ∈ Ann(M ), but
f g(x) 6= 0.
Ann(M 0 ) Ann(M 00 ) ⊆ Ann(M ) ⊆ Ann(M 0 ) ∩ Ann(M 00 ).
Theorem 3.29. M f.g. graded module over S = k[x0 , . . . , xn ] then HM (d) is
eventually equal to a polynomial PM (d) of degree dim(Supp(M )).
Proof. ∃ filtration 0 = M 0 ⊂ . . . ⊂ M r = M . such that M i /M i−1 ' (S/Pi )(`i )
where Pi ⊆ S is a homogeneous
prime. P
Pr
r
Note: HM (d) = i=1 HM i /M i−1 (d) = i=1 HS/Pi (d + `i ).
r
i
i−1
r
Supp(M ) = ∪i=1 Supp(M /M ) = ∪i=1 V+ (Pi ). WLOG, M = S/P , P a
homogeneous prime. Induction of V+ (P ):
If P = (x0 , . . . , xn ), then the theorem is true if we take dim ∅ = deg 0 = −1.
Otherwise, some xi ∈
/ P . Set I = P + (xi ). Then V+ (I) ( V+ (P ), so
dim V+ (I) = dim V+ (P ) − 1 by the projective dimension theorem.
By induction, HS/I (d) is eventually polynomial and deg(PS/I ) = dim V+ (P )−
1
0 → (S/P )(−1) → S/P → S/I → 0, so ∆HS/P (d − 1) = HS/P (d) −
HS/P (d − 1) = HS/I (d), so HS/P (d) is eventually polynomial of degree equal to
dim V+ (P ).
z
z
Note: PM (z) = c0 + c1
+ . . . + cr
∈ Q[z] is a numeric polynomial,
1
r
so ci ∈ Z.
r = dim Supp(M ), so r!(lead coef of PM (z)) ∈ Z ≥ 0
35
Definition 3.19 (Hilbert Polynomial of a Variety). Let X ⊆ Pn be a closed
subvariety of dimension r, then PX (z) = PS/I(X) (z) is the Hilbert Polynomial
for X.
We now define deg(X) = r!(lead coef of PX (z)).
Examples
1. deg V+ (f ) = deg f
2. V ⊆ Pn linear subspace. WLOG,
V = V+ S(xr+1 , . . . , xn ). S/I(V ) '
r+d
k[x0 , . . . , xr ], so PS/I(X) (d) =
, so lead coef is 1/r!, so we get
r
degree 1.
Proposition 3.30. X1 , X2 ⊆ Pn closed, dim X1 = dim X2 = r, no components
in common, then deg(X1 ∪ X2 ) = deg X1 + deg X2
Proof. I1 = I(X1 ), X2 = I(X2 ).
I(X1 ∪ X2 ) = I1 ∩ I2 , so 0 → S/(I1 ∩ I2 ) → S/I1 ⊕ S/I2 → S/(I1 + I2 ) → 0.
The first takes f 7→ (f, f ) and the second takes (f, g) 7→ f − g. They
are injective and surjective, and so we see this this is a short exact sequence.
Thus PX1 (d) + PX2 (d) = PX1 ∪X2 (d) + PS/(I1 +I2 ) (d), so Supp(S/(I1 + I2 )) =
V+ (I1 + I2 ) = X1 ∩ X2 . dim < r, so LC(PX1 ) + LC(PX2 ) = LC(PX1 ∪X2 ).
Corollary 3.31. If X ⊆ Pn has dim zero, then deg(X) =the number of points
in X.
Definition 3.20 (Simple Module). If R is a ring and M an R-module, then
M is simple if M has no nontrivial submodules and M 6= 0. This is equivalent
to M ' R/P where P ⊆ R is a maximal ideal.
Definition 3.21 (Decomposition Series). A decomposition series for M is a
filtration 0 = M0 ( M1 ( . . . ( Mr = M such that Mi /Mi−1 is simple for all i.
Definition 3.22 (Artinian Module). If there exists a decomposition series, then
M is an Artinian module and we define length(M ) = r.
Assume that R is Nötherian and that M is a finitely generated R-module.
Then there exists a filtration 0 = M0 ( M1 ( . . . ( Mr = M . such that
Mi /Mi−1 is isomorphic to R/Pi where Pi ⊆ R is maximal.
Note: Ann(M ) ⊆ Pi for all i.
Lemma 3.32. If P ⊆ R is a minimal prime over Ann(M ) then MP is an
Artinian RP -module which has lengthRP (MP ) = |{i : Pi = P }| in the filtration
of M .
Proof. 0 = (M0 )P ⊆ (M1 )P ⊆ . . . ⊆ (Mr )P = MP , and (Mi )P /(Mi−1 )P =
(Mi /Mi−1 )P = (R/Pi )P .
If P = Pi , then we get RP /P RP , else we get 0.
36
X ⊆ Pn is a closed subvariety, S = k[x0 , . . . , xn ], then dimk (S/I(X))d =
PX (d) for all d >> 0. dim(X) = deg(PX ) and deg(X) = (dim X)! LC(PX )
Assume X ⊆ Pn has pure dimension r. That is, all components of X have
dimension r.
Y = V+ (f ) ⊆ Pn a hypersurface such that no component of X is contained
in Y . Also assume that f ∈ S is square-free.
Let Z ⊆ X ∩ Y be a component. Then dim Z = r − 1. Set M = S/(I(X) +
(f )). Supp(M ) = X ∩Y , do if P = I(Z), then P is minimal prime over Ann(M ).
Definition 3.23 (Intersection Multiplicity). I(X · Y ; Z) = lengthSP (MP )
P
Theorem 3.33. deg(X) deg(Y ) = Z⊆X∩Y I(X · Y ; Z) deg(Z).
Proof. Set d = deg(X), e = deg(Y ).
f
Then we have a short exact sequence 0 → S/I(X) → S/I(X) → M → 0.
This gives us 0 → (S/I(X))`−e → (S/(I(X)))` → M` → 0 will be a short
exact sequence for each `, where this is breaking it into homogeneous parts.
Thus PM (`) = PX (`) − PX (` − e). Thus LC(PM ) = r · e · LC(PX ) = red/r! =
de
(r−1)! .
Take filtration 0 = M0 ( M1 ( . . . ( Mt = M , Mi ⊆ M is homogeneous
and Mi /Mi−1 is isomorphic to S/Pi where Pi ⊆ P is a homogeneous prime
ideal.
Pt
P
So PM (`) = i=1 PMi /Mi−1 (`) = Q⊆S hom prime PS/Q (`)|{i : Q = Pi }|,
P
and from this we get Z⊆X∩Y ;component |{i : Pi = I(Z)}|PZ (`) + LOT . So we
P
1
have LC(PM ) = (r−1)!
Z⊆X∩Y I(X · Y ; Z) deg(Z).
Corollary 3.34 (Bezout’s Theorem). P
Let X, Y ⊆ P2 be curves of degree d and
e such that X ∩ Y is a finite set, then P ∈X∩Y I(X · Y ; P ) = de.
Exercise: P ∈ X ∩ Y ⊆ P2 then I(X · Y ; P ) = 1 ⇐⇒ P is a nonsingular
point of both X and Y and X and Y have different tangent directions at P .
Bezout’s Theorem for Pn
Idea: If P
X ⊆ Pn is closed and irreducible and Y ⊆ Pn is a hypersurface, then
[X] · [Y ] = Z⊆X∩Y I(X · Y ; Z) · [Z].
Suppose that Y1 , Y2 , . . . , Yn ⊆ Pn are hypersurfaces suchPthat their interN
section is finite. Then (. . . ([Pn ] · [Y
2 ] . . .) · [Yn ] =
i=1 ci [Pi ] where
P1 ]) · [YQ
n
Y1 ∩ . . . ∩ Yn = {P1 , . . . , PN }. Then
ci = j=1 deg(Yj ).
PNm (m)
In fact: (. . . ([Pn ] · [Y1 ]) . . .) · [Ym ] = i=1
ci [Zi ] where Y1 ∩ . . . ∩ Ym =
Qm
PNm (m)
Z1 ∪ . . . ∪ ZNm components, then j=1 deg(Yj ) = i=1
ci deg(Zi ).
(m)
Fact: ci , ci
do not depend on the order of multiplication.
Corollary 3.35. Y1 ∩ . . . ∩ Yn finite implies that |Y1 ∩ . . . ∩ Yn | ≤
Q
deg(Yj ).
Useful Fact: X1 , . . . , Xm ⊆ Pn irreducible closed, d ∈ N then there exists
irreducible hypersurface Y ⊆ Pn of degree d such that Xi ⊂
6 Y for all i.
37
vd : Pn → PN , N =
n+d
− 1 the veronese map, then Y = Pn ∩ H,
n
H ⊆ PN hyperplane.
Intrinsic: {hypersurfaces of degree d in Pn } ↔ P(Sd )−nonsquarefree by
V+ (f ) ↔ [f ]
{irreducible hypersurfaces} ↔ P(Sd )\∪p+q=d ϕp,q (P(Sp )×P(Sq )) where ϕp,q :
P(S
p ) × P(S
q ) →
P(Sp+q
) :[f ] × [g] 7→[f g].
n+p
n+q
n+p+q
+
≤
so U ⊆ P(Sd ) is a dense open subset.
n
n
n
n
Now, if X ⊆ P is closed, then X ⊆ V+ (f ) iff f ∈ I(X)d . Thus
{hypersurfaces6⊇ X} ↔ WX = P(Sd ) \ P(I(X)d ). Now, we can conclude
that {irreducible hypersurfaces Y in Pn of degree d such that Xi 6⊆ Y for all i
correspond to U ∩ WX1 ∩ . . . ∩ WXm ⊆ P(Sd ) is still a dense open subset.
4
Sheaves
Definition 4.1 (Presheaf). Let X be a topological space. A presheaf F of
abelian groups on X is an assignment U 7→ F (U ) of an abelian group F (U ) to
each open subseteq U of X plus group homomorphisms ρU V : F (U ) → F (V )
whenever V ⊆ U ⊆ X open with
1 ρU U = id
2 ρV W ◦ ρU V = ρU W when W ⊆ V ⊆ U
Notation: elements of F (U ) are called sections of F over U , ρU V are
called restriction maps and s ∈ F (U ), V ⊆ U , then s|V = ρU V (s). Also,
say Γ(U, F ) = F (U ).
Definition 4.2 (Sheaf). A sheaf is a presheaf F such that
3 If s ∈ F (U ) and U = ∪Vi an open cover, then s|Vi = 0 iff s = 0.
4 If U = ∪Vi an open cover and have sections si ∈ F (Vi ) and si |Vi ∩Vj =
sj |Vi ∩Vj for all i, j, then there exists s ∈ F (U ) such that s|Vi = si .
Note: The section s of axiom 4 is unique by axiom 3.
Remark: We can easily define sheaves of rings, sets, modules, etc.
Examples:
1. X is a SWF, then we can define OX of k-algebras by OX (U ) = k[U ]. This
sheaf is called the structure sheaf of X.
2. Let M be a manifold, OM (U ) = {C ∞ f : U → R}. Then there is π :
T M → M the tangent bundle, π −1 (x) = Tx M . Then T M ↔ a sheaf T
of OM -modules. T (U ) = {s : U → T M such that π(s(x)) = x for all
x ∈ U }. T (U ) is a OM (U )-module.
38
Definition 4.3 (Morphism of Sheaves). A morphism ϕ : F → G of (pre)sheaves
consists of homomorphisms ϕU : F → G for all U ⊆ X open such that s ∈ F ,
V ⊆ U open, then ϕU (s)|V = ϕV (s|V ) ∈ G (V ) and
G (V ) .............................................. G (U )
...
...
...
...
...
...
...
...
...
..
.........
.
...
...
...
...
...
...
...
...
...
..
.........
.
ϕV
ϕU
F (V ) ......................................... F (U )
If ϕ : F → G and ψ : G → H are morphisms, we define ψ ◦ ϕ : F → H by
(ψ ◦ ϕ)U = ψU ◦ ϕU . An isomorphism is a morphism with an inverse morphism.
Definition 4.4 (The stalk of F at x). If F is a presheaf on X, x ∈ X then
the stalk of F at x is Fx = limU 3x F (U ). This means
−→
1. Fx is an abelian group (or whatever)
2. If x ∈ U we have homomorphism ρU x : F (U ) → Fx , if x ∈ V ⊆ U then
ρU x = ρV x ◦ ρU V .
3. If G is any abelian group with homomorphisms ΘU : F (U ) → G such
that ΘU = ΘV ◦ ρU V for all x ∈ V ⊆ U , then there exists a unique group
homomorphism Θx : Fx → G such that ΘU = Θx ◦ ρU x
Example: X a variety, x ∈ X. Then OX,x = limU 3x OU =local ring of X at
−→
L
Construction Fx =
U 3x F (U ) /h(0, . . . , 0, s, 0, . . . , 0, −s|V , 0, . . . , 0) : ∀s ∈
F (U ), x ∈ V ⊆ U i.
Notation: If s ∈ FU and x ∈ U write sx = ρU x (s) ∈ Fx .
Exercise Let F be a presheaf.
x.
1. All elements of Fx can be written as sx for some s ∈ F (U ), x ∈ U .
2. s ∈ F (U ), x ∈ U , sx = 0 ∈ Fx iff s|V = 0 ∈ F (V ) for x ∈ V ⊂ U .
Exercise: F is a sheaf. s ∈ F (U ). s = 0 ⇐⇒ sx = 0 for all x ∈ U .
Note: A morphism ϕ : F → G gives a homomorphism ϕx : Fx → Gx ,
ϕx (sx ) = ϕU (s)x , s ∈ F (U ), x ∈ U .
x ∈ U:
Gx ................................................... G (U )
...
...
...
...
...
...
...
...
...
..
.........
.
Fx .......
...
...
...
...
...
...
...
...
...
..
.........
.
....... ....... ...........
F (U )
Proposition 4.1. ϕ : F → G is a morphism of sheaves. ϕ is an isomorphism
iff ϕx : Fx → Gx are isomorphisms for all x ∈ X.
39
Proof. ⇒: Clear
⇐: We must show that ϕ : F (U ) → G (U ) are isomorphisms. ϕU is injective
as s ∈ F (U ), if ϕU (s) = 0 ∈ G (U ) then ϕx (sx ) = ϕU (s)x = 0, but ϕx is
injective, and so sx = 0 for all x ∈ U , so s = 0.
To see that ϕU is surjective, take t ∈ G (U ). ϕx surjective implies that
tx = ϕx (s(x)x ) ∈ Gx for some s(x) ∈ F (Vx ) where Vx ⊆ U are open subsets
containing x.
Now tx = ϕVx (s(x))x . We can make Vx smaller such that t|Vx = ϕVx (s(x)) ∈
G (Vx ). ϕ(s(x)|Vx ∩Vy ) = t|Vx ∩Vy ϕ(s(y)|Vx ∩Vy ). Thus s(x)|Vx ∩Vy = s(y)|Vx ∩Vy .
Patch: There exists a unique s ∈ F (U ) such that s|Vx = s(x) ∈ F (Vx ) for all
x ∈ U . ϕU (s)|Vx = ϕVx (s|Vx ) = ϕVx (s(x)) = t|Vx . Thus ϕU (s) = t ∈ G (U ).
Remark: F is a sheaf on X, U ⊆ X is open. Define F |U to be the sheaf on
U by Γ(V, F |U ) = Γ(V, F )
Example: Let S 1 = {(x, y) ∈ R2 : x2 + y 2 = 1}. Define sheaves F , G by
F (U ) = {f : U → R : f is locally constant} and G (U ) = {g : U → R : ∂g
∂θ = 1}.
If U ( S 1 open then F |U ' GU by f (x, y) 7→ f (x, y) + Arg(x, y), so Fx ' Fx
for all x ∈ S 1 but F and G are not isomorphic as F (S 1 ) = R and G (S 1 ) = ∅.
Sheafification
Let F be a presheaf on X, define a sheaf `
F + as follows: U ⊆ X open set
F + (U ) to be the set of all functions s : U → x∈U Fx such that
1. s(x) ∈ Fx for all x
2. ∀x ∈ U there exists an open set V with x ∈ V ⊆ U and t ∈ F (V ) such
that s(y) = ty ∈ Fy forall y ∈ V .
Definition 4.5. We define a morphism Θ : F → F + by t ∈ F (U ), ΘU (t) =
[x 7→ tx ] ∈ F + (U ).
Exercise: Θx : Fx → Fx+ .
Proposition 4.2. Let F be a presheaf and G be a sheaf. ϕ : F → G is any
morphism. Then there exists a unique ϕ+ : F + → G such that ϕ = ϕ+ ◦ Θ.
`
Proof. Let s ∈ F + (U ). i.e. s : U → x∈U Fx . For x ∈ U choose t(x) ∈ F (Vx )
s.t. x ∈ Vx ⊂ U open and s(y) = t(x)y for all y ∈ Vx .
Set τ (x) = ϕVx = ϕVx (t(x)) ∈ G (Vx ).
If y ∈ Vx , then τ (x)y = ϕ(t(x))y = ϕy (t(x)y ) = ϕy (s(y)) ∈ Gy . Thus,
τ (x1 )|Vx1 ∩Vx2 = τ (x2 )|Vx1 ∩Vx2 , which gives that {τ (x)} glue to τ ∈ G (U ). Set
ϕU (s) = t. Exercise: Check the details!
Definition 4.6. ϕ : F 0 → F is a morphism of sheaves.
We say ϕ is injective if ϕU is injective for all U ⊆ X. If ϕ is injective, then
we say F 0 ⊆ F is a subsheaf as F 0 (U ) ⊆ F (U ).
Exercise: ϕ injective iff ϕx : Fx0 → Fx is injective for all x ∈ X.
Consequence: If ϕ : F → G is a morphism of presheaves such that ϕU :
F (U ) → G (U ) injective for all U ⊆ X open, then ϕ+ : F + → G + is injective.
40
Proof. Check: ϕp : Fp → Gp injective for all p ∈ X. If ϕp (sp ) = 0 ∈ Gp
for s ∈ F (U ), p ∈ U then ϕU (s)p = 0 so ϕU (s)|V = 0 for p ∈ V ⊆ U , so
ϕV (s|V ) = 0. Thus s|V = 0, so sP = 0.
In particular: If a presheaf F is a subpresheaf of a sheaf G then F + is a
subsheaf of G .
Definition 4.7 (Kernel and Image). Let ϕ : F → G be a morphism of sheaves
of abelian groups.
Then ker ϕ = the sheaf U 7→ ker(ϕU ) ⊆ F (U ).
Im ϕ = the sheafification of the presheaf U 7→ Im(ϕU ) ⊂ G (U ).
Note, ker ϕ ⊂ F , Im ϕ ⊂ G are subsheaves, and ϕ injective iff ker ϕ = 0.
Definition 4.8 (Surjective). ϕ is surjective if Im(ϕ) = G .
WARNING: ϕ surjective DOES NOT IMPLY that ϕU is surjective.
Exercise: ϕ surjective iff ϕp : Fp → Gp surjective for all p ∈ X.
Definition 4.9 (Quotient Sheaf). F 0 ⊂ F a subsheaf, then F /F 0 = the
sheafification of [U 7→ F (U )/F 0 (U )]
We have a surjective morphism F → F /F 0 which has kernel F 0 .
φi
φi+1
Notation: A sequence of sheaves → F i → F i+1 → F i+2 → . . . is a
complex if φi+1 ◦ φi = 0 for all i and is exact if Im φi = ker φi+1 for all i.
Equivalently, complexes and exact sequences of the stalks in the category of
abelian groups.
Example, 0 → F 0 → F → F 00 → 0 is exact iff F 0 ⊂ F and F 00 ' F /F 0 .
Definition 4.10. f : X → Y a continuous map, F a sheaf on X, then f∗ F is
a sheaf on Y defined by f∗ F (V ) = F (f −1 (V )).
Example: X a variety, Y ⊆ X closed, i : Y → X the inclusion, U ⊆ X open,
then IY (U ) = {f ∈ OX (U )|f (y) = 0, ∀y ∈ U ∩ Y }, IY ⊆ OX is a subsheaf of
ideals. Then OX (U )/IY (U ) are the regular functions U ∩ Y → k which can be
extended to all of U .
OX /IY ' i∗ OY because we can extend locally. We have exact sequence
0 → IY → OX → i∗ OY → 0, which will often be written 0 → IY → OX →
OY → 0.
Example, f : X → y a morphism of SWFs, we get morphism f ∗ : OY →
f∗ OX by OY (V ) → f∗ (OX (V )) = OX (f −1 (V )), h 7→ h ◦ f = f ∗ h.
Exercise: Find a morphism f : X → Y of varieties such that f ∗ : OY →
f∗ OX is an isomorphism, but f is NOT an isomorphism.
Definition 4.11 (Inverse Image Sheaf). Let f : X → Y continuous, G a sheaf
on Y . U ⊆ X open, define pre − f −1 G (U ) = limV ⊇f (U ) G (V ).
−→
We have maps G (V ) → pre − f −1 G (U ), s 7→ f −1 s, f (U ) ⊆ V . pre −
f −1 G (U ) = {f −1 s|s ∈ G (V ), V ⊇ f (U )}. f −1 s = 0 ⇐⇒ s|W = 0 where
f (U ) ⊆ W ⊆ V . This is a presheaf on X. We define f −1 G to be the sheafification of this presheaf.
41
Special Case: X ⊆ Y is a subset, i : X → Y the inclusion, G |X = i−1 G .
Example: X ⊆ Y open, pre − i−1 G (U ) = limV ⊇U G (V ) = U . It is already
−→
a sheaf, and so the special case just mentioned above holds.
Exercise: (f −1 G )p = Gf (p) .
Adjoint Property
f : X → Y continuous, F a sheaf on X, G a sheaf on Y , let ϕ : G → f∗ F
ϕV
be a morphism of sheaves on Y . U ⊆ X open, V ⊇ f (U ), then G (V ) →
F (f −1 (V )) → F (U ) and maps are compatible with restrictions of G , so this
induces ψU : pre−f −1 G (U ) → F (U ), which gives a morphism ψ : pre−f −1 G →
F , we sheafify to get ψ + : f −1 G → F .
Exercise: hom(G , f∗ F ) → hom(f −1 G , F ) : ϕ 7→ ψ + is an isomorphism of
abelian groups.
Category Theory Interpretation: f −1 is a left adjoint functor to f∗ and f∗
is a right adjoint functor to f −1 .
Let X be a variety (ringed-space)
Definition 4.12 (OX -module). An OX -module is a sheaf F on X such that
F (U ) is an OX (U )-module for all open U ⊆ X, such that if V ⊆ U open
then F (U ) → F (V ) is OX (U )-homomorphism (ie, f ∈ OX (U ), m ∈ F (U ),
(f · m)|V = f |V · m|V .)
An OX -homomorphism ϕ : F → G is a morphism of sheaves such that
ϕU : F (U ) → G (U ) is an OX (U )-homomorphism for all U ⊆ X open.
Note ker ϕ ⊆ F and Im ϕ ⊆ G are sub OX -modules, F is an OX -module
implies that FP is an OX,P -module.
Definition 4.13 (Tensor Product). F , G , OX -modules, F ⊗OX G = [U 7→
F (U ) ⊗OX (U ) G (U )]+
Exercise: (F ⊗OX G )p = Fp ⊗OX,P Gp
Definition 4.14 (Locally Free). An OX -module F is locally free if ∃
X an open cover such that F |Uα ' OU⊗r
.
α
S
α
Uα =
Example: π : An+1 \ {0} → Pn , let m ∈ Z.
Definition 4.15. A sheaf O(m) = OPn (m) of OPn -modules is Γ(U, O(m)) =
{h ∈ k[π −1 (U )] : h(λx) = λm h(x)∀x ∈ π −1 (U ), λ ∈ k ∗ }.
Note: f ∈ S = k[x0 , . . . , xn ] homogeneous of degree > 0. k[π −1 (D+ (f ))] =
k[D(f )] = Sf . So Γ(D+ (f ), O(m)) = (Sf )m .
O(m) is an invertible OPn -module (invertible means locally free of rank 1).
On Ui = D+ (xi ) : OUi → O(m)|Ui , h 7→ xm
i h. Note that OPn (0) = OPn .
Γ(Pn , O(m)) = Sm when m ≥ 0 and 0 else.
Lemma 4.3. O(m) ⊗Pn O(p) ' O(m + p).
Proof. U ⊆ Pn open. Γ(U, O(m)) ⊗ Γ(U, O(p)) → Γ(U, O(m + p)), f ⊗ g 7→ f g.
Sheafification gives a map O(m) ⊗ O(p) → O(m + p), restricting to U we have
OUi ⊗ OUi → OUi .
42
Consequence: O(m) ' O(p) iff m = p. If m ≤ p and O(m) ' O(p) implies
that O(m − p) ' O(p − p), on the right we have nonzero global sections, on the
left we only do if m − p nonnegative, so m = p.
Later: Any invertible sheaf on Pn is isomorphic to O(m), m ∈ Z.
Coherent Sheaves
Let X be an affine variety and A = k[X], let M be an A-module.
Definition 4.16 (Quasi-Coherent Sheaf). M̃ = [U 7→ M ⊗A OX (U )]+ is called
a quasi-coherent OX -module.
Note: M ⊗A OX (D(f )) = M ⊗A Af = Mf .
Examples à = OX , Y ⊆ X closed, I = I(Y ) ⊂ A. I˜ = IY ⊂ OX .
˜ = i∗ OY .
Exercise: A/I
Claim: M̃p = MI(p) for p ∈ X, I(p) = I({p}) ⊂ A.
MI(p) → M̃p by m/f 7→ (m ⊗ 1/f )p , this tensor product is in M̃ (D(F )).
Surjectivity is clear.
Injective: if m/f 7→ 0 ∈ M̃p then (m ⊗ 1/f )|D(h) = 0 for some h ∈ A \ I(p),
so m⊗1/f = 0 ∈ M ⊗A Ah = Mh , thus hn m = 0 ∈ M , and so m/f = 0 ∈ MI(p) .
`
Consequence M̃ (U ) =set of functions s : U → p∈U MI(p) such that ∀p ∈ U ,
there exists p ∈ V ⊂ U and m ∈ M, f ∈ A such tthat s(p) = m/f ∈ MI(q) for
all q ∈ V .
Proposition 4.4. M̃ (D(f )) ' Mf .
Corollary 4.5.
1. OX (D(f )) = Af
2. Γ(X, M̃ ) = M .
We now prove the proposition.
Proof. ψ : Mf → M̃ (D(f )) the obvious m
It is injective, as if ψ(m/f n ) = 0 then m/f n = 0 ∈ MI(p) for all p ∈ D(f ).
Thus,
for all p ∈ D(F ), there exists hp ∈ A \ I(p) with hp m = 0 ∈ P
M . D(f ) ⊆
S
D(hp ) ⇒ V (f ) ⊇ V ({hp }), so f ∈ I(V ({hp })) means f N =
p ap hp for
P
ap ∈ A. f N m = ap hp m = 0, thus m/f n = 0 ∈ Mf .
Sr
Surjectivity: Let s ∈ M̃ (D(f )), there exists a cover D(f ) = i=1 Vi , mi ∈
M , hi ∈ A such that s = mi /hi on Vi . WLOG Vi = D(gi ) and Vi = D(hi )
(replace mi 7→ gi mi by hi 7→ gi hi ).
On D(hi hj ) = D(hi )∩D(hj ), s = mi /hi = mj /hj ∈ M̃ (D(hi hj )), injectivity
for D(hi hj ) : mi /hi = mj /hj ∈ Mhi hj . So (hi hj )N (hj mi − hi mj ) = 0 ∈ M .
+1
N
Replace mi 7→
hN
i P, hj mi = hi mj .
S hi mi , hi 7→
P
n
D(f ) ⊆ D(hi ) so f =
ai hi where ai ∈ A. Set m =
ai mi ∈ M .
Claim: s = ψ(m/f n ).P
P
For all j, hj m = i hj ai mi = ( i ai hi ) mj = f n mj .
So m/f n = mj /hj = s on D(hj ).
43
Definition 4.17 (Quasi-Coherent and Coherent). Let X be any variety.SAn
OX -mdule F is quasi-coherent if there exists an open affine cover X = Ui
and k[Ui ]-modules Mi such that F |Ui ' M̃i as OUi -modules.
F is coherent if Mi finitely generated k[Ui ]-modules for all i.
Examples:
1. All locally free OX -modules are coherent, U = Spec −m(A), A˜⊗r = OU⊗r .
2. Y ⊆ X closed, IY and i∗ OY are coherent OX -modules.
/U
OA1 (U ) 0 ∈
. This is called the extenExample: X = A1 , F (U ) =
0
0∈U
sion of the OA1 \{0} by zeros. F is an OX -module, but is not quasi-coherent, as
if U ⊆ X is open affine and 0 ∈ U , then Γ(U, F ) = 0 but F |U is not the zero
sheaf.
Exercises:
˜ ) for all open affine U ⊆ X.
1. F is quasi-coherent iff F |U ' F (U
2. F is coherent implies F (U ) is finitely generated k[U ]-module for all affine
open U .
3. f : X → Y a morphism of varieties.
(a) f affine implies that f∗ OX quasi-coherent.
(b) f finite f∗ OX coherent.
Example: M a finitely generated A-module, X = Spec −m(A). Then M̃ is
locally free OX -module of rank r iff M is a projective A-module of const. rank
r.
S
⊗
(M̃ loc free iff X = D(fi ), M̃D(fi ) ' OD(f
r iff Mfi = A⊗r
fi iff M projeci)
tive.)
Recall: X is complete implies Γ(X, OX ) = k. More general fact: F a
coherent sheaf on a complete variety X then dimk Γ(X, F ) < ∞. We will use
this without proof. (Projective case in Hartshorne, General in EGA.)
Note that Γ(A1 , OA1 ) = k[t] which has infinite dimension.
Pushforward, Pullback
Let f : X → Y be a morphism of varieties. F an OX -module, then f∗ F is
a f∗ OX -module, we have ring homomorphism f ∗ : OY → f∗ OX , and so f∗ F is
an OY -module.
Let G be a OY -module f −1 G is an f −1 OY -module. (f −1 h · f −1 s = f −1 (hs))
OY → f∗ OX is the same as f −1 OY → OX by the adjoint property.
Definition 4.18 (Pullback). Define f ∗ G = f −1 G ⊗f −1 OY OX , we call it the
pullback.
Examples
44
1. f ∗ OY = OX .
2. (f ∗ G )p = (f −1 (G))p ⊗(f −1 OY )p OX,p = Gf (p) ⊗OY,f (p) OX,p .
3. U ⊆ Y open, i : U → Y inclusion, i∗ G = G |U ⊗OY |U OU = G |U .
Adjoint Property
f : X → Y a morphism of SWFs. G an OY -module. We have f −1 OY homomorphism f −1 G → f ∗ G by s 7→ s⊗1. This gives an α : OY -homomorphism
F → f∗ f ∗ G .
Lemma 4.6. G is an OY -module, F is an OX -module. Then homOX (f ∗ G , F ) '
homOY (G , f∗ F ).
α
f∗ ψ
Proof. Given ψ : f ∗ G → F we obtain φ : G → f∗ f ∗ G → f∗ F .
Given φ : G → f∗ F , we obtain φ̃ : f −1 G → F which is an f −1 OY -hom.
Take ψ : f ∗ G = f −1 F ⊗f −1 OY OX → F by s ⊗ h 7→ h · φ̃(s).
f
g
Functoriality X → Y → Z morphism of SWFs. If F is a sheaf on X,
g∗ (f∗ F ) = (gf )∗ F .
Proposition 4.7.
1. G on Z implies that (gf )−1 G = f −1 (g −1 G )
2. G an OZ -module implies that (gf )∗ G = f ∗ (g ∗ G ).
Proof. We will prove case 2.
id : (gf )∗ G → (gf )∗ G gives G → (gf )∗ (gf )∗ G = g∗ f∗ (gf )∗ G gives g ∗ G →
f∗ (gf )∗ G which gives f ∗ (g ∗ G ) → (gf )∗ G .
We have a global homomorphism, so enough to check stalks. f ∗ (g ∗ G )p =
∗
(g G )f (p) ⊗OY,f (p) OX,p . This is (Gg(f (p)) ⊗OZ,gf (p) OY,f (p) ) ⊗OY,f (p) OX,p =
Ggf (p) ⊗OZ,gf (p) OX,p = ((gf )∗ G )p
Let f : X → Y be a morphism of SWFs. G is an OY -module, and f −1 G is
an f −1 OY -module.
Definition 4.19 (Pullback). f ∗ G = f −1 G ⊗f −1 OY OX is an OX -module.
f ∗ : OY → f∗ OX induces a map f −1 OY → OX .
Some sections: σ ∈ G (V ), we set f ∗ σ = f −1 σ ⊗ 1 ∈ Γ(f −1 (V ), f ∗ G ). The
stalks (f ∗ G )p = Gf (p) ⊗OY,f (p) OX,p .
g
f
If Z → X → Y , then (f g)∗ G = g ∗ (f ∗ G ).
Corollary 4.8. If G is a locally free OY -module, then f ∗ G is a locally free
OX -module of the same rank.
S
Proof. Let Y =
Vi be an open cover such that G |Vi ' OV⊕r
. Set Ui =
i
−1
f (Vi ) ⊂ X.
45
X
f
........................................................
..
.........
....
...
..
...
...
...
...
...
.
q
p
Ui
Y
f0
..
.........
....
...
..
...
...
...
...
...
.
........................................................
f G |Ui
∗
Vi
= p∗ f ∗ G = f 0∗ q ∗ G ' f 0∗ q ∗ (OV⊕r
) = OU⊕r
i
i
Lemma 4.9. f : X → Y a morphism, 0 → G 0 → G → G 00 → 0 is a short exact
sequence of OY -modules. Then f ∗ G 0 → f ∗ G → f ∗ G 00 → 0 is an exact sequence
of OX -modules. If G 00 is locally free, then the first map is injective.
Proof. On stalks we start with 0 → Gf0 (p) → Gf (p) → Gf00(p) → 0 exact. Tensor
produce is right exact and gets us to f ∗ , ad so we have the first part of the
theorem immediately.
If G 00 is locally free, then Gf00(p) is a free OY,f (p) -module, and so the original
sequence is split-exact.
Definition 4.20 (Generated by Finitely Many Global Sections). The OX module F is generated by finitely many global sections iff ∃ a surjective map
⊕m
OX
→ F.
Equivalently, ∃s1 , . . . , sm ∈ Γ(X, F ) such that Fp generated by (s1 )p , . . . , (sm )p
as an OX,p -module.
Example: Any quasi-coherent OX -module, if X is affine (this is just generated by global sections, requires coherent to be generated by finitely many)
Example: OPn (1) is generated by x0 , . . . , xn ∈ Γ(Pn , O(1)).
Suppose that f : X → Pn is a morphism, then OP⊕n+1
→ O(1) → 0 exact
n
⊕n+1
implies that OX
→ f ∗ O(1) → 0 is exact, so f ∗ O(1) is generated by global
sections f ∗ (x0 ), . . . , f ∗ (xn ).
Proposition 4.10. X a variety, L invertible OX -module generated by global
sections s0 , . . . , sn ∈ Γ(X, L ). Then ∃!f : X → Pn such that f ∗ O(1) ' L and
f ∗ (xi ) ↔ si .
Proof. Set Ui = {p ∈ X|(si )p ∈
/ mp Lp }.
Ui open: If V ⊆ X open with L |V ' OV , then L |V is generated by
t ∈ Γ(V, L ) so we write si = hi t on V with hi ∈ k[V ]. Then Ui ∩ V = {p ∈
V |hi (p) 6= 0} is open in V .
Sn
Note: L is generated by s0 , . . . , sn implies that X = i=1 Ui , and OUi '
L |Ui implies that 1 7→ si |Ui .
On Ui , we can write si = hij sj for some hij ∈ k[Ui ]. We define a map
g : Ui → Pn by f (p) = (hi0 (p) : . . . : hin (p)).
The maps are compatible: On Ui ∩ Uj , h`j s` = sj = hij si = hij h`i s` , so
h`j = h`i hij . The map on U` : p 7→ (h`0 (p) : . . . : h`n (p)) = h`i (p)hi0 (p) : . . . :
h`i (p)hin (p). Thus, we have a morphism f : X → Pn .
Claim: ∃ isomorphism L → f ∗ O(1) by si 7→ f ∗ (xi ). On Ui , we define
L |Ui → f ∗ O(1)|Ui by hsi 7→ hf ∗ (xi ). This means that s` = hi` si 7→ hi` f ∗ (xi ),
46
we must check that f ∗ (x` ) = hi` f ∗ (xi ). The definition of f implies that (x` /xi )◦
= hi` , so f ∗ (x` ) = f ∗ ( xx`i xi ) = f ∗ ( xx`i )f ∗ (xi ) = hi` f ∗ (xi ).
f = hhi`
ii
And now we prove uniqueness: If f : X → Pn is any morphism such
that L ' f ∗ O(1) and si ↔ f ∗ (xi ) on Ui , hi` si = f ∗ (x` ) = f ∗ ( xx`i xi ) =
f ∗ ( xx`i )f ∗ (xi ) = f ∗ ( xx`i )si , so f ∗ (x` /xi ) = hi` on Ui .
Definition 4.21 (Very Ample Sheaf). Let L be an invertible sheaf on X.
L is very ample iff L is generated by (finitely many) global sections and the
map f : X → Pn given by generators s0 , . . . , sn ∈ Γ(X, L ) is an isomorphism
f : X → W ⊆ Pn locally closed.
Exercise: OPn (m) is very ample iff m ≥ 1.
Definition 4.22 (PGL). P GL(n) = GL(n + 1)/k ∗ .
Exercise: P GL(n) is an affine algebraic group.
FACT: Every invertible sheaf on Pn is isomorphic to O(m) for some m.
Corollary 4.11. Aut(Pn ) ' P GL(n).
Proof. P GL(n) ≤ Aut(Pn ) is trivial.
Let f : Pn → Pn be anautomorphism.
The fact implies that f ∗ O(1) '
n+m
O(m) for some m. In fact,
= dimk Γ(Pn , O(m)) = dim Γ(f ∗ O(1)) =
m
dim Γ(O(1)) = n + 1, so m = 1.
Pn
f ∗ x0 , . . . , f ∗ xn ∈ Γ(Pn , O(1)) form a basis. We write f ∗ (xi ) = i=0 aij xj
for aij ∈ k. Then A = (aij ) ∈ GL(n + 1).
P
Define ϕ : Pn → Pn by ϕ(x0 : . . . : xn ) = ( a0j xj : . . . : anj xj ). ϕ∗ ( xx`i ) =
P
P a`j xj
aij xj
= f ∗ ( xx`i ).
Thus, f = ϕ ∈ P GL(n).
Corollary 4.12. Pn is not an algbriac group.
Normal Varieties
Definition 4.23 (Normal Variety). X is irreducible. Then X is normal iff
OX,P is normal (integrally closed) for all p.
Example: Nonsingular varieties.
Note: X affine, then X is normal iff k[X]m normal for all maximal m iff k[X]
is normal.
Exercise: If A is a domain, S ⊆ A multiplicative, then S −1 A = S −1 A.
Definition 4.24 (Normalization). If f is an affine variety, k[X] ⊂ k(X),
then k[X] ⊆ k(X) is the integral closure. The normalization of X is X =
Spec −m(k[X]).
As we have the inclusion k[X] → k[X], we get a projection map X → X
which is finite.
47
Lemma 4.13. ϕ : U → X morphism of affines, ϕ an open embedding iff
∃f1 , . . . f − n ∈ k[X] such that (ϕ∗ f1 , . . . , ϕ∗ fn ) = (1) ⊂ k[U ] and ϕ∗ : k[X]fi →
k[U ]ϕ∗ fi is an isomorphism for all i.
Sr
Proof. ⇒: Take open
Sr cover U = i=1 ∗D(fi ), fi ∗∈ k[X].
⇐: Set V = i=1 D(fi ) ⊆ X. (ϕ f1 , . . . , ϕ fr ) = (1) ⊆ k[U ] implies that
ϕ(U ) ⊆ V . Thus, we have that ϕ : ϕ−1 (D(fi )) → D(fi ) is an isomorphism, so
ϕ : U → V isomorphism.
Lemma 4.14. Assume X affine, U ⊆ X is an open affine, then Ū ⊆ X̄ open
affine.
Proof. k[X] ⊆ k[U ] ⊆ k(X). Thus, k[X] ⊆ k[U ], so we have a morphism
ϕ : Ū → X̄. Take f1 , . . . , fn ∈ k[X] as in lemma wrt U ⊂ X.
(f1 , . . . , fn ) = (1) ⊆ k[U ], and k[Ū ]fi = k[U ]fi = k[X]fi = k[X̄]fi . And so,
the lemma implies that ϕ is an open embedding.
Exercise: Given pre-varieties X1 , . . . , Xn , open subsets Uij ⊆ Xi and isomorphisms ϕij : Uij → Uji such that Uii = Xi , ϕii = id, for all i, j, k,
ϕij (Uij ∩ Uik ) = Uji ∩ Ujk and ϕik = ϕjk ◦ ϕij on Uij ∩ Uik , then ∃! prevariety X with morphisms
Sn ψi : Xi → X such that ψi : Xi → open⊂ X is an
isomorphism, X = i=1 ψi (Xi ), ψi (Uij ) = ψi (Xi ) ∩ ψj (Xj ) and ψi = ψj ◦ ϕij
on Uij .
Definition 4.25 (Normal). A variety X is normal iff X is irreducible and OX,P
are normal for all p.
Construction: X an irreducible variety, X = X1 ∪ . . . ∪ Xn open affine cover,
set Uij = Xi ∩ Xj . Then Uij is affine, have ψij : Uij → Uji nbe the identity.
Now Ūij ⊆ X̄i and still have φij : Ūij → Ūji is the identity, so the satisfy the
hypotheses of the exercise. Thus, there exists a prevariety X̄ = X̄1 ∪ . . . ∪ X̄n .
We call this the normalization of X.
Note: k[Xi ] is a finitely generated k[Xi ]-module, so we have finite π : X̄i →
Xi , which we can glue to a morphism π : X̄ → X.
Exercise: π : X̄ → X is finite. (Check that π −1 (Xi ) = X̄i )
Exercise: ϕ : X → Y affine morphism of pre-varieities. Then Y is separated
implies that X is serpated, and so X̄ is an irreducible normal separated variety.
Example: X irred curve implies π : X̄ → X resoluton of singularities.
Definition 4.26 (Local Ring along Subvariety). Let X be a variety, V ⊆ X
irreducible and closed. Then OX,V = limU ⊆X,U ∩V 6= OX (U ).
−→
If X is irreducible, then OX,V = {f ∈ k(X) : f is defined at one point of
V }.
General Case: U ⊆ X open affine, U ∩ V 6= ∅, P = I(U ∩ V ) ⊆ k[U ],
OX,V = k[U ]P .
Definition 4.27 (Regular along V ). X is regular along V if OX,V is a regular local ring. i.e., the maximal ideal in OXV is generated by dim(OX,V ) =
codim(V ; X) elements. This happens iff V 6⊆ Xsing .
48
5
Divisors
Let X be a normal variety.
Definition 5.1 (Prime Divisor). A prime divisor on X is a closed irreducible
subvariety of codimension 1.
Note: Y a prime divisor implies that OX,Y is a normal Nötherian local ring
of dimension 1. That is, OX,Y is a DVR.
Consequences
1. codim(Xsing ; X) ≥ 2.
2. Have valuation map vY : k(X)∗ → Z for each prime divisor Y ⊆ X.
Lemma 5.1. Let f ∈ k(X)∗ . Then vY (f ) = 0 for all but finitely many Y .
Proof. Show vY (f ) < 0 for finitely many Y . Set U ⊆ X open set where f
defined. Z = X \ U . vY (f ) < 0 iff f ∈
/ OX,Y iff f is not defined at any point of
Y iff Y ⊆ Z component.
Definition 5.2 (Divisor Group). Define Div(X) =free abelian group generated
by all prime divisors.P
An element D =
ni [Yi ] is a finite sum and is called a Weil Divisor.
P
Definition 5.3. For f ∈ k(X)∗ , set (f ) = Y vY (f ) · [Y ] ∈ Div(X).
Note (f −1 ) = −(f ), (f g) = (f ) + (g) Thus k(X)∗ → Div(X) by f 7→ (f ) is
a group homomorphism.
Definition 5.4 (Class Group). Define C`(X) = Div(X)/{(f ) : f ∈ k(X)∗ }.
Example: C`(An ) = 0, as every hypersurface corresponds to a prime divisor.
Remark: X complex, dim(X) = n, then C`(X) ↔ H2n−2 (X; Z).
Remark: X irreducible but not normal, we can still define C`(X), use
vY (f /g) = lengthOX,Y (OX,Y /(f )) − lengthOX,Y (OX,Y /(g)).
Divisors on Pn
Note: All prime divisors are hypersurfaces Y = V+ (h) where h ∈ S =
k[x0 , . . . , xn ] is an irreducible form.
P
Definition
5.5. Degree of a Divisor deg : Div(Pn ) → Z by deg( mi [Yi ]) =
P
mi deg Yi .
Qr
mi
n ∗
P Let f ∈ k(P ) , mi ∈ Z, g = i=1 nhi , hi ∈ S irreducible form. Then
mi deg(hi ) = 0, so Yi = V+ (hi ) ⊆ P is a prime divisor, so vYi (hi ) = 1,
vYi (f ) = mP
i.
P
r
(f ) =
mi deg(hi ) = 0, thus, deg :
i=1 mi [Yi ] implies that deg(f ) =
C`(Pn ) → Z is well-defined.
Claim: Isomorphism.
49
n
P Surjective: Hn ⊆ P hyperplane, deg(m[H]) = m. Injective: Let D =
m
0, then Yi = V+ (hi ), hi ∈ S irred form,
Pi [Yi ] ∈ Div(P ), suppose deg(D) =Q
i
∈ k(Pn )∗ and D = (f ).
so
mi deg(hi ) = deg(D) = 0, so f = hm
i
Later: X nonsingular implies C`(X) ' Pic(X), thus Pic(Pn ) = Z.
X normal, Y ⊆ X prime divisor implies that OX,Y is a DVR.
Theorem 5.2. R normal Nötherian domain implies R = ∩ht p=1 Rp where
ht p = dim Rp , the max m such that ∃0 ( p1 ( . . . ( pm = p.
Corollary 5.3. If X is normal, f ∈ k(X)∗ , then f ∈ k[X] iff vY (f ) ≥ 0 for all
Y ⊆ X prime divisors.
Lemma 5.4. R Nötherian, then R is a UFD iff all prime ideals of height one
are principal.
Proof. ⇒: Assume R a UFD. P ⊆ R prime of height 1, let x ∈ P be an
irreducible element, 0 ( (x) ⊆ P , so P = (x).
⇐: Exercise: R any Nötherian domain then every element of R is a product
of irreducible elements.
Unique Factorization: Show x ∈ R irred and x|f g implies that x|f or x|g,
ie, (x) ⊆ R is prime, let P ⊇ (x) min prime, PIT implies ht(P ) = 1 implies
P = (y), x = ay, a ∈ R a unit.
Proposition 5.5. X irreducible affine variety, k[X] a UFD iff X normal and
C`(X) = 0.
Proof. ⇒: UFD implies normal. Let Y ⊆ X a prime divisor, P = I(Y ) ⊆ k[X]
prime of height 1. P = (h) ⊆ k[X], h ∈ k[X]. So (h) = [Y ] implies [Y ] = 0 ∈
C`(X).
⇐: Let P ⊆ k[X] prime of height 1, Y = V (P ) ⊆ X a prime divisor,
[Y ] = 0 ∈ C`(X) so [Y ] = (h) ∈ Div(X), h ∈ k(X)∗ . vZ (h) ≥ 0 for all Z ⊆ X
prime divisors implies h ∈ k[X]. Claim: P = (h) ⊆ k[X]. ⊇ is clear. Let g ∈ P ,
then vY (g) ≥ 1, so vZ (g/f ) ≥ 0 for all Z, so g/f ∈ k[X], and g = af ∈ (f ).
Proposition 5.6. X normal, Z ⊆ X is a proper closed subset, U = X \ Z.
Then
1. C`(X) → C`(U ) by [Y ] 7→ [Y ∩ U ] if Y ∩ U 6= ∅ and 0 else is surjective
2. If codim(Z; X) ≥ 2, then C`(X) = C`(U ).
3. If Z prime divisor, then Z → C`(X) → C`(U ) → 0 is exact.
Proof.
1. Well defined Div(X) → Div(U ). f ∈ k(X)∗ , (f ) 7→ (f |U ) (because
if Y ⊆ X is a prime divisor, Y ∩ U 6= ∅ then OX,Y = OU,U ∩Y ).
Thus, C`(X) → C`(U ) is well defined. Surjective: If V ⊆ U a prime
divisor, V̄ ⊆ X is a prime divisor [V̄ ] 7→ [V ].
2. Div(X) = Div(U ), so (f ) = (f |U ).
50
P
3. If D =
nY [Y ] 7→ 0 ∈ C`(U ), then ∃f ∈ k(X)∗ : vY (f ) = nY for all
Y 6= Z. D − (f ) = m[Z] ⇒ D = m[Z] ∈ C`(X).
Example: X = V (xy − z 2 ) ⊂ A3 .
Exercise: X (above) is normal. L = V (y) ∩ X = V (y, z) is a prime divisor
on X.
2
Max ideal of OX,L is generated by z, y = zx ∈ OX,L .
Set U = X \ L affine, k[U ] = (k[x, y, z]/(xy − z 2 ))y = k[y, z]y is a UFD, so
C`(U ) = 0. Z → C`(X) → C`(U ) = 0. So C`(X) = {m[L] : m ∈ Z}, y ∈ k(X)∗ .
(y) = vL (y)[L] = vL (z 2 )[L] = 2[L]. Thus, C`(X) = Z/2Z or C`(X) = 0. As
k[x, y, z]/(xy − z 2 ) is not a UFD, C`(X) = Z/2Z.
Picard Group
Invertible OX -module = line bundle.
Let X be any variety, L1 and L2 are line bundles, then L1 ⊗OX L2 is
a line bundle. L is invertible implies that we can define L −1 = [U 7→
homOU (L |U , OU )].
Exercise: L −1 is an invertible OX -module and L ⊗ L −1 ' OX .
Definition 5.6 (Picard Group). Pic(X) = {isomorphism classes of invertible
sheaves on X}.
This is a group under tensor product.
Notation: X irreducible variety, L an invertible sheaf on X and s ∈ L (U ),
t ∈ L (V ) are nonzero sections. Take W ⊂ U ∩ V open such that L |W ' OW
generated by u ∈ L (W ). Then s|W = f u and t|W = gu, f, g ∈ k[W ]. Define
s/t = f /g ∈ k(X)∗ .
Example: s0 , . . . , sn ∈ Γ(X, L ). Define f : X 99K An ⊂ Pn . f (x) =
(s1 /s0 (x), . . . , sn /s0 (x)) = (s1 /s0 (x) : . . . : sn /s0 (x) : 1). If s0 , . . . , sn generate
L , then f extends to a morphism f : X → Pn .
X is a normal variety, s ∈ L (U ), s 6= 0, Y ⊆ X is a prime divisor, take
V ⊆ X open such that L |V ' OV generated by t ∈ L (V ) and V ∩ Y 6= ∅.
Definition 5.7. VY (s) = VY (s/t).
Well defined, if V 0 ⊆ X, V 0 ∩ Y 6= ∅, L |V 0 generated by y 0 ∈ L (V 0 ) implies
that t/t0 nowhere vanishing function on V ∩ V 0 , so t/t0 is a unit in OX,Y .
Thus, VY (s/t0 ) = VY (s/t · t/t0 ) = VY (s/t) + 0.
P
Definition 5.8. (s) = Y VY (s)[Y ] ∈ Div(X).
Note: If s0 ∈ L (U 0 ) then (s0 ) = (s) + (s0 /s). Then (s0 ) = (s) ∈ C`(X), so
all nonzero sections of a line bundle are equivalent in the class group.
Thus, we have a well defined map Pic(X) → C`(X) by L 7→ (s).
Check: s1 ∈ L1 (U ), s2 ∈ L2 (U ), then s1 ⊗ s2 ∈ L1 ⊗ L2 (U ) and (s1 ⊗ s2 ) =
(s1 ) + (s2 ), so this map is a group homomorphism.
Cartier Divisors
X normal.
51
P
Definition 5.9 (Cartier Divisor). A Cartier is a Weil Divisor
ni [Yi ]
Sn D =
which is locally principal. I.E. there exists an open covering i=1 Ui = X such
that D|Ui = 0 ∈ C`(Ui ) for all i.
Note: D|Uj = (fj ) where fj ∈ k(Uj )∗ . We can think about D as the
collection {fj } of these generators.
Definition 5.10 (Cartier Class Group). CaC`(X) = {Cartier Divisors on
X}/{(f ) : f ∈ k(X)∗ }.
Recall:
P L is an invertible OX -module, s ∈ L (U ) nonzero section, then
(s) = Y prime vY (s) · [Y ] ∈ Div(X) where vY (s) = vY (s/t) for t ∈ L (V )
generator of L |V ' OV , Y ∩ V 6= ∅.
Note: (s) is Cartier.
Thus, we have a group homomorphism Pic(X) → CaC`(X) ⊂ C`(X).
Line Bundles
P from Divisors
Let D =
nY [Y ] ∈ Div(X).
Definition 5.11. OX -module L (D) or OX (D) Γ(U, L (D)) = {f ∈ k(X)∗ |vY (f ) ≥
−nY for all prime divisors Y such that Y ∩ U 6= ∅} ∪ {0}.
Example: L (0) = OX (0) = OX .
Example: X = P1 , Q = (a : b) ∈ X, D = n[Q], Q = V+ (h), h = bx0 − ax1 ∈
k[x0 , x1 ]. So OP1 (n[Q]) ' OP1 (n) by f 7→ hn f .
Note: If h ∈ k(X)∗ then OX (D + (h)) → OX (D) by f 7→ hf is an isomorphism. vY (f ) ≥ −nY − vY (h) ⇐⇒ vY (f h) ≥ −nY .
Consequence: D is a Cartier Divisor implies that OX (D) is an invertible
OX -module.
If D|U = (h) ∈ Div(U ), then OX (D)|U = OU ((h)) ' OU . Note, this says we
have a map CaC`(X) → Pic(X) by D 7→ OX (D).
WARNING: If f ∈ Γ(U, OX (D)) then f a rational function. The notation
(f ) means two things!
Proposition 5.7. Pic(X) ' CaC`(X) as abstract groups.
←
Proof. Will check
P that Pic(X) → CaC`(X) are inverse maps.
Let D =
nY [Y ] a Cartier Divisor. Set V = X \ (∪nY <0 Y ) ⊆ X open.
Then 1 ∈ k(X)∗ is a section of OX (D) over V . (vY (1) ≥ −nY ⇐⇒ nY ≥ 0).
Claim: (1) = D ∈ Div(X). If D|U = (g) ∈ Div(U ), then OX (D)|U ' OU
gneerated by h−1 ∈ Γ(U, OX (D)), Y ∩ U 6= ∅ implies that vY (1) = vY (1/h−1 ) =
vY (h). Thus, (1)|U = (h)|U = D|U .
Thus, CaC`(X) → Pic(X) → CaC`(X) is the identity.
Let L be a line bundle on X. t ∈ Γ(U, L ) a non-zero section.
Note: If 0 6= s ∈ L (V ), then Y ∩ V 6= ∅ implies vY (s/t) = vY (s) − vY (t) ≥
−vY (t), and so s/t ∈ Γ(V, OX ((t))).
Claim: L ' OX ((t)) by s 7→ s/t. If L |V ' OV gneerated by u ∈ L (V )
then (t)|V = (t/u) ∈ Div(V ) implies that OX ((t))|V is generated by u/t as
u 7→ u/t we get L |V ' OX ((t))|V .
52
Examples:
1. Pic(An ) = CaC`(An ) ⊂ C`(An ) = 0, so all line bundles on An are trivial.
Fact: Any locally free OAn -module of finite rank is trivial.
Sn
2. Pn = i=0 D+ (xi ), C`(D+ (xi )) = 0, so all Weil divisors are Cartier, thus
Pic(Pn ) = CaC`(Pn ) = C`(Pn ) = Z.
By the maps we have, any line bundle is isomorphic to OPn (m[H]), where
H ⊂ Pn is a hyperplane. I(H) = (h), OPn (m[H]) ' OPn (m) by f 7→ hm f .
Thus, Pic(Pn ) = {O(m)}.
3. X = V (xy − z 2 ) ⊂ A3 . L = V (y) ∩ X, I(L) = (y, z) ⊂ k[A3 ]. Claim:
[L] is not Cartier. Otherwise there exists open affine U ⊂ X such that
P = (0, 0, 0) ∈ U with [L ∩ U ] = (f |U ) ∈ Div(U ). Thus f ∈ k[U ] and
I(L ∩ U ) = (f ) ⊂ k[U ], so I(L) · OX,P = (y, z) ⊂ OX,P is principal.
But P ∈ X is a singular point, so dimk (mP /(m)2P ) = 3, so {x, y, z} is a
basis, and so dim((y, z) + m2P /m2P ) = 2, so (y, z) ⊂ OX,P is not principal,
which is a contradiction. Thus CaC`(X) = Pic(X) = 0 6= C`(X).
Note: OX,P is not a UFD, as (y, z) ⊂ OX,P is height 1 prime but not
principal.
Definition 5.12 (Locally Factorial). An irreducible variety X is locally factorial if OX,P is a UFD for all p ∈ X.
Example: Nonsingular implies locally factorial implies normal.
Proposition 5.8. X locally factorial implies Pic(X) = C`(X).
Proof. Show that any prime divisor [Y ] is Cartier. First: U = X \ Y , [Y ]|U = 0.
Let P ∈ Y , then I(Y ) · OX,P ⊂ OX,P is a height 1 prime, so I(Y ) · OX,P =
(f ) ⊂ OX,P , f ∈ OX,P ⊂ k(X).
Note: vY (f ) = 1, if Z 6= Y prime divisor, p ∈ Z, then f ∈ OX,Z (defined
at
P
P ) and f ∈
/ I(Z) · OX,P . Thus, vZ (f ) = 0, and we have (f ) = [Y ] + ni [Zi ]
where p ∈
/ Zi for all i.
Set U = X \ (∪Zi ) open in X, p ∈ U . Then [Y ]|U = (f )|U ∈ Div(U )
principal, so [Y ] is Cartier.
Example: X = V (xy − z 2 ) ⊂ A3 , X0 = X \ {0, 0, 0}, X0 is nonsingular, so
Pic(X0 ) = C`(X0 ) = C`(X) = Z/2Z, so there exists a unique nontrivial line
bundle on X0 which is NOT equal to the restriction of a line bundle on X.
Definition 5.13 (Affine, Finite). Let f : X → Y be a morphism of varieties.
f is affine if f −1 (V ) ⊂ X is affine for all V ⊂ Y is open affine.
f is finite if it is affine and k[f −1 (V )] is a finitely generated k[V ]-module.
53
Exercise: Enough that this is true for an open affine cover of Y .
Examples: X, Y affine, f : X → Y morphism is affine.
X ⊂ Y closed, then the inclusion is finite.
Divisors on Non-Singular Curves
Recall that X nonsing complete curve implies that X is projective. X ⊂ CK
open, K = k(X), X = CK .
Lemma 5.9. Let X be a complete, nonsingular curve, then any nonconstant
morphism f : X → Y is finite.
Proof. WLOG: Y is a curve. Thus, f ∗ : k(Y ) ⊂ k(X) is a finite field extension.
Take V ⊆ Y open affine, k[V ] ⊂ k(Y ). Set A = k[V ] ⊂ k(X), then A is a
finitely generated k[V ]-module.
U = Spec −m(A) a nonsingular curve, k(U ) = k(X) ⇒ we have diagram
f .
X ....................................................... Y
...
...
...
...
...
...
...
...
...
..
.........
.
...
...
...
...
...
...
...
...
...
..
.........
.
⊂
U
⊂
........................................................
V
Claim: f −1 (V ) = U . x ∈ f −1 (V ) ⇒ k[V ] ⊆ OX,x , so A ⊂ OX,x , thus
OX,x = AP for some P ⊂ A prime.
Thus, x = P ∈ U = Spec −m(A).
Definition 5.14 (Degree of f ). Let f : X → Y be a finite, dominant morphism,
then deg(f ) = [k(X) : k(Y )].
Pullback of Divisors on Curves
f : X → Y a finite morphism of nonsingular curves. Q ∈ Y , mQ = (t) ⊆
OY,Q . If f (P ) = Q then f ∗ : OY,Q → OX,P , f ∗ t ∈ mP .
P
Definition 5.15. f ∗ : Div(Y ) → Div(X) : [Q] → P ∈f −1 (Q) vP (t)[P ].
Alternatively, if D ∈ Div(Y ), set V = Y −Supp(D), then s = 1 ∈ Γ(V, L (D)).
Note: (s) = D. Then f ∗ s ∈ Γ(f −1 (V ), f ∗ L (D)) is the pullback.
Exercise: f ∗ D = (f ∗ s) ∈ Div(X).
Definition 5.16 (Torsion Free). Let R be a domain, M an R-module. M is
torsion free if ∀a ∈ R, x ∈ M , then ax = 0 implies a = 0 or x = 0.
Fact: Any f.g. torsion-free module over a PID is free.
Definition
5.17 (Degree of a Divisor).
X a nonsingular curve, D =
P
P
ni Pi ∈ Div(X). Set deg(D) = ni
P
ni [Pi ] =
Warning: If X is not complete, then deg is not defined on C`(X).
Proposition 5.10. f : X → Y is a finite morphism of nonsingular curves,
D ∈ Div(Y ). Then deg(f ∗ D) = deg(f ) deg(D).
54
Proof. ETS if Q ∈ Y a point, then deg(f ∗ Q) = deg(f ). V ⊆ Y open affine with
Q ∈ V . Then f −1 (V ) = Spec −m(A) ⊂ X, A = k[V ] ⊂ k(X).
Q ⊂ k[V ] a max ideal, set B = AQ = (k[V ] \ Q)−1 A. A finitely generated
k[V ]-module implies B f.g. k[V ]Q = OY,Q -module. OY,Q a DVR, B torsion free,
so B is free OY,Q -module.
rankOY,Q (B) = dimk(Y ) k(X) = deg(f ). mQ = (t) ⊂ OY,Q . OY,Q /tOY,Q =
k. Thus, dimk (B/tB) = deg(f ).
Note: points in f −1 (Q) correspond to max ideals in P ⊂ A such that P ∩
k[V ] = Q, which correspond to max ideals in AQ = B.
Write f −1 (Q) = {P1 , . . . , Ps }, Pi ⊆ A max ideals, B = ∩si=1 BPi ⇒ tB =
s
∩i=1 tBPi = ∩si=1 (tBPi ∩ B).
By the Chinese Remainder Theorem, B/tB ' ⊕si=1 B/(tBPi ∩ B).
Injective: clear
Surjective: t ∈ Pi for all i, BPi DVR, so tBPi = (Pi BPi )ni , so tBPi ∩B ⊆ Pi B
and tBPi ∩ B 6⊆ Pj B for j 6= i.
Thus, this is an isomorphism after ⊗BPi (only the ith summand survives).
Now: B/(tBPi ∩ B) = (B/(tBPi ∩ B))Pi = (B/tB)Pi = BPi /tBPi =
OX,Pi /(t). Thus dimk B/(tB
Pi ∩ B) = vPi (t).
P
Thus, deg(f ∗ Q) = vPi (t) = dimk (B/tB) = deg(f ).
Lemma 5.11. h ∈ k(Y )∗ implies f ∗ ((h)) = (f ∗ h). f ∗ h = h ◦ f ∈ k(X).
Proof. Let P ∈ X, Q = f (P ) ∈ Y . mP = (s) ⊆ OX,P . mQ = (t) ⊂ OY,Q .
h = utm , u ∈ OY,Q a unit. f ∗ t = vsn , v ∈ OX,P a unit.
Coef of [P ] in f ∗ ((h)) = vQ (h)vP (t) = mn.
f ∗ h = f ∗ (utm ) = (f ∗ u)(f ∗ t)m = (f ∗ u)v m snm , so coef of [P ] in (f ∗ h) is
nm.
So, we have a group homomorphism f ∗ : C`(Y ) → C`(X).
Corollary 5.12. X complete nonsing curve, f ∈ k(X)∗ implies deg((f )) = 0.
Proof. f is defined on open U ⊂ X. Then f : U → A1 ⊂ P1 is a regular
function. As P1 is complete, f extends to f : X → P1 . As X is complete, f is
finite.
k[A1 ] = k[t], f ∗ (t) = f ∈ k(X). (f ) = (f ∗ t) = f ∗ ((t)), so deg((f )) =
deg(f ∗ ((t))) = deg(f ) deg(t).
(t) = [0] − [∞] ∈ Div(P1 ) so deg((t)) = 0.
So if X is a complete nonsingular curve, there exists deg : C`(X) → Z.
Notation: D, D0 ∈ Div(X), D ∼ D0 iff D = D0 ∈ C`(X).
Proposition 5.13. If X complete nonsingular curve, then X rational iff ∃P 6=
Q ∈ X such that P ∼ Q.
Proof. ⇒: X = P1 , then P ∼ Q for all P, Q ∈ P1 .
⇐: ∃f ∈ k(X)∗ such that (f ) = [P ] − [Q] ∈ Div(X). f : X → P1 a
morphism. Then (f ) = (f ∗ t) = f ∗ ((t)) = f ∗ ([0] − [∞]).
This tells us that f ∗ ([0]) = [P ] and f ∗ ([∞]) = [Q]. So deg(f ∗ [0]) = 1 =
deg(f ) · 1, so f is degree 1, so it is birational. Thus isomorphism.
55
Definition 5.18 (Elliptic Curve). An elliptic curve is a nonsingular closed
plane curve E ⊂ P2 such that deg(E) = 3.
Example: V+ (zy 2 − x3 + z 2 x) ⊂ P2 .
Claim: No elliptic curve is rational.
Exercise: Set OE (1) = OP2 (1)|E . Then Γ(E, OE (1)) = (S/I(E))1 , S =
k[x, y, z].
Therefore, dimk Γ(E, OE (1))P
= 3.
Let L ⊂ P2 be a line. L.E = P 0 ∈L∩E I(L·E; P 0 )[P 0 ] = P +Q+R ∈ Div(E).
Exercise: OE (1) = OP2 ([L])|E = OE ([L.E]), so ∃D(= L.E), D ∈ Div(E)
such that deg(D) = 3 and dimk Γ(E, OE (D)) = 3.
Let D ∈ Div(P1 ) such that deg(D) = 3. Then OP1 (D) = OP1 (3), this gives
us that Γ(P1 , OP1 (D)) = Γ(P1 , OP1 (3)) = k[x0 , x1 ]3 . The dimension of this is 4.
Conclude: E is not rational.
Let X ⊂ P2 be any nonsingular curve. deg : C`(X) → Z, [P ] 7→ 1.
Definition 5.19. C`0 (C) = ker(deg). So we have a short exact sequence 0 →
C`0 (X) → C`(X) → Z → 0 that splits, so C`(X) = C`0 (X) ⊕ Z.
Fact: C`0 corresponds to a nonginular complete abelian algebraic group, the
Jacobi Variety of X.
Let L = P
V+ (f ), M = V+ (g) be lines in P2 .
X.L =
P I(X · L; P )P = P1 + . . . + Pn where n = deg(X). X.M =
Q1 + . . . + Qn .
Exercise: f /g ∈ k(X)∗ and (f /g) = X.L − X.M ∈ Div(X).
X.L − X.M = P1 + P2 + P3 − Q1 − Q2 − Q3 = 0 ∈ C`(X) for X = E.
Theorem 5.14. Let P0 ∈ E be any point, then E → C`0 (E) by P 7→ P − P0 is
bijective.
Proof. Injective: If P − P0 = Q − P0 ∈ C`0 (E) then P ∼ Q so P = Q as E is
not rational.
Surjective: Let M ⊂ P2 bePtangent line to E at P0 . M.E = 2P0 + R, R ∈ E.
Let D ∈ C`0 (E). Write D =
ni (Qi − P0 ) for Qi ∈ E, ni ∈ Z.
Assume that ni < 0. Then L =line through Qi and R, L.E = Qi + R + Q0i ,
0 = L.E − M.E = Qi + R + Q0i − 2P0 − R so Qi − P0 = −(Q0i − P0 ) ∈ C`0 (E).
Replace Qi by Q0i , ni 7→ −ni , WLOG, ni ≥ 0.
P
0
Claim:
ni .
P D = P − P0 ∈ C` (E), P ∈ E. Induction on
If
ni ≥ 2, then Q1 − P0 , Q2 − P0 have positive coefficients. L =line
through Q1 , Q2 , L.E = Q1 + Q2 + Q0 ∈ Div(E).
Let L0 be the line through Q0 and P0 . Then L0 .E = Q0 + P0 + Q00 . L.E −
0
L .E = Q1 +Q2 −P0 −Q00 = 0, so (Q1 −P0 )+(Q2 −P0 ) = (Q00 −P0 ) ∈ C`0 (E).
Example: char k 6= 2, λ ∈ k, λ 6= 0, 1. Eλ = V+ (zy 2 −x(x−z)(x−λz)) ⊂ P2 .
Take P0 = (0 : 1 : 0) ∈ E Eλ corresponds to C`0 (Eλ ) by P 7→ P − P0 . Let ⊕
be a group op on W . Q1 ⊕ Q2 = Q00 (Picture omitted)
Fact: Any elliptic curve is isomorphic to Eλ by P0 ↔ (0 : 1 : 0).
56
Theorem 5.15. E is an algebraic group.
Proof. (char k 6= 2): WLOG, E = Eλ , P0 = (0 : 1 : 0). Define ϕ : E × E → E
by ϕ(P, Q) = R the unique point such that ∃ a line L with L.E = P + Q + R.
It is enough to show that ϕ : E × E → E is a morphism.
P ⊕ Q = ϕ(P0 , ϕ(P, Q)), −P = ϕ(P1 , P0 ). Set U1 = D+ (z) and U2 = D+ (y)
subsets of E. E = U1 ∪ U2 .
Show that (Ui ×Uj )∩ϕ−1 (U` ) → U` by ϕ is a morphism for all i, j, ` ∈ {0, 1}.
U1 = V (y 2 −x(x−1)(x−λ)) ⊂A2 . (U1 ×U1 )∩ϕ−1 (U1 ) → U1 → k is the regu(y2 −y1 )2

x1 6= x2
(x2 −x1 )2 − (x1 + x2 ) + 1 + λ
2
2
lar function (x1 , y1 )×(x2 , y2 ) 7→
2
x
+x
x
+x
+λ−(1+λ)(x
+x
)
1
2
1
2
1
2

+ 1 + λ − (x1 + x2 ) y1 + y2 6= 0
y1 +y2
Differentials
R is a ring, S is a commutative R-algebra, M an S-module.
Definition 5.20 (R-derivation). A function D : S → M is an R-derivation if
D(f g) = f D(g)+gD(f ) for all f, g ∈ S, D(f +g) = D(f )+D(g), and D(f ) = 0
for all f ∈ R.
Remark: the third conditioon holds iff D is an homomorphism of R-modules.
⇒: f ∈ R ⇒ D(f g) = f D(g) and ⇐ is an exercise (use D(1) = 0).
Definition 5.21 (Module of Kähler differentials). F =free S-module with basis
{d(f )|f ∈ S} = ⊕f ∈S S · d(f ). F 0 =submodule generated by d(f ) for f ∈ R,
d(f g) − f d(g) − gd(f ), d(f + g) − d(f ) − d(g).
We define ΩS/R = F/F 0 is the module of Kähler differentials of S over R
We define d = dS = dS/R : S → ΩS/R by f 7→ d(f ) + F 0 . This is the
universal R-derivation of S.
It has the universal property that given any R-derivation D : S → M , there
exists a unique map S-homomorphism D̃ : ΩS/R → M such that D = D̃ ◦ dS .
Exercise: Let P (x1 , . . . , xn ) ∈ R[x1 , . . . , xn ] and
. . , fn ∈ S, and D : S →
Pnf1 , . ∂P
(f1 , . . . , fn )D(fi ).
M is an R-derivation. Then D(P (f1 , . . . , fn )) = i=1 ∂x
i
Consequence: If S generated by f1 , . . . , fn as an R-algebra, then ΩS/R is
gererated by dS (f1 ), . . . , dS (fn ) as an S-module.
Proposition 5.16. S = R[x1 , . . . , xn ]. Then ΩS/R is the free S-module on
dx1 , . . . , dxn .
Proof. Have a surjectve S-hom from S n → ΩS/R which sends
ei 7→ dxi . This
∂P
∂P
n
is surjective. We define D : S → S by P (x1 , . . . , xn ) 7→ ∂x
, . . . , ∂x
. By
1
n
the universal property, there is a unique S-homomorphism D̃ : ΩS/R → S n , by
definition, d(xi ) 7→ D(xi ) = ei , so this is an inverse.
Proposition 5.17. Given ring homomorphisms R → S → Y then we have an
exact sequence of T -modules ΩS/R ⊗S T → ΩT /R → ΩT /S → 0.
57
d
T
ΩT /R is an R-deriv of S. So we get S-hom ϕ : ΩS/R → ΩT /R
Proof. S → T →
ϕ̃→
via ϕ(dS (f )) = dT (f ). Thus, we have a T -hom ΩS/R ⊗ T Ω T /R by ω ⊗ h 7→
hϕ(ω).
Note: Image(ϕ̃) =submodule of ΩT /R generated by dT (f ) for f ∈ S. Thus
ΩT /R / Im(ϕ̃) = ΩT /S .
Note: I ⊂ S an ideal, T = S/I, then I/I 2 is a T -module, T × I/I 2 → I/I 2
by (f + I) · (h + I 2 ) = f h + I 2 ∈ I/I 2 .
Proposition 5.18. T = S/I. We have an exact sequence of T -modules I/I 2 →
ΩS/R ⊗S T → ΩT /R → 0, where the first map is given by h + I 2 7→ ds (h) ⊗ 1.
Proof. Set M equal to the image of I/I 2 in ΩS/R ⊗S T . Then M is generated
by {ds (h) ⊗ 1|h ∈ I}.
We define D : T → (ΩS/R ⊗ T )/M by D(f + I) = (dS (f ) ⊗ 1) + M . This is
an R-derivation.
Thus, there is a unique T -hom D̃ : ΩT /R → (ΩS/R ⊗S T )/M by dT (f + I) 7→
(ds (f ) ⊗ 1) + M
Example: S = R[x1 , . . . , xn ], I = (f1 , . . . , fp ) ⊂ S. T = S/I. ΩS/R ⊗S T =
⊕ni=1 T dxi = T ⊕n .
¯i
¯i
∂f
∂f
,
.
.
.
,
The image of fi under I/I 2 → T ⊕n : ds (fi ) ⊗ 1 = ∂x
∂xn . Set
1
¯i
∂f
J = ∂xj ∈ M at(p × n; T ) is the Jacobi matrix.
J
So the image of fi is ei J, so ΩT /R = coker(I/I 2 → ΩS/R ⊗ T ) = coker(T p →
T ).
e.g. T = k[x, y]/(y 2 − x3 + x), so J = [1 − 3x2 , 2y], so ΩT /k = T ⊕ T /h(1 −
2
3x )e1 + (2y)e2 i.
n
Proposition 5.19. S an R-algebra, U ⊆ S multiplicatively closed subset, then
ΩU −1 S/R = U −1 ΩS/R
Proof. S → U −1 S → ΩU −1 S/R is an R-derivation. Thus, it induces an Shomomorphism ΩS/R → ΩU −1 S/R , dS (f ) 7→ d(f ), where d is the universal
derivation of U −1 S.
This induces U −1 S-hom U −1 ΩS/R → ΩU −1 S/R by ds (f )/u 7→ u−1 d(f ).
. Exercise: D is
We define D : U −1 S → U −1 ΩS/R by D(s/u) 7→ ud(s)−sd(u)
u2
well defined R-derivation.
This induces D̃ : ΩU −1 S/R → U −1 ΩS/R is the inverse map.
Let X be a topological space. R, S sheaves of rings on X, R → S a ring
hom.
Definition 5.22. pre − ΩS /R (U ) = ΩS (U )/T (U ) for U ⊆ X open. For V ⊂ U
d
open, S (U ) → S (V ) → pre − Ω(V ) is an R(U )-derivation. So, we get S (U )hom pre − Ω(U ) → pre − Ω(V ).
We define ΩS /R = (pre − ΩS /R )+ , the sheafification.
58
Let ϕ : X → Y morphism of varieties, then we have ring hom ϕ∗ : ϕ−1 OY →
OX .
Definition 5.23 (Relative cotangent sheaf). ΩX/Y = ΩOX /ϕ−1 OY is called the
relative cotangent sheaf
Special case: X → {pt}, ΩX = ΩX/k = ΩX/{pt} . This is called the cotangent
sheaf.
Proposition 5.20. ϕ : X → Y a morphism of affine varieties, then ΩX/Y =
˜
Ωk[X]/k[Y
]
Proof next time.
As a consequence, ΩX/Y is always coherent.
Lemma 5.21. If (A, m) is a local Nötherian domain, N a finitely generated
A-module, then we set r = dimA/m (N/mN ). If r ≤ dimA0 (N0 ), then N is free
of rank r.
Proof. Nakayama’s Lemma implies that N can be generated by r elements.
Thus, there exists an exact sequence 0 → K → Ar → N → 0, localization
is exact, so 0 → K0 → Ar0 → N0 → 0 is exact, so the last morphism is an
isomorphism of vector spaces, so Ar0 ' N0 , so K0 = 0. Thus, K = 0 as it is
torsion free and localizes to zero, so Ar ' N .
Recall: Let X ⊂ An be a closed irreducible variety. Let I = I(X) =
(f1 , . . . , fs ). Let P ∈X. Set M = I({p}) ⊂ k[An ], M/M 2 ' k n via h + M 2 7→
∂h
∂h
∂x1 (P ), . . . , ∂xn (P )
So we have an exact sequence (I + M 2 )/M 2 → M/M 2 → mP /m2P → 0.
mP ⊂ OX,P a max ideal, therefore k s → k n → mP /m2P → 0 is also exact where
the first map is J(P ), and we call the second
φ.
If h ∈ M , then φ
∂h
∂h
∂x1 (P ), . . . , ∂xn (P )
s J
n
= h + m2P .
Note: rank(k(X) → k(X) ) ≤ c = codim(X; An ). (If h is any (c + 1) × (c +
1)-minor of J, then h ∈ k[X], and h(P ) = (c + 1) × (c + 1)-minor. J(P ) = 0 for
all P ∈ X. So h = 0 ∈ k[X].)
Theorem 5.22. Assume X is an irreducible variety of dimension r, let P ∈
⊕r
X. Then P is a nonsingular point iff ΩX,P ' OX,P
. If mP is generated by
h1 , . . . , hr ∈ mP , then dh1 , . . . , dhr ∈ ΩX,P is a basis for ΩX,P .
∂fi
Proof. WLOG: X ⊆ An affine. I(X) = (f1 , . . . , fs ), J = ∂x
.
j
J
J
s
n
k[X]s → k[X]n → Ωk[X]/k → 0 yields OX,P
→ OX,P
→ ΩX,P → 0, which
we will call (∗).
J(P )
We mod out by mP , and get k s → k n → ΩX,PP
/mP ΩX,P → 0.
n
∂h
Thus, mP /m2P ' ΩX,P /mP ΩX,P by h + m2P 7→ j=1 ∂x
(P )dxj .
j
2
Assume that ΩX,P is free of rank r, then dimk (mP /mP ) = r, thus P is a
nonsingular point.
59
J
Assume that dimk (mP /m2P ) = r. (∗) ⇒ k(X)s → k(X)n → (ΩX,P )0 → 0 is
exact.
Note: r = dimk (ΩX,P /mP ΩX,P ) ≤ dimk(X) ((ΩX,P )0 ).
⊕r
The lemma implies that ΩX,P ' OX,P
.
Lemma 5.23. ϕ : X → Y is a morphism of affine varieties. Then Γ(X, pre −
ΩX/Y ) = Ωk[X]/k[Y ]
Proof. S = Γ(X, ϕ−1 OY ), ring homomorphisms k[Y ] → S → k[X]. Thus,
ΩS/k[Y ] ⊗S k[X] → Ωk[X]/k[Y ] → Ωk[X]/S → 0 where the last is Γ(X, pre−ΩX/Y ),
so enough to show that the first map is zero.
Let f ∈ Im(S → k[X]). We must show that df = 0 ∈ Ωk[X]/k[Y ] . There
exists open cover X = ∪ni=1 Ui such that f |Ui ∈ image of Γ(Ui , pre − ϕ−1 OY ) =
limV ⊃ϕ(U ) OY (V )
−→
i
WLOG, Ui = Xgi where gi ∈ k[X]. Enough to show that df = 0 in
Ωk[X]/k[Y ] g for each i, since giN df = 0 ∈ Ωk[X]/k[Y ] and (giN , . . . , gnN ) =
i
(1) = k[X].
But Ωk[X]/k[Y ] g = Ωk[Ui ]/k[Y ] . So we replace X with Ui , we may assume
i
that f ∈ image of Γ(X, pre − ϕ−1 OY ) = limV ⊃ϕ(X) OY (V ). I.E. there exists
−→
V ⊂ Y open, f 0 ∈ OY (V ) such that ϕ(X) ⊂ V and f = ϕ∗ (f 0 ) ∈ k[X]. Now
N +1
0
N 0
V = ∪m
df =
i=1 Yhi , hi ∈ k[Y ], f ∈ k[Y ]hi , so hi f ∈ k[Y ] for all i, so hi
N +1
d(hi f ) = 0 ∈ Ωk[X]/k[Y ] .
N
Now, X = ∪Xhi ⇒ (hN
1 , . . . , hm ) = (1) ⊂ k[X], so df = 0.
˜
Proposition 5.24. ϕ : X → Y morphism of affines. Then ΩX/Y = Ωk[X]/k[Y
]
Proof. Set Ω = Ωk[X]/k[Y ] . We have Ω ' Γ(X, pre − ΩX/Y ) → Γ(X, ΩX/Y ), this
gives an OX -homomorphism Ω̃ → ΩX/Y .
Let f ∈ k[X]. Γ(Xf , Ω̃) = Ωf = Ωk[Xf ]/k[Y ] = Γ(Xf , pre − ΩX/Y ).
{Xf } is a basis for the top, and so they have the same stalks.
Corollary 5.25. ϕ : X → Y any morphism of varieties. Then ΩX/Y is coherent.
Proof. Let Y = ∪Vi open affine cover. ϕ−1 (Vi ) = ∪Uij ⊆ X is an open affine
cover of X. ΩX/Y |Uij = Ω̃k[Uij ]/k[Vi ] .
Corollary 5.26. If X irreducible, then X is nonsingular iff ΩX is a locally free
OX -module.
Example: X = P1 , ΩP1 is a line bundle. The projective coordinate ring is
k[x0 , x1 ]. Set t = xx01 ∈ k(P1 )∗ . t ∈ OP1 (D+ (x0 )), dt ∈ Γ(D+ (x0 ), ΩP1 ). Find
(dt) ∈ Div(P1 ).
Ui = D+ (xi ), U0 = A1 ⊂ P1 . If p ∈ U0 , then t − p generated mp , so ΩP1 ,p is
generated by d(t − p) = dt, so vp (dt) = 0 for all p ∈ U0 .
k[U1 ] = k[s], s = t−1 , dt = d(s−1 ) = −s−2 ds, v∞ (dt) = v∞ (s−2 ) = −2.
Thus (dt) = −2[∞] ∈ Div(P1 ), and so ΩP1 ' OP1 (−2[pt]) = OP1 (−2).
60
Example: E = Eλ ⊂ P2 an elliptic curve. Then ΩE ' OE .
Linear Systems
P
Let X be a nonsingular projective variety. D =
nY [Y ] ∈ Div(X). We say
that D is effective if nY ≥ 0 for all Y . If D is effective, we write D ≥ 0.
Definition 5.24 (Complete Linear System of D). Given any D ∈ Div(X),
define |D| = {D0 ∈ Div(X) : D0 ∼ D and D0 ≥ 0}.
Theorem 5.27. If X is projective and F is a coherent OX -module, then
dimk Γ(X, F ) < ∞.
Definition 5.25. Let `(D) = dimk Γ(X, OX (D)).
Theorem 5.28. P(Γ(X, OX (D))) → |D| by s 7→ (s) is bijective.
Proof. Let s ∈ Γ(X, OX (D)), then (s) ≥ 0 and (s) ∼ D. So the map is well
defined.
Injective: Suppose s1 , s2 ∈ Γ(X, OX (D)), assume that (s1 ) = (s2 ) ∈ Div(X).
Then (s1 /s2 ) = (s1 ) − (s2 ) = 0, so s1 /s2 ∈ k[X] = k.
Surjective: Let D0 ∈ |D|. Then D0 ∼ D, so D0 = D + (f ) where f ∈ k(X)∗ .
We define s to be the section given by f ∈PΓ(X, OX (D)). This is a global section,
because vY (f ) ≥ −nY for all Y , (D =
nY [Y ]). Set s0 = 1 ∈ Γ(U, OX (D)).
(s) = (f · s0 ) = (f ) + (s0 ) = (f ) + D = D0 .
Lemma 5.29. X is a complete nonsingular curve, D ∈ Div(X), if `(D) 6= 0
then deg(D) ≥ 0 and if `(D) 6= 0 and deg(D) = 0 then D ∼ 0.
Proof. `(D) 6= 0, then |D| =
6 ∅, so D ∼ D0 ≥ 0. So deg(D) = deg(D0 ) ≥ 0.
If deg(D) = 0, then deg(D0 ) = 0, but as D0 is effective, D0 = 0.
Riemann-Roch Theorem
Let X be a complete nonsingular curve.
Definition 5.26 (Canonical Divisor). K ∈ Div(X) is a canonical divisor on X
if ΩX ' O(K).
Definition 5.27 (Genus). The genus g = `(K) = dimk Γ(X, ΩX )
Example: X = P1 , ΩP1 = O(−2), so g = 0.
Example: E = Eλ ⊂ P2 elliptic curve, ΩE ' OE . So g = 1.
Theorem 5.30 (Riemann-Roch). For any D ∈ Div(X) where X is a complete
nonsingular curve, we have `(D) + `(K − D) = deg(D) + 1 − g.
Example: X = P1 , K = −2P for some P ∈ P1 . The RRT theorem says that
`(nP ) + `(−2P − nP ) = n + 1 − 0, so if n ≥ 0, we have that `(nP ) = n + 1. If
n = −1, then 0 + 0 = −1 + 1 = 0. If n ≥ −2, we also see that it works.
Example: Set D = K, then `(K) − `(K − K) = deg(K) + 1 − g, so g − 1 =
deg(K) + 1 − g, so deg(K) = 2g − 2.
Corollary 5.31. A nonsingular curve is either affine or projective.
61
Proof. C nonsingular curve implies that C = CK \ {P1 , . . . , Pn } where K =
k(C), X = CK . If m >> 0, then `(mPi ) = m+1−g ≥ 2. 1, fi ∈ Γ(X, OX (mPi )),
fi ∈
/ k.
Pn
(fi ) = −ri [Pi ] effective divisor in Div(X). Set f = i=1 fi ∈ k(X)∗ . f is
defined exactly on C ⊆ X, so C ' Spec −m(k[f ]) is affine.
Corollary 5.32. X is rational iff g = 0.
Proof. ⇒: genus of P1 is 0.
⇐: Let P 6= Q ∈ X. Set D = P − Q ∈ Div(X). By Riemann-Roch,
`(D) ≥ deg(D) + 1 − g, so `(D) ≥ 1. Thus |D| =
6 ∅, so there is D0 ≥ 0 such that
0
0
0
D ∼ D , but deg(D ) = 0, so D = 0.
Corollary 5.33. X complete nonsingular curve of g = 1, P0 ∈ X, char(k) 6= 2.
Then ∃ isomorphism X ' Eλ = V+ (zy 2 − x(x − z)(x − λz)) ⊂ P2 for some λ
not 0 or 1, that sends P0 7→ (0 : 1 : 0).
Proof. deg(K) = 2g − 2 = 0, so Riemann-Roch implies `(nP0 ) = n + 1 − 1 = n
for all n ≥ 1, k = Γ(OX ) = Γ(OX (P0 )) ( Γ(OX (2P0 )) ( . . . ( k(X).
Take x ∈ Γ(OX (2P0 )) \ k. vP0 (x) = −2 (x) = A + B − 2P0 for A, B ∈ X.
x : X → P1 is a morphism, x∗ ([0] − [∞]) = A + B − 2P0 , so x∗ ([0]) = A + B,
thus [k(X) : k(x)] = deg(x) = 2.
Take y ∈ Γ(OX (3P0 )) \ Γ(OX (2P0 )). vP0 (y) = −3, but as this is odd,
y∈
/ k(x). Thus k(X) = k(x, y).
{1, x, y, x2 , xy} is a basis for Γ(OX (5P0 )), 1, x, y, x2 , xy, x3 , y 2 ∈ Γ(OX (6P0 )),
dim = 6.
So there exists a linear relations. Rescale x, y: y 2 + b1 xy + b0 y = x3 + a2 x2 +
a1 x + a0 . Replace y with y + 21 (b1 x + b0 ), y 2 = (x − a)(x − b)(x − c) where
a, b, c ∈ k.
Claim: a 6= b 6= c 6= a.
ϕ : X \ {P0 } → C = V (y 2 − (x − a)(x − b)(x − c)) ⊂ A2 , P 7→ (x(P ), y(P )).
This is birational, as k(X) = k(x, y).
Assume a = b = c. Then C is a curve with a cusp, and is rational, so X
would be rational, contradiction.
Assume a = b 6= c, then C is the nodal curve, which is also rational.
c−a
2
Replace x with x−a
b−a , rescale y, y − x(x − 1)(x − λ) where λ = b−a 6= 0, 1.
2
X \ {P0 } → C ⊂ A extends to an isomorphism X → Eλ .
Lemma 5.34. X complete nonsingular curve P
of genus g.P
Let P0 , Q0 , . . . , Qg ∈
g
g
X, then there exists P1 , . . . , Pg ∈ X such that i=0 Pi ∼ i=0 Qi .
Proof. WLOG
P P0 6= Qi for all i.
Set D =
Qi . `(D) ≥ deg(D) + 1 − g = 2. Thus, ∃h ∈ Γ(X, OX (D)) such
that h ∈
/ k. Set f = h − h(P0 ) ∈ k(X)∗ . (f ) = −D + P0 + P1 + . . . + Pg = 0 ∈
C`(X).
0
g
Corollary 5.35.
PgThe Map X = X × . . . × X with g factors to C` (X) by
(P1 , . . . , Pg ) 7→ i=1 (Pi − P0 ) is surjective.
62
Proof. Note: Let Q ∈ X, the lemma implies that there
P1 , . . . , Pg ∈ X such
Pare
g
0
that (g + 1)PP
∼
Q
+
P
+
.
.
.
+
P
,
so
−(Q
−
P
)
=
(P
0
1
g
0
i − P0 ) ∈ C` (X).
i=1
0
Let D =
nQ (Q − P0 ) ∈ C` (X). The note implies
P that we can assume
nQ ≥ 0, and the lemma implies that we may assume
nQ ≤ g.
Blow-Up of Varieties
Y affine variety, X ⊆ Y closed, I = I(X) = (f0 , . . . , fn ) ⊂ k[Y ]. Let
ϕ : Y \ X → Pn by ϕ(y) = (f0 (y) : . . . : fn (y)).
Definition 5.28 (Blowup of Y along X.). B`X (Y ) = (y, ϕ(y)) : y ∈ Y \ X ⊂
Y × Pn .
Note: If Y \ X ⊆ Y is dense, then π : B`X (Y ) → Y is surjective because Pn
is complete. And π : π −1 (Y \ X) → Y \ X is an isomorphism.
Point: If Y is singular along X, then usually B`X (Y ) is less singular.
Example: Y = V (y 2 − x2 (x + 1)) ⊂ A2 . I(X) = (x, y) ⊂ k[Y ]. ϕ : Y \ {0} →
1
P , P 7→ line through O and P .
B`X (Y ) = {(P, ϕ(P )), (0, (1 : −1)), (0, (1 : 1))}.
Note: π −1 (X) ⊂ B`X (Y ) is an effective Cartier divisor. i.e. codim = 1 and
the ideal of π −1 (X) is locally generated by a single element.
∗
In fact: L = πP∗n (OPn (1)). Define si =
PnπPn (zi ) ∈ Γ(B`X (Y ), L ), zi ∈
n
Γ(P , OPn (1)) for 0 ≤ i ≤ n. We define s = i=0 fi si ∈ Γ(B`X (Y ), L ). Then
π −1 (X) = Z(s) ⊂ B`X (Y ).
Note: B`X (Y ) is independent of the generators fi of I(X). B`X (Y ) ⊂
Y × Pn , set J = I(B`X (Y )) ⊂ k[Y ][z0 , . . . , zn ], a graded ideal.
It is a fact that one can recover B`X (Y ) from k[Y ][z0 , . . . , zn ]/J.
Claim: k[Y ][z0 , . . . , zn ]/J ' ⊕d≥0 I d and this is by definition ⊕I d td ⊂ k[Y ][t]
and it is the subring of k[Y ][t] generated by k[Y ] as well as tf0 , . . . , tfn .
Morphism ψ : Y × A1 → Y × An+1 by ψ(y, t) 7→ (y, (tf0 (y), . . . , tfn (y))).
ψ(Y × A1 ) =cone over B`X (Y ), so J = I(ψ(Y × A1 )) ⊂ k[Y ][z0 , . . . , zn ]. ψ ∗ :
k[Y ][z0 , . . . , zn ] → k[Y ][t] with zi 7→ tfi .
J = ker(ψ ∗ ), so k[Y ][z0 , . . . , zn ]/J ' Im(ψ ∗ ) = k[Y ][tf0 , . . . , tfn ] = ⊕d≥0 I d .
Now let Y be any variety, X ⊂ Y any closed subset. Take and open
affine cover Y = ∪Yi . B`Yi ∩X (Yi ) can be glued together to get B`X (Y ) =
∪i B`Yi ∩X (Yi ).
IX ⊂ OY a sheaf of ideals. ⊕d≥0 IXd is a sheaf of graded OY -algebras, which
can be turned into a variety, B`X (Y ) (see next semester).
6
Schemes
We will try to state definitions and theorems from commutative algebra, but
will not prove many of them, as our focus is geometry.
Let A be a ring
Definition 6.1 (Spec A). We define Spec A = {prime ideals P ⊂ A}.
If I ⊆ A, we set V (I) = {P ∈ Spec A : I ⊆ P }.
63
Lemma 6.1.
1. V (IJ) = V (I ∩ J) = V (I) ∪ V (J)
P
2. V ( Iα ) = ∩V (Iα )
3. V (0) = Spec A, V (A) = ∅
√
√
4. V (I) ⊆ V (J) ⇐⇒ I ⊇ J
√
Proof.
We begin with 4: This follows from the fact that V (I) = V ( I) and
√
I = ∩P ∈V (I) P .
1: IJ ⊂ I ∩ J ⊆ I, so V (IJ) ⊇ V (I ∩ J) ⊇ V (I) ∪ V (J). Let P ∈ V (IJ).
Assume it is not in V (I). Then there exists a ∈ I with a ∈
/ P . Then for any
b ∈ J, ab ∈ IJ ⊆ P , so b ∈ P , thus J ⊂ P , so P ∈ V (J).
2 and 3 are straightforward exercises.
Topology: Let U ⊂ Spec(A) be open iff Spec(A) \ U = V (I) for some I ⊂ A.
We note that P ∈ Spec(A), then {P } = V (P ), so P is a closed point iff P is
maximal.
Example: Take k = k̄ an alg closed field, A = k[x, y]. Then Spec(A) =
{(x − a, y − b) : a, b ∈ k} ∪ {(f ) : f (x, y) irreducible} ∪ {0}, and these correspond
to {(a, b) ∈ k × k} ∪ {irred curves ⊂ k × k} ∪ {k × k}. The points (a, b) are
the closed points, the others are called generic points for curves or the plane,
because their closure is either everything (generic point of the plane) or are all
of the points that lie on the curve.
Structure Sheaf: For P ∈ Spec(A), set AP =`{a/f : a, f ∈ A, f ∈
/ P }. For
U = Spec(A), open define, O(U ) = {s : U → P ∈U AP : s(P ) ∈ AP and s
is locally a quotient}, that is, ∀P ∈ U , there exists an open neighborhood V ,
P ∈ V ⊂ U and a, f ∈ A such that s(Q) = a/f ∈ AQ for all Q ∈ V .
Note: (1) O is a shead of rings on Spec(A)
(2) The O(U ) are not ”really” functions, so care must be taken.
Definition 6.2 (Spectrum of a Ring). The spectrum of A is (Spec(A), O).
Definition 6.3. For f ∈ A, set D(f ) = {P ∈ Spec(A) : f ∈
/ P } = Spec(A) \
V (f ).
Note: {D(f )} is a basis for the topology on Spec(A). Let U ⊆ Spec(A) be
open, P ∈ U . Write Spec(A) \ U = V (I). Then P ∈
/ V (I), so I 6⊂ P , thus
∃f ∈ I, f ∈
/ P , so P ∈ D(f ) ⊂ U
Proposition 6.2.
1. OP = AP for all P ∈ Spec(A)
2. O(D(f )) = Af , for all f ∈ A.
3. O(Spec(A)) = A.
Proof.
1. For P ∈ U open, we have a ring homomorphism O → AP by
s 7→ s(P ). This induces ϕ : OP → AP by ϕ(sP ) = s(P ). This is surjective,
as we can let a/f ∈ AP . Then a/f ∈ O(D(f )) and ϕ(a/f ) = a/f . It is
injective, as we can assume that ϕ(sP ) = 0 ∈ AP , where s ∈ O(U ), U
64
open, P ∈ U . WLOG, s = a/f for all Q ∈ U , a/f = 0 ∈ AP , so f ∈
/ P
and there exists g ∈ A \ P such that ga = 0 ∈ A. s|U ∩D(g) = fagg = 0 ∈
O(U ∩ D(g)) ⇒ sP = 0 ∈ OP .
2. Define ψ : Af → O(D(f )) by ψ(a/f n ) = [P 7→ a/fn ∈ AP ]. For injectivity, assume that ψ(a/f n ) = 0 ∈ O(D(f )). Set I = Ann(a) ⊂ A, for
P ∈ D(f ), we have a/f n = 0 ∈ AP , so there exists H ∈ A \ P such that
ha = 0. Thus h ∈ I, h ∈
/√P , so P ∈
/ V (I). Therefore, D(f ) ∩ V (I) = ∅, so
V (I) ⊂ V (f ), thus f ∈ I, so f m ∈ Ann(a), so a/f n = 0 ∈ Af .
For surjectivity, let s ∈ O(D(f )). There exists an open cover D(f ) = ∪Vi
such that s = ai /gi on Vi for all i. WLOG, Vip
= D(hi ) for hi ∈ A. D(hi ) ⊂
D(gi ) implies that V (hi ) ⊇ V (gi ), so hi ∈ (gi ). Thus hni i = ci gi with
ci ∈ A. Note that ai /gi = ci ai /ci gi = ci ai /hni i . So we replace ai with ci ai
and hi with hni i , and WLOG, s = ai /hi pm D(hi ) for all i. We claim that
D(f ) =union ofp
finitely many D(hi ). D(f ) ⊆ P
∪D(hi ) iff V (f ) ⊇ ∩V (hi ) =
V (hhi i) iff f ∈ hhi i iff f m ∈ hhi i iff f m =
bi hi for a finite sum with
bi ∈ A. Note that s = ai /hi = aj /hj on D(hi ) ∩ D(hj ) = D(hi hj ). ψ
injective implies ai /hi = aj /hj ∈ Ahi hj , so (hi hj )n (hj ai − hi aj ) = 0 That
is, hn+1
(hni ai ) = hn+1
(hnj aj ) for all i, j. Replace ai with hni ai and hi with
j
i
hn+1
. WLOG, hj ai = hi aj ∈ A.
i
P
P
P
P
m
f = bi hi , set a =
bi ai , hj a = i bi ai hj =
bi aj hi = f m aj , so s =
m
aj /hj = a/f on D(hi ), the D(hi ) cover D(f ), so s = a/f m ∈ O(D(f )).
3. Follows from part 2.
Definition 6.4 (Ringed Space). A Ringed Space is a pair (X, OX ) where X is
a topological space and OX is a sheaf of rings on X. (Usually just denoted by
X).
A morphism of ringed spaces f : X → Y is a pair f = (f, f ] ) where f : X →
Y is continuous and f ] : OY → f∗ OX a ring homomorphism.
Let f : X → Y be a morphism of ringed spaces. P ∈ X. For V ⊂ Y open,
f]
f (P ) ∈ V , then we have OY (V ) → f∗ OX (V ) = OX (f −1 (V )) → OX,P induces
fP] : OY,f (P ) → OX,P .
Definition 6.5 (Locally Ringed Space). X = (X, OX ) is a locally ringed space
if it is a ringed space such that OX,P is a local ring for all P ∈ X. We call the
maximal ideal mP ⊆ OX,P .
f : X → Y is a morphism of locally ringed spaces iff it is a morphism of
ringed spaces such that fP] : OY,f (P ) → OX,P is a local homomorphism for all
P ∈ X, that is, fP] (mf (P ) ) ⊆ mP . (iff (fP] )−1 (mP ) = mf (P ) ).
Example: (Spec A, O) is a locally ringed space.
Why do we want locally ringed spaces? Look at all the morphisms Spec(A) →
Spec(B), there are a lot of morphisms of ringed spaces, however, the morphisms
of locally ringed spaces are in correspondence with ring homomorphisms B → A.
65
Set X = Spec(A), Y = Spec(B). Let f : A → B be a ring homomorphism.
Define ϕ : Y → X by ϕ(Q) = f −1 (Q).
Continuous: I ⊂ A an ideal, ϕ−1 (V (I)) = {Q ∈ Y : f −1 (Q) ⊃ I} =
V (f (I)B) ⊂ Y
For Q ∈ Y , let fQ : Af −1 (Q) → BQ be a map on local rings. Let U ⊂ X open,
`
OX (U ) = {s : U → p∈U Ap |s is locally a quotient}. Define ϕ] : OX (U ) →
ϕ∗ OY (U ) = OY (ϕ−1 (U )) by s 7→ [Q 7→ fQ (s(ϕ(Q))) ∈ BQ ]
Check: ϕ]Q = fQ : OX,ϕ(Q) → OY,Q .
Proposition 6.3. There is a bijective correspondence between ring homomorphisms f : A → B and morphisms of LRS ϕ : Y → X by f 7→ ϕ.
Proof. Note: f = ϕ] : Γ(X, OX ) = A → Γ(Y, OY ) = B.
Must Show: Any morphism of LRS ϕ : Y → X is determined by ϕ] : A → B.
Let Q ∈ Y , set P = ϕ(Q) ∈ X. The following commutes:
ϕ] .
A ....................................................... B
...
...
...
...
...
...
...
...
...
..
.........
.
...
...
...
...
...
...
...
...
...
..
.........
.
ϕ.]Q
AP = OX,P ............ OY,Q = BQ
ϕ]Q is a local ring homomorphism so mP = (ϕ]Q )−1 (mQ ), so P = (ϕ] )−1 (Q)
Remark: If X is an LRS, U ⊆ X open, then (U, OX |U ) is an LRS and U → X
the inclusion is a morphism of LRS.
Definition 6.6 (Affine Scheme). An affine scheme is an LRS (X, OX ) such
that (X, OX ) ' Spec(A) for some ring A.
Definition 6.7 (Scheme). A scheme is an LRS (X, OX ) such that there is an
open cover X = ∪Uα such that (Uα , OX |Uα ) is an affine scheme for all α.
Example: If k is a field, then Ank = Spec k[x1 , . . . , xn ], AnZ = Spec Z[x1 , . . . , xn ],
and the dimension of AnZ is n + 1.
Example: Assume that char(k) = p > 0, and A any k-algebra. Then there
is a ring homomorphism A → A : a 7→ ap . This gives the Absolute Frobenius
Morphism F : Spec(A) → Spec(A), F (P ) = {a ∈ A : ap ∈ P } = P , so it is the
identity on points, but is NOT the identity of schemes.
Example: Suppose k = k̄, X a (pre)variety over k. Define X̃ = {closed
irreducible subvarieties in X}. Then i : X → X̃ is the inclusion. Open subsets
of X̃ are Ũ for U ⊆ X open. Define OX̃ = i∗ OX .
Exercise: (X̃, OX̃ ) is a scheme.
Note: There exists a unique morphism of varieties X → {pt} = Spec(k),
and this gives a structure morphism from X̃ → Spec(k). (X̃ is a scheme over
Spec(k))
66
ϕ
X̃
...................................................................................................................................
.......
.........
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
....
...
.....
.....
.....
.....
.
.
.
.
.....
....
.....
.....
.....
.
.
.
.
.....
.....
.......
.......
Ỹ
Spec(k)
If ϕ : X → Y is a morphism of varieties, then
commutes.
Exercise: Morphisms of varieties X → Y correspond to morphisms of schemes
as above.
Example: The absolute Frobenius map F : X̃ → X̃ is NOT a morphism of
varieties.
Definition 6.8 (Reduced Scheme). A scheme X is reduced if OX (U ) is a reduced ring for all U ⊂ X open.
Definition 6.9 (Finite Type). A morphism ϕ : X → Y of schemes is of
finite type if for all open affine V ⊂ Y there exists a finite open affine cover
ϕ−1 (V ) = ∪Uα such that OX (Uα ) is a finitely generated OY (V )-algebra for all
α.
Exercise: The category of prevarieties over an algebraically closed field k is
equivalent to the reduced schemes of finite type over Spec(k).
Projective Schemes
S = ⊕d≥0 Sd graded ring, Sd · Se ⊆ Sd+e . Exercise: 1 ∈ S0 .
I ⊆ S a homogeneous ideal, if I = ⊕(I∩Sd ) iff I is generated by homogeneous
elements. EG S+ = ⊕d>0 Sd .
Definition 6.10. Proj(S) = {P ⊆ S homogeneous prime such that S+ 6⊂ P }
For I a homogeneous ideal, set V (I) = {P ∈ Proj(S)|P P
⊇ I}.
Check V (I) ∪ V (J) = V (IJ) = V (I ∩ J). ∩V (Iα ) = V ( Iα )
Topology: U ⊂ Proj(S) open iff Proj(S) \ U = V (I).
Let P ∈ Proj(S), set T = {homogeneous elements of S \ P }, T −1 S =
{a/f : a ∈ S, f ∈ T } = ⊕d∈Z (T −1 S)d , deg(a/f ) = deg a − deg f . Define
S(P ) = (T −1 S)0
`
For U ⊂ Proj(S) open, define O(U ) = {s : U → p∈U S(p) |s(p) ∈ S(p) and
is locally a quotient }.
ie, for all P ∈ U , there is an open nbhd P ∈ V ⊂ U and homogeneous
elements a, f ∈ S of the same degree such that s(Q) = a/f ∈ S(Q) for all
Q∈V.
Note: S(P ) is a local ring, S(P ) ⊂ T −1 S → SP , max ideal S(P ) ∩ P SP .
Proposition 6.4. S graded ring, (Proj(S), O) as above, OP = S(P ) for all
P ∈ Proj(S).
Proof. OP → S(P ) , tP 7→ t(P ), and this is injective and surjective, following the
proof for Spec A.
67
For f ∈ S+ homogeneous, define D+ (f ) = {P ∈ Proj(S) : f ∈
/ P} =
Proj(S) \ V (f )
Exercise: {D+ (f )} is a basis for the topology on Proj(S).
Define: S(f ) = (Sf )0 = {a/f n : deg(a) = n deg(f )}.
Proposition 6.5. (D+ (f ), O|D+ (f ) ) ' Spec(S(f ) ).
Proof. S → Sf ⊇ S( f ). Define ϕ : D+ (f ) → Spec(S(f ) ) by P 7→ P Sf ∩ S(f ) .
p
Exercise: For Q ∈ Spec(S(f ) ) we have hQi ⊂ Sf prime ideal.
p
Inverse map ϕ−1 (Q) = hQi ∩ S.
ϕ
Homeomorphism: D+ (h) ⊆ D+ (f ) → Spec(S(f ) ) sends D+ (h) to D(hdeg(f ) /f deg(h) ).
Note: (S(f ) )ϕ(P ) ' S(P ) .
−1
−1
` For V ⊂ Spec(S(f ) ) open, then
` ϕ∗ OProj (V ) = OProj (ϕ (V )) = {s : ϕ (V ) →
P ∈ϕ−1 (V ) S(P ) } = {s : V →
Q∈V (S(f ) )Q } = OSpec S(f ) (V ).
Example: A a ring, PnA = Proj A[x0 , . . . , xn ], covered by
D+ (xi ) = Spec A[x0 , . . . , xn ](xi ) = Spec A[x0 /xi , . . . , xn /xi ] = AnA
Definition 6.11 (Properties of Schemes). A scheme X is
1. connected if it is connected as a topological space
2. irreducible if it is irreducible as a topological space
3. reduced if OX (U ) is a reduced ring for all open U ⊂ X iff OX,P reduced
for all P ∈ X.
4. integral if OX (U ) a domain for all U ⊂ X open
5. noetherian if X = ∪ Spec(Ai ) where Ai Noetherian and the cover is finite
6. locally noetherian if X = Spec(Ai ) for Ai noetherian, not necessarily a
finite cover
√
√ Example: X = Spec(A) is irreducible iff 0 ⊂ A is prime, it is reduced iff
0 = 0 and it is integral iff A is a domain.
Proposition 6.6. A scheme X is integral iff X is irreducible and reduced.
Note: An open subscheme of (locally) Noetherian scheme is (locally) Noetherian
Proposition 6.7. Spec(A) is locally noetherian iff notherian iff A is noetherian
Proof. Assume that Spec(A) is locally noetherian. Let U = Spec(B) ⊂ Spec(A)
be open, B is noetherian. There exists f ∈ A such that D(f ) ⊆ U , f |U ∈
O(U ) = B. Spec(Af ) = Spec(Bf |U ) = D(f ). Af = Bf |U Noetherian.
We can write Spec(A)
P = ∪i Spec(Afi ), Afi Noetherian, so ∅ = V ({fi }) ⇒
(fi ) = (1) = A ⇒ 1 = ai fi with ai ∈ A a finite sum.
68
Thus, there are f1 , . . . , fr ∈ A such that Afi noetherian and (f1 , . . . , fr ) = A.
Let I ⊆ A ideal, choose g1 , . . . , gm ∈ I such that Ifi = IAfi = (g1 , . . . , gm ) ⊆
Afi for all i. Claim: I = (g1 , . . . , gm ). b ∈ I, fiN b ∈ (g1 , . . . , gm ) for all i. So
(f1N , . . . , frN ) = (1) ⊆ A, so A is Noetherian.
Definition 6.12 (Properties of Morphisms). A morphism f : X → Y is
1. locally of finite type if for all open affine V ⊆ Y there exists open affine
cover f −1 (V ) = ∪Ui s.t. OX (Ui ) is finitely generated OY (V )-algebra.
2. of finite type if ∃ a finite cover of f −1 (V ) as above.
3. affine if for all open affine V ⊆ Y , f −1 (V ) ⊆ X is open affine.
4. is finite if affine and OX (f −1 (V )) is a finitely generated OY (V )-module
whenever V is affine.
Exercise: In all cases, it is enough to know the property on a single open
affine cover.
Definition 6.13 (Open Immersion). f : X → Y is an open immersion if it can
'
be factored f : X → U ⊆ Y open.
Definition 6.14 (Closed Immersion/Closed Subscheme). f : X → Y is a closed
immersion if
1. f is a homeomorphism of X with closed subset of Y .
2. f ] : OY → f∗ O∗ is surjective.
Example: Y = Spec(A), I ⊆ A ideal, A → A/I gives a closed immersion
Spec(A/I) → Spec(A) with image V (I).
Exercise: All closed immersions X → Spec(A) have this form.
Exercise: Y scheme, V ⊆ Y a closed subset, then there exists a unique closed
immersion X → Y with image V such that X is reduced.
Definition 6.15 (Dimension). dim(X) =supremum of n such that ∃ ∅ 6= Z0 (
Z1 ( . . . ( Zn ⊆ X with Zi closed irreducible subset.
Z ⊆ X closed irreducible. Then codim(Z; X) is the supremum of n such
that ∃Z = Z0 ( Z1 ( . . . ( Zn ⊆ X with Zi closed irreducible.
Y ⊆ X is any closed subset, then codim(Y ; X) = inf{codim(Z; X) : Z ⊆ Y
closed irreducible}.
WARNING: Z ⊆ X closed and irreducible, dim(Z)+codim(Z; X) = dim(X)
does not always hold!
Products
69
Definition 6.16 (Product). Let X, Y, S be schemes with morphisms α : X → S
and β : Y → S. A product of X and Y over S is a scheme X ×S Y with
morphisms
X ×S Y
.....
....
..... q
.....
.....
....
......
...........
.........
...... p
X ..
Y
.....
...
..... α
.....
.....
....
.
.
.
.
......
.......
........
....... β
S
along with the universal property that for any scheme Z with morphisms f, g
to X and Y s.t. αf = βg, there exists a unique morphism ϕ : Z → X ×S Y
such that f = pϕ and g = qϕ.
Exercise: X ×S Y is unique up to unique isomorphism.
Exercise: X any scheme, A is a commutative ring, then there is a correspondence between {morphisms X → Spec(A)} and {ring hom A → OX (X)}.
Consequence: X = Spec(A), Y = Spec(B) and S = Spec(R), then α, β
make A and B into R-algebras, and X ×S Y = Spec(A ⊗R B).
Observe that if S ⊆ T is an open subscheme, then X ×T Y = X ×S Y , as if
j : S → T is the inclusion, then αf = βg ⇐⇒ jαf = jβg. Also observe that if
U ⊆ X open, then U ×S Y = p−1 (U ) ⊆ X ×S Y , and so it is an open subscheme
of X ×S Y , because if h : Z → X is a morphism, so that h(Z) ⊆ U , then we
can factor h : Z → U ⊆ X.
Consequence: If X 0 ⊆ X and Y 0 ⊆ Y , S 0 ⊆ S are all open such that
α(X 0 ), β(Y 0 ) ⊆ S 0 , then X 0 ×S 0 Y 0 = X 0 ×S Y 0 and this is p−1 (X 0 ) ∩ q −1 (Y 0 ) ⊆
X ×S Y .
Construction
Assume thatSS is affine. Take open affine covers X = ∪Xi , Y = ∪Yj . We
glue X ×S Y := i,j Xi ×S Yj by (Xi ×S Yj )∩(Xk ×S Y` ) = (Xi ∩Xk )×S (Yj ∩Y` ).
If S is any scheme, we take an open affine cover S = ∪Si . Xi = α−1 (Si ),
Yi = β −1 (Si ) in X and Y are open sets. And here we glue X ×S Y := ∪i Xi ×Si Yi
by (Xi ×Si Yi ) ∩ (Xj ×Sj Yj ) = (Xi ∩ Xj ) ×Si ∩Sj (Yi ∩ Yj )
Examples: X is a scheme
1. U, V ⊆ X open. U ×X V = U ∩ V .
2. Y, Z ⊆ X closed subschemes. We define Y ∩ Z := Y ×X Z, the scheme
theoretic intersection. This is still a closed subscheme of X.
X = A2k = Spec k[x, y], Y = V (y − x2 ) = Spec k[x, y]/(y − x2 ), Z = V (y) =
Spec k[x, y]/(y). We have a diagram of commutative rings
k[x, y]
...
.....
.........
........
.....
.....
.......
........
k[x, y]/(y − x2 ) k[x, y]/(y)
.....
.....
......
........
...
.....
........
.......
k[x, y]/(y − x2 , y)
70
So Y ×X Z = Y ∩Z = Spec k[x, y]/(x2 , y) = Spec k[x]/(x2 ). dimk k[x]/(x2 ) =
2.
Let X be a scheme and p ∈ X a point.
Definition 6.17 (Residue Field at p). k(P ) = OX,P /mP is called the residue
field at P .
If P ∈ U = Spec(A) ⊆ X open, then k(P ) = AP /P AP , the field of fractions
of A/P .
Note: A → k(P ) gives a morphism Spec(k(P )) → Spec(A) → X with image
{P }.
Examples: X an irreducible algebraic variety over k. If P ∈ X is a closed
point, we get k(P ) = k. If P0 ∈ X is a generic point, we get k(P0 ) = k(X).
For p ∈ U ⊆ X, f ∈ OX (U ), set f (P ) =the image of f under OX (U ) →
OX,P → k(P ).
Note: V (f ) = {P ∈ U : f (P ) = 0 ∈ k(P )} is relatively closed in U .
Let ϕ : Y → X a morphism, ϕ(Q) = P , P ∈ U . Then ϕ]Q : OX,P → OY,Q
is a local ring homomrophism so it induces ϕ̄]Q : k(P ) → k(Q) a field extension.
ϕ] (f ) ∈ OY (ϕ−1 (U )), ϕ] (f )(Q) = ϕ̄]Q (f (P )) = ϕ̄]Q (f (ϕ(Q))) ∈ k(Q).
E.G. X, Y varieties over k = k̄, Q ∈ Y a closed point implies that ϕ] (f )(Q) =
f (ϕ(Q)) ∈ k.
Exercise: A rational map of irreducible varieties f : X 99K Y is the same as
a morphism of schemes over k f : Spec k(X) → Y
P ∈ X a closed point, then {rational maps X 99K Y defined at P } correspond
to {morphismsSpec OX,P → Y over k}.
Note: Spec(k(X)) → Spec(OX,P ) → X.
Examples of Products
X, Y varieties over k, the product X × Y from last semester corresponds to
X ×k Y = X ×Spec(k) Y .
Fibers: ϕ : X → Y morphism of schemes, Q ∈ Y , then XQ = ϕ−1 (Q) :=
X ×Y Spec k(Q) with the following diagram:
ϕ .
X ....................................................... Y
.
..........
..
...
...
...
...
...
...
...
...
.
..........
..
...
...
...
...
...
...
...
...
XQ ................................... Spec(k(Q))
Example: ϕ : A1 → A1 , ϕ(t) = t2 . For a ∈ A1 a closed point, ϕ−1 (a) =
2
2
2
A ×A1 {a} = Spec(k[t]
− a)), if a 6= 0, we
√
√ ⊗k[t2 ] k[t ]/(t−1− a)) = Spec(k[t]/(t
−1
2
get ϕ (a) = { a, − a}, if a = 0, ϕ (0) = Spec k[t]/(t ).
Example: P2k ×k A1k = Proj k[x, y, z] × Spec k[t] = Proj A[x, y, z] where A =
k[t].
E = V (zy 2 − x(x − z)(x − tz)) ⊆ P2 × A1 , ϕ : E → A1 projection, then for
λ ∈ A1 a closed point, Eλ = V (zy 2 − x(x − z)(x − λz)) ⊆ P2k .
Separated Morphisms
1
71
Let f : X → Y a morphism, ∆ : X → X ×Y X the diagonal morphism (the
unique map into this product which, when composed with either projection, is
the identity)
Definition 6.18 (Separated). f is separated if ∆ : X → X ×Y X is a closed
immersion.
X is separated if X → Spec(Z) is separated.
Note: A a ring, then there exists a unique Z → A.
Example: X = (A1 \ {0}) ∪ {01 , 02 } the doubled line to Spec(k) is not
separated (See last semester).
Proposition 6.8. Any morphism f : X → Y of affine schemes is separated.
Proof. X = Spec(A), Y = Spec(B), f ] : B → A. Then X×Y X = Spec(A⊗B A).
∆ : X → X ×Y X corresponds to ∆] : A ⊗B A → A, a1 ⊗ a2 7→ a1 a2 .
∆] is surjective, so A = A ⊗B A/I, so X = V (I) ⊂ X ×Y X.
Corollary 6.9. f : X → Y separated iff ∆(X) ⊆ X ×Y X closed.
Proof. Assume that ∆(X) is closed. It is a homeomorphism as ∆ : X → ∆(X)
has continuous inverse ∆(X) → X × X → X by p1 .
We must now check that ∆] : OX×Y X → ∆∗ OX is surjective.
If Q ∈ X ×Y X \ ∆(X), then OX×Y X,Q → (∆∗ OX )Q = 0 is surjective. Let
P ∈ X. Choose V ⊆ Y open affine such that f (P ) ∈ V . Choose U ⊆ f −1 (U )
open affine such that P ∈ U . Then ∆(P ) ∈ U ×V U ⊆ X ×Y X so ∆] is
surjective in a nbhd of ∆(P ), because ∆ : U → U ×V U is separated.
Let (G, ≤) be a totally ordered abelian group. IE, g1 ≤ g2 implies g1 + g3 ≤
g2 + g3 . K a field, and K × = K \ {0}.
Definition 6.19 (Valuation). A valuation of K with values in G is a map
v : K × → G s.t. v(xy) = v(x) + v(y), v(x + y) ≥ min{v(x), v(y)}.
e.g. v : k(t) → Z by v(tm f (t)) = m if f is defined at 0 and f (0) 6= 0
Note: {x ∈ K × : v(x) ≥ 0} ∪ {0} ⊆ K is a subring.
Definition 6.20 (Valuation Ring). R is a valuation ring if R is a domain and
there exist a valuation v : K(R)× → G for some G such that R = {x ∈ K(R)× :
v(x) ≥ 0} ∪ {0}.
R ⊆ K(R) gives us a morphism i : Spec K(R) → Spec R.
Theorem 6.10. f : X → Y is a morphism, then f is separated iff the following
condition holds: For any valuation ring R and morphisms α : Spec R → Y and
β : Spec K(R) → X such that αi = f β
f .
X ....................................................... Y
β
.
..........
....
...
...
...
...
...
...
...
.
.
..........
....
...
...
...
...
...
...
...
.
α
i
Spec K(R) ....................... Spec(R)
72
Then there is at most one γ : Spec R → X making this all commute, that is,
f γ = α and γi = β.
Intuition: X, Y are prevarieties, C a curve and P ∈ C a nonsingular point.
R = OC,P is a DVR. K = k(C). Then β : C 99K X is a rational map, and
α : C 99K Y is also a rational map defined at P ∈ C. Then α = f β is the
commutativity condition on the square. If f is separated, then there is at most
one possible value of β(P ).
e.g. If X is the doubled affine line, Y is Spec k then β : A1 99K X defined
on A1 \ {0}. There are two ways to define β(0), so this is not separated.
Corollary 6.11.
1. open and closed immersions are separated
2. compositions of separated morphisms are separated.
3. separated morphisms are stable under base extensions
X
f
........................................................
Y
..
.........
....
...
...
...
...
...
...
...
.
..
.........
....
...
...
...
...
...
...
...
.
f0
X 0 = X ×Y Y 0 ........................ Y 0
Definition 6.21 (Base Extension). Then f 0 is the base extension of f by Y 0 .
Proof. We will prove (b).
f
g
Let X → Y → Z are separated morphisms. Let R be a valuation ring, take
β : Spec K(R) → X and α : Spec(R) → Z.
Assume that γ1 , γ2 : Spec R → X are morphisms satisfying the diagram.
Proof by diagram chasing.
Exercise: If g is separated, then gf is separated iff f is separated.
Let X be a scheme, x1 ∈ X.
Definition 6.22 (Specialization). x0 ∈ X is a specialization of x1 if x0 ∈ {x1 }.
This gives a ring homomorphism OX,x0 → OX,x1 which is NOT local.
Note: Spec k(x0 ) → X has image {x0 }, Spec k(x0 ) → Spec OX,x0 → X has
image {x1 ∈ X : x0 is a specialization of x1 }.
Assume R is a local domain, K = K(R) i : Spec K → Spec R, t0 = mR ∈
Spec R and t1 = 0 ∈ Spec R. Let β : Spec K → X a morphism, x1 = β(0),
β ] : k(x1 ) → K
Lemma 6.12. There is a correspondence {γ : Spec R → X|γi = β} to {x0 ∈
β]
{x1 }|image OX,x0 → OX,x1 → k(x1 ) → K is a subset of R} by γ 7→ γ(t0 ).
73
Proof. Assume γi = β. t0 ∈ {t1 } implies that γ(t0 ) ∈ {x1 }. mx0 7→ mR . We get
the following commutative square:
γt]
OX,x0 ...................0.................. Ot0 = R
...
...
...
...
...
...
...
...
...
..
.........
.
...
...
...
...
...
...
...
...
...
..
.........
.
γt]
OX,x1 ...................1................. Ot1 = K
Assume that x0 ∈ {x1 }. Then define γ : Spec R → Spec OX,x0 → X.
Recall: A domain R is a valuation ring of K if K = K(R) and ∃ valuation
v : K × → G such that R = {a ∈ K|v(a) ≥ 0} ∪ {0}.
Note: R is a local ring, mR = {a ∈ K|v(a) > 0}.
Fact: Every local ring R0 ⊆ K is dominated by some valuation ring R of K
(that is, R0 ⊆ R with mR0 ⊆ mR ). R0 ⊆ K a local subring is a valuation ring of
K iff R0 is not domainated by strictly larger R ⊆ K.
Theorem 6.13. X Nötherian. f : X → Y a morphism, ∆ : X → X ×Y X = P .
Then ∆(X) ⊆ P is closed iff for all valuation rings R with morphisms α, β with
αi = f β there exists at most one γ sch that f γ = α and γi = β.
Proof. ⇒: Let γ1 , γ2 : Spec R → X be given, with γj i = β, f γj = α. Define
ϕ : Spec R → P by πj ϕ = γj . Then γj i = β imply that ϕ(t1 ) = ϕ(i(0)) ∈ ∆(X)
and ϕ(t0 ) ∈ ϕ(t1 ) ⊆ ∆(X) = ∆(X). Thus, π1 (ϕ(t0 )) = γ1 (t0 ) = γ2 (t0 ) =
π2 (ϕ(t0 )), so γ1 = γ2 by the lemma.
⇐: Let z1 ∈ ∆(X) and z0 ∈ {z1 } ⊆ P . Claim: z0 ∈ ∆(X).
Set K = k(z1 ), R0 = Im(OP,z0 → OP,z1 → K) ⊆ K. The fact implies that
there is a valuation ring R of K such that OP,z0 → R ⊆ K is a local ring
homomorphism. β : Spec K → P , πk : P → X.
Exercise: z1 ∈ ∆(X) implies that π1 β = π2 β : Spec K → X.
Now the lemma implies that there is ϕ : Spec R → P such that β = ϕi,
ϕ(t0 ) = z0 . Set γj = πj ϕ : Spec R → X. Then f γ1 = f π1 ϕ = f π2 ϕ = f γ2 . So
the algebraic assumption says that γ1 = γ2 , so ϕ = ∆π1 ϕ, thus z0 = ϕ(t0 ) ∈
∆(X), so the claim holds.
So now X Nötherian implies that X = X1 ∪ . . . ∪ Xn where Xi is closed and
irreducible.
So Xi = {xi }, xi ∈ X, we set zi = ∆(xi ) ∈ P . So X ⊆ {x1 , . . . , xn }, thus
∆(X) ⊆ {z1 , . . . , zn } = {z1 } ∪ . . . ∪ {zn } ⊆ ∆(X) by the claim, and so the
theorem holds.
Definition 6.23 (Properness). f : X → Y is proper if f is separated, of finite
type, and f is universally closed, which means that any base extension of f is a
closed map. (takes closed sets to closed sets.)
Definition 6.24 (Redefinition of a Variety). Let k be a field. A prevariety over
k is a reduced scheme of finite type over k. A variety is a separated pre-variety
X, that is, the structure morphism X → Spec(k) is separated.
74
Definition 6.25 (Complete). A variety is complete if X → Spec(k) is proper.
ie, X ×k Y → Y is closed for all Y .
Theorem 6.14. X Nötherian, f : X → Y of finite type. Then f is proper iff
∀ valuation rings R with a commutative diagram
f .
X ....................................................... Y
β
.
.........
.....
...
..
...
...
...
...
....
.
.......
.
.........
.........
.....
.....
....
.....
...
.....
...
.....
.....
...
.....
...
.....
.....
...
.....
...
.....
.....
..
..... ..
.
.
.
........................
∃!γ
α
i
Spec R
Spec K(R)
there ∃!γ such that f γ = α, γi = β
Corollary 6.15. X a complete variety, C a curve, P ∈ C a nonsingular point,
then f : C \ {P } → X is a morphism of varieties, then we can extend f to
C → X.
This follows by setting Y = Spec(k), R = OC,P a DVR, and K(R) = K(C)
the function field of C.
Corollary 6.16.
1. Closed immersions are proper.
2. Compositions of proper morphisms are proper.
3. Base extensions of proper maps are proper
4. Properness is determined locally. That is, f : X → Y a morphism it is
proper iff for all V ⊆ Y open subscheme, f : f −1 (Y ) → Y is proper.
Exercise: Assume that P is a property of morphisms such that closed embeddings are P , compositions of P -morphisms are P and base extensions of P
f
g
morphisms are P . Then products of P -morphisms are P , X → Y → Z, if gf is
P and g is separated then f is P , and if f : X → Y is P then fred : Xred → Yred
is P .
Projective Morphisms
PnA = Proj A[x0 , . . . , xn ] = PnZ ×Spec Z Spec A.
Definition 6.26 (Projective Space over a Scheme). For any scheme Y , set
PnY = PnZ × Y
Definition 6.27 (Projective Morphism). f : X → Y is projective if there exists
a closed immersion i : X → PnY s.t. f = πY ◦ i.
f is quasi-projective if there is an open immersion j : X ⊂ X 0 and a projective morphism f 0 : X 0 → Y s.y. f = f 0 ◦ j.
Example: A variety X is quasi-projective iff X → Spec k is quasiprojective,
and likewise projective.
Example: S is a graded ring, gneerated bny finitely many elements of S1 as
an S0 -algebra. Then Proj(S) → Spec(S0 ) is projective.
75
Theorem 6.17. A projective morphism of Nötherian schemes is proper.
Proof. Enough to show that PnZ → Spec Z is proper.
Separated: Show that ∆ : PnZ → Pn ×Pn is a closed immbedding. D+ (xi xj ) →
D+ (xi )×D+ (xj ) gives Z[x0 , . . . , xn ](xi xj ) ← Z[x0 , . . . , xn ](xi ) ⊗Z Z[x0 , . . . , xn ](xj ) .
Properness; R a valuation ring, K = K(R). We want γ such that πγ = α
and γi = β. Let t0 = mR , t1 = 0 ∈ Spec(R). Set p1 = β(t1 ) ∈ Pn . If p1 ∈ V (xi ),
then induction imples that there is a γ : Spec R → V (xi ) ' Pn−1 ⊆ Pn .
WLOG: p1 ∈ D+ (x0 x1 . . . xn ) ⇒ xi /xj ∈ OP,p1 for all i, j.
β ] : k(p1 ) → K, set fij = β ] (xi /xj ) ∈ K.
A valuation v : K → G, R = {v ≥ 0}. Choose m such that v(fm0 ) is
minimal. v(fim ) = v(fi0 /fm0 ) = v(fi0 ) − v(fm0 ) ≥ 0, so fim ∈ R for all i. So
we have a ring homomorphism Z[x0 /xm , . . . , xn /xm ] → R by xi /xm 7→ fim .
γ
Thus, Spec R → D+ (xm ) ⊆ Pn is as desired.
Exercise:
nm+n+m
1) There exists a closed Segre Embedding PnZ × Pm
.
Z → PZ
2) Composition of projective morphisms is projective.
af f ine
Example: X → Y → Spec k, X a prevariety, Y a variety, then X is a
variety.
OX modules
X is a scheme, F , G are sheaves of OX -modules.
homOX (F , G ) = {homomorphisms of OX -modules F → G }.
We define a sheaf of OX -modules H omOX (F , G ) by U 7→ homOX |U (F |U , G |U ).
F ⊗OX G by the sheafification of U 7→ F (U ) ⊗OX (U ) G (U ).
Let f : X → Y be morphisms of schemes, H an OY -module. f∗ F is an
f∗ OX -module, f ] : OY → f∗ OY a ring homomorphism implies that f∗ F is an
OY -module.
The adjoint property: homOY (H , f∗ F ) ⇐⇒ homf −1 OY (f −1 H , F ).
Definition 6.28 (Pullback). f ∗ H = f −1 H ⊗f −1 OY OX , this is an OX -module.
Continued...see Sheaves from last semester
Proposition 6.18. M̃P = MP , M̃ (D(a)) = Ma .
Corollary 6.19. Γ(X, M̃ ) = M and M 7→ M̃ is an exact, fully faithful functor
from A-mod to OX -mod with inverse the global section functor, ⊕i˜Mi = ⊕i M̃i
and M ⊗˜A N = M̃ ⊗OX Ñ
We have done this several times before, and so will not prove them again.
Proposition 6.20. f : Spec A = X → Spec B = Y a morphism, M an Amodule and N a B-module. Then f∗ M̃ = M̃B , f ∗ Ñ = N ⊗˜B A where MB is
M as a B-module using f ] : B → A.
Proof. Let b ∈ B, Γ(D(b), f∗ M̃ ) = Γ(f −1 (D(b)), M̃ ) = Γ(D(f ] (b)), M̃ ) =
Mf ] (b) = (MB )b .
Let P ∈ X. Q = (f ] )−1 (P ) = f (P ) ∈ Y . (f ∗ Ñ )P = Ñf (P ) ⊗OY,f (P ) OX,P =
NQ ⊗BQ AP = (N ⊗B A)P .
76
Definition 6.29 (Quasicoherent and Coherent Sheaves). Let X be a scheme.
An OX -module F is quasi-coherent if there exists an open cover X = ∪Ui ,
Ui = Spec Ai and Ai modules Mi such that F |Ui ' M̃i .
F is coherent if the Mi are ginitely generated.
Example: i : Y → X a closed subscheme implies that i∗ OY is a coherent OX ˜ = A/I.
˜
module. U = Spec A ⊆ X open, i−1 (U ) = Spec(A/I), i∗ OY = i∗ (A/I)
Lemma 6.21. X = Spec A, f ∈ A. F a quasi-coherent OX -module.
1. If s ∈ Γ(X, F ) and s|D(f ) = 0 ∈ Γ(D(f ), F ), then ∃n > 0 such that
f n s = 0 ∈ Γ(X, F )
2. If t ∈ Γ(D(f ), F ) then ∃s ∈ Γ(X, F ) and m > 0 so that s|D(f ) = f n t.
Proof. P ∈ X, there exists an open nbhd P ∈ U = Spec(B) ⊆ X and B-module
M such that F |U ' M̃ . So there exists g ∈ A such that P ∈ D(g) ⊆ U .
F |D(g) = M̃ |D(g) = M ⊗˜B Ag . Thus, we can write X = D(g1 ) ∪ . . . ∪ D(gm )
such that F |D(gi ) = M̃i , with Mi a Agi -module.
1. Let s ∈ Γ(X, F ), with s|D(f ) = 0. Set mi = s|D(gi ) ∈ Mi . Then
mi |D(f gi ) = 0 ∈ (Mi )f , so f n mi = 0 ∈ Mi for all i, so f n s = 0 ∈ Γ(X, F ).
2. X = D(g1 ) ∪ . . . ∪ D(gr ). F |D(gi ) = M̃i , where Mi is an Agi -module.
t|D(f gi ) ∈ (Mi )f . ∃ti ∈ M i such that ti |D(f gi ) = f n t|D(f gi ) . So then
(ti − tj )|D(f gi gj ) = (f n t − f n t)|D(f gi gj ) = 0. By part 1, f m (ti − tj ) =
0 ∈ Γ(D(gi gj ), F ). That is, f m ti = f m tj on D(gi ) ∩ D(gj ). We flue
s ∈ Γ(X, F ) s.t. s|D(gi ) = f m ti and s|D(f ) = f n+m t, as s|D(f gi ) = f m ti =
f n f m t|D(f gi ) .
Corollary 6.22. Γ(D(f ), F ) = Γ(X, F )f .
Proposition 6.23. X a scheme, F an OX -module.
1. F is quasi-coherent iff for all open Spec A = U ⊆ X, there exists an
A-module M such that F |U = M̃
2. X Nötherian: F coherent iff the same is true but M finitely generated.
Proof.
1. Assume that F is quasicoherent, U = Spec A ⊆ X is open. Then
F |U is a quasicoherent OX -module. Write U = D(gi ) ∪ . . . ∪ D(gr ),
gi ∈ A. F |D(gi ) ' M̃i . Set M = F (U ). M 7→ Γ(U, F ) gives an OU homomorphism α : M̃ → F . The lemma implies that Mi = F (D(gi )) =
Mgi . Therefore, α is an isomorphism over D(gi ) implies that it is one over
U.
77
2. Assume X Nötherian, F is coherent. Then A is Nötherian, so Mi = Mgi
is a finitely generated module over Agi for all i. We must show that M
is a finitely generated A-module: take m1 , . . . , mN ∈ M such that Mgi is
gneerated by {mj /1} for all i. Then M is generated by {m1 , . . . , mN } as
an A-module.
Corollary 6.24. X = Spec A. Then there is an equivalence M 7→ M̃ and
F → Γ(X, F ) between A-modules and quasi-coherent OX -modules.
Recall: 0 → F 0 → F → F 00 → 0 is exact iff 0 → Fp0 → Fp → Fp00 → 0 is
exact for all p ∈ X. We only get automatically that 0 → Γ(U, F 0 ) → Γ(U, F ) →
Γ(U, F 00 ) is exact.
Proposition 6.25. X is affine scheme, 0 → F 0 → F → F 00 → 0 exact
sequence, if F 0 is quasi-coherent, then Γ(U, F ) → Γ(U, F 00 ) is surjective, so
Γ(U, −) becomes an exact functor.
Proposition 6.26. Any kernel, cokernel or image of an OX -homomorphism
of quasicoherent OX -modules is quasicoherent. An extension of quasi-coherent
OX -modules is quasi-coherent.
If X is Nötherian, then the same holds for coherent modules.
Proof. WLOG, X = Spec A. ϕ : M̃ → Ñ . This gives 0 → K → M → N →
C → 0 an exact sequence, as −̃ is exact, 0 → K̃ → M̃ → Ñ → C̃ → 0 is exact.
So the kernel and cokernel must be quasicoherent. The image is coker(K̃ → M̃ ),
and so is also quasicoherent.
We now assume that 0 → M̃ → F → Ñ → 0 is a short exact sequence
with M̃ , Ñ quasicoherent and F an OX -module. Define F = Γ(X, F ). The
proposition states that 0 → M → F → N → 0 is a short exact sequence of
A-modules. .. Then we obtain
0 ...................................................... M̃ ........................................................ F̃ ........................................................ Ñ ........................................................ 0
...
...
...
.
.
.
..........'
..........'
..........
.
.
.
0
M̃ ...................................................... F ...................................................... Ñ
So by the five lemma, F̃ ' F .
........................................................
...
...
........................................................
0
Definition 6.30 (Ideal Sheaf). Let X be a scheme, i : Y → X a closed immeri]
sion. Then the ideal sheaf of Y is IY = ker(OX → i∗ OY ) ⊆ OX .
Proposition 6.27. {closed subschemes of X} correspond to {quasicoherent
sheaves of ideals ⊆ OX }.
Proof. Assume that i : Y → X is a closed immersion, then we have 0IY →
i]
OX → i∗ OY → 0. i∗ OY is quasicoherent (we will prove this next, it is true
because i is separated and quasicompact). Then IY is quasicoherent as well.
Assume that I ⊆ OX is quasicoherent. For p ∈ X, Ip ⊆ OX,p . Define
Y = {P ∈ X : IP ⊆ mP ( OX,P } = Supp(OX /I ) = {P ∈ X|(OX /I )P 6= 0}.
Let i : Y → X be the inclusion.
78
Define OY = i−1 (OX /I ). Then (Y, OY ) is a locally ringed space. OX,P =
id
(OX /I )P = OX,P /IP . Then i−1 (OX /I ) → OY , which corresponds to a map
ϕ
OX /I → i∗ OY by the adjoint property.
ϕ
Then we define i] : OX → OX /I → i∗ OY , so (i, i] ) : Y → X is a morphism
of locally ringed spaces. Claim: This is a closed immersion.
WLOG X = Spec A. Then I = I˜ ⊆ OX , with I ⊆ A an ideal. Then
Y = V (I) ⊆ X is closed, and i : Y → X is a homeomorphism onto its image.
We must check that Y = Spec A/I.
OY,P = (OX /I )P = AP /IP = (A/I)P = OSpec A/I,P .
Corollary 6.28. All closed subschemes of Spec A are in correspondence with
ideals of A.
All closed subschemes of an affine scheme are affine.
Corollary 6.29. If U1 , U2 ⊆ X open affine subschemes, and if X is separated,
then U1 ∩ U2 is affine.
X
.
..........
....
..
...
...
...
...
...
...
..
∆
..............................................
X ×X
.
..........
...
....
..
...
...
...
...
...
..
⊆
U1 ∩ U2 ................................ U1 × U2
Proof. As ∆ is a closed embedding, the bottom map must be as well. The
product of affines over an affine is affine, and so then U1 ∩ U2 is affine, as it is
a subscheme of U1 × U2 .
Proposition 6.30. Let f : X → Y be a morphism.
1. G a quasi-coherent OY -module, then f ∗ G is a quasi-coherent OX -module.
2. X, Y both Nötherian, same for coherent.
3. Assume X is Nötherian or that f is separated and quasicompact, then if
F is quasicoherent OX -module, then f∗ F is a quasicoherent OY -module
Proof.
1. Let P ∈ X, Let Spec B = V ⊆ Y open affine such that f (P ) ∈ V .
Take U = Spec A ⊆ X open such that P ∈ U ⊆ f −1 (V ). G |V = M̃ with
M a B-module. Then (f ∗ G )|U = f¯∗ (G |V ) = f¯∗ (M̃ ) = M ⊗˜B A.
2. Similar, but noting finite generation everywhere.
3. WLOG, Y is affine. Last time, we showed that (c) is true when X is
affine. Assumptions imply that there exists a finite open affine cover
X = ∪Ui such that Ui ∩ Uj has a finite open affine cover Ui ∩ Uj = ∪Uijk
finite. As F is a sheaf, we obtain an exact sequence of OY -modules
0 → f∗ F → ⊕i f∗ F |Ui → ⊕i,j,k f∗ F |Uijk . So f∗ F is the kernel of a map
of quasicoherent OY -modules.
79
Previously, we’ve shown that the following properties are all equivalent
1. Closed subschemes of affine schemes are affine
2. If X is separated and U1 , U2 ⊆ X open affine, then U1 ∩ U2 affine.
3. Pushforward of a quasicoherent sheaf is quasicoherent
4. The ideal sheaf of a closed subscheme is quasi-coherent.
We now prove number 1:
Proof. If i : Y → X is a homeomorphism onto i(Y ), and i(Y ) is closed, then
(i∗ OY )P is OY,P if P ∈ Y and 0 else.
For f ∈ Γ(X, OX ) write Xf = D(f ) = {P ∈ X : f ∈
/ mP ⊆ OX,P }.
Assume that i : Y → X = Spec A is a closed subscheme. Let P ∈ Y ,
then there exists open affine V ⊆ Y . Note: {Yf = i−1 (Xf )} is a basis for the
topology on Y .
Thus, there exists f ∈ A such that P ∈ Yf ⊆ V , Yf = Vf is affine. So we can
write X = Xf1 ∪ . . . ∪ Xfr such that Yfi affine for all i. (f1 , . . . , fr ) = (1) ⊆ A,
thus (i] (f1 ), i] (f2 ), . . . , i] (fr )) = (1) ⊆ Γ(Y, OY ).
By exercise 2.17b in Hartshorne, Y is affine iff ∃f1 , . . . , fr ∈ Γ(Y, OY ) such
that Yfi is affine for all i and (f1 , . . . , fr ) = 1 ⊆ Γ(Y, OY ).
S = ⊕d≥0 Sd a graded ring, X = Proj(S), and M = ⊕d∈Z Md a graded Smodule. Sd · Me ⊆ Me+d for P ∈ X defin an S(P ) -module M(P ) = {m/f : m ∈
M, f ∈ S homogeneous of the same degree}.
`
For U ⊆ X open, define OX -module M̃ by Γ(U, M̃ ) = {s : U → P ∈U M(P ) |s(P ) ∈
M(P ) ∀P ∈ U and s is locally a quotient}.
Proposition 6.31.
1. M̃P = M(P )
2. M̃ |D+ (f ) = M˜(f ) for f ∈ S+ , deg(f ) > 0.
Corollary 6.32. M̃ is quasicoherent. If S is Nötherian and M finitely generated S-module, then M̃ is coherent.
Definition 6.31 (Twisted Modules). Define M (n) = M as an S-module, but
with grading M (n)e = Me+n .
˜
If X = Proj(S), define OX (n) = S(n).
For any OX -module, F , define
F (n) = F ⊗OX OX (n).
Example:
1. OX (0) = S̃ = OX .
2. S = A[x0 , . . . , xm ], X = Proj(S) = Pm
A . Then Γ(D+ (xi ), OX (n)) =
S(n)xi = {f /xri |f ∈ Sr+n } = {elements of degree n} ⊆ Sxi
80
Claim: Γ(X, OX (n)) = Sn . We have a map Sn = S(n)0 , Let t ∈ Γ(X, OX (n)),
set ti = t|D+ (xi ) ∈ Sxi element of degree n, if ti = tj on D+ (xi xj ), then
ti = tj ∈ Sxi xj , not that Sxi ⊆ Sxi xj , Sxi ∩ Sxj = S ⊆ Sxi xj , thus ti = tj ∈ Sn
for all i, j.
Proposition 6.33. Let X = Proj(S), S generated by S1 as an S0 -algebra.
1. OX (n) is an invertible OX -module.
2. M̃ ⊗OX Ñ = M ⊗˜S N .
·f −n
Proof.
1. Let f ∈ S1 . Γ(D+ (f ), OX (U )) = S(n)(f ) → S(f ) , which says that
˜
OX (n)|D+ (f ) = S(n)
(f ) ' S̃(f ) = OD+ (f ) .
2. M ⊗S N is a graded S-module, by (M ⊗S N )e =submodule generated by
m ⊗ n where m ∈ Mr , n ∈ Nt with r + t = e. Let P ∈ X. Then we have
a map M(P ) ⊗S(P ) N(P ) → (M ⊗S N )(P ) by m/f ⊗ n/g 7→ m ⊗ n/f g. S
is generated by S1 , so this is an isomorphism. Why? Because P 6⊇ S+ ⇒
P 6⊇ S1 ⇒ ∃h ∈ S1 such that h ∈
/ P . Let m ⊗ n/f ∈ (M ⊗ N )(P ) , go to
m/hr ⊗ hr n/f ∈ M(P ) ⊗ N(P ) where m ∈ Mr , n ∈ Nt and s ∈ Sr+t .
˜ N by s ∈ Γ(U, M̃ ⊗ Ñ ) 7→
Construct an isomorphism ϕ : M̃ ⊗OX Ñ → M ⊗
[P 7→ sp ∈ M̃P ⊗OX,P ÑP = (M ⊗ N )(P ) ], which is an isomorphism.
˜ = (M ⊗S S(n))˜= M˜(n).
Corollary 6.34. M̃ (n) = M̃ ⊗OX OX (n) = M̃ ⊗ S(n)
This says that OX (m) ⊗ OX (n) = OX (m + n).
Definition 6.32. Let X = Proj(S), F an OX -module. Define Γ∗ (F ) =
⊕n∈Z Γ(X, F (n)). This is a graded S-module.
Let s ∈ Sd , t ∈ Γ(X, F (n)) gives s ∈ Γ(X, OX (d)), so t ⊗ s ∈ Γ(X, F (n)) ⊗
OX (d) = Γ(X, F (n + d))
Note: X = Proj S where S = A[x0 , . . . , xn ]. Γ∗ (OX ) = S. This is NOT
always true!
(Side Note: Look at the operation Γ∗ (Proj S). Is it a functor? What properties does it have? Filling in lower degrees?)
Let X be a scheme, I an invertible LX -module, and f ∈ Γ(X, L ).
Definition 6.33. Xf = {P ∈ X|fP ∈
/ mP LX }. Xf ⊆ X is open.
Lemma 6.35. Let X be a quasi-compact scheme, L an invertible OX -module,
f ∈ Γ(X, L ) and F a quasi-coherent OX -module.
1. Let s ∈ Γ(X, F ), s|Xf = 0, then s ⊗ f n = 0 ∈ Γ(X, F ⊗ L ⊗n ).
Proof.
1. X = U1 ∪ . . . ∪ Ur where Ui = Spec Ai and L |Ui ' OUi . F |Ui =
M̃i with Mi an Ai -module. si |Ui ∈ Mi . fi = f |Ui ∈ Γ(Ui , L ) ' Ai ,
F ⊗ L ⊗n |Ui ' M̃i ⊗ OU⊗n
= M̃i and s ⊗ f n 7→ fin si = 0 for n >> 0, so
i
n
⊗n
s ⊗ f = 0 ∈ Γ(X, F ⊗ L ).
81
X a schemem L an invertible OX -module.
Definition 6.34. L −1 = H om(L , OX ).
If L |U ' OU then L −1 |U ' H om(OU , OU ) ' OU .
Note: Global OX -hom: L ⊗ L −1 → OX by s ⊗ α 7→ α(s) is an isomorphism
on U .
Let f ∈ Γ(X, L ), Xf = {P ∈ X : fP ∈
/ mP LP } ⊆ X open. L −1 |Xf →
OXf , α 7→ α(f ).
Define f −1 ∈ Γ(Xf , L −1 ) to be the inverse image of 1 by this isomorphism.
So f ⊗ f −1 = 1, adn L ⊗ L −1 ' OX .
Lemma 6.36. Let X be a quasi-compact scheme, L an invertible OX -module,
f ∈ Γ(X, L ) and F a quasicoherent OX -module.
1. s ∈ Γ(X, F ), s|Xf = 0 ⇒ s ⊗ f n = 0 ∈ Γ(X, F ⊗ L ⊗n )
2. Assume X = U1 ∪ . . . ∪ Ur with Ui open affine and L |Ui ' OUi for all i,
and Ui ∩ Uj quasicompact. t ∈ Γ(Xf , F ) ⇒ ∃s ∈ Γ(X, F ⊗ L ⊗n ) such
that s|Xf = t ⊗ f n .
Corollary 6.37. S = ⊕d≥0 Γ(X, L ⊗d ) is a graded ring, f ∈ S1 . Γ∗ F =
⊕e∈Z Γ(X, F ⊗ L ⊗e ) is a graded Smodule. Then (Γ∗ F )(f ) = Γ(Xf , F )
Proof. (Γ∗ F )(f ) = {s/f d |s ∈ Γ(X, F ⊗ L ⊗d )}
f −d ∈ Γ(Xf , L ⊗−d ). (Γ∗ F )(f ) → Γ(Xf , F ) by s/f d 7→ s ⊗ f −d . This is
injective and surfective.
Let S = ⊕d≥0 Sd be a graded ring, X = Proj S.
˜
Recall, OX (n) = S(n).
f ∈ Sd implies f ∈ S(d)0 ⇒ f ∈ Γ(X, OX (d)). F
an OX -module, Γ∗ F = ⊕e∈Z Γ(X, F (e)). Construct an OX -hom β : Γ∗˜F → F
˜ (f ) → F |X , which corresponds to a module homomorphism
by β|Xf : Γ∗ F
f
(Γ∗ F )(f ) → Γ(Xf , F ) by s/f m 7→ s ⊗ f −m .
Proposition 6.38. If F is quasi-coherent, and S is finitely generated by S1 as
an S0 -algebra, then β : Γ∗˜F → F is an isomorphism.
Proof. Assume S0 [f1 , . . . , fr ] → S is surjective, fi ∈ S. Ui = Xfi affine. X =
˜(x ) = S(x
˜ ) = OU , Ui ∩ Uj = Xf f affine, so the
U1 ∪ . . . ∪ Ur . O(1)|Ui = S(1)
i
i j
i
i
lemma implies that β|Xfi is an isomorphism for all i.
Let I ⊆ S be a homogeneous ideal. Natural inclusion i : Y = Proj(S/I) →
˜ (exercise)
X = Proj(S) i(Y ) = V (I), i∗ OY = S/I
˜
So OX → S̃ → S/I = i∗ OY , so Y ⊆ X is a closed subscheme. Note that
IY = I˜ ⊆ OX .
Corollary 6.39.
1. Assume Y ⊆ PrA a closed subscheme with PrA = Proj S,
S = A[x0 , . . . , xr ]. Then there exists a homogeneous ideal I ⊆ S such that
Y = Proj(S/I).
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2. A morphism ϕ : Y → Spec A is projective iff Y = Proj(S), S0 = A and S
is finitely generated by S1 .
Proof.
1. IY ⊆ OPrA a quasicoherent subsheaf, IY ⊗ O(d) ⊆ O(d) implies
that I = Γ∗ IY ⊆ Γ∗ OPrA = S a homogeneous ideal.
Proj(S/I) ⊆ PrA has ideal sheaf I˜ = Γ∗˜IY = IY ⊆ OPrA . Thus, Y =
Proj(S/I).
2. ϕ is projective iff it factors through PrA as a closed immersion for some r
iff Y = Proj S, with S = A[x0 , . . . , xr ]/I.
Definition 6.35 (Twisting Sheaf). Let Y be any scheme. The twisting sheaf
π
of PrY = PrZ × Y → PrZ is O(1) = π ∗ OPr (1).
Definition 6.36 (Immersion). A morphism i : X → Z is an immersion if we
can factor it as i : X → Z1 → Z with X → Z1 an open immersion and Z1 → Z
a closed immersion.
Exercise: A composition of immersions is an immersion.
Definition 6.37 (Very Ample). Let X be a scheme over Y and L an invertible
OX -module. L is very ample relative to Y if ∃ an immersion i : X → PrY such
that L = i∗ O(1).
Note: If ϕ : X → Y is projective, then ϕ is proper, and X has a very ample
invertible sheaf relative to Y .
Suppose that ϕ : X → Y is proper and X has a very ample invertible L
relative to Y . We have an immersion i : X → PrY , ϕ proper implies that i is
proper, so i(X) ⊆ PrY is closed, and so i is a closed immersion.
Definition 6.38 (Generated by global sections). Let X be a scheme, F an
α
OX -module, F is generated by globa section if ∃ ⊕i∈I OX → F a surjective
OX -hom.
IE, ∃si ∈ Γ(X, F ) such that FP is generayed by {(si )P } as an OX,P -module
for all P ∈ X.
Examples:
1. X = Spec A, F = M̃ .
2. X = Proj S, S generated by S1 as an S0 -module. Then OX (1) is generated
by global sections.
Definition 6.39 (Ample Sheaf). An invertible OX -module L on a Nötherian
scheme X is ample if ∀ coherent OX -modules F , ∃n0 > 0 such that F ⊗ L ⊗n
generatd by global sections for all n ≥ n0 .
Later: If X is of finite type over Spec A, A Nötherian, then L is ample iff
L ⊗m is very ample relative to A for some m > 0.
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Theorem 6.40. Let X be projective over Spec A, A Nötherian. If O(1) is very
ample relative to A, then O(1) is ample.
Proof. i : X → PrA is a closed immersion, O(1) = i∗ OPr (1). F a coherent
OX -module. Then i∗ F is coherent on PrA . For P ∈ X, (i∗ F )P = FP .
Exercise: Γ(X, F ⊗ O(1)⊗m ) = Γ(PrA , (i∗ F )(m)).
WLOG: X = PrA = Proj A[x0 , . . . , xr ].
F coherent implies that F |D+ (xi ) ' M̃i , with Mi a finitely generated Bi
module, where Bi = A[x0 /xi , . . . , xr /xi ]. Let si1 , . . . , siN generate Mi as a Bi module. Then the lemma implies that there exists tij ∈ Γ(X, F ⊗ OX (n)) such
that tij |D+ (xi ) = sij ⊗ xni .
Take n large enough for all i, j. Claim: F (n) = F ⊗ OX (n) is generated by
{tij } ⊂ Γ(X, F (n)).
'
Why? Because F |D+ (xi ) → F (n)|D+ (xi ) by s 7→ s ⊗ xni .
Corollary 6.41. X projective over Nötherian A, F a coherent OX -module.
Then there exists ⊕OX (ni ) → F surjective with a finite sum.
Proof. The theorem says that F ⊗O(n) = F (n) is generated by global sections.
⊕N
Thus, OX
→ F (U ) → 0 exact, tensor with O(−n), and we get the result.
Definition 6.40 (*-scheme). X is a *-scheme if it is Nötherian, separated,
integral and OX,P is a DVR whenever dim OX,P = 1.
Examples: nonsing alg variety
X normal, Nötherian, separated and integral.
Definition 6.41 (Prime Divisor). Assume X is a *-scheme, a prime divisor is
a closed integral subscheme of codimension 1.
Note that OX,Y is a DVR, we have a valuation vY : k(X)∗ → Z by OX,Y =
{f ∈ k(X)|vY (f ) ≥ 0}. We call vY (f ) the order of vanishing of f along Y .
v( f g) = vY (f ) + vY (g) and vY (f + g) ≥ min(vY (f ), vY (g)).
We set Div(X) to be the free abelian
P group on the prime divisors. A principal
divisor is one of the form (f ) = Y vY (f )[Y ] with f ∈ k(X)∗ . For D, D0 ∈
Div(X) write D ∼ D0 ⇐⇒ D − D0 is principle.
Define the divisor class group C`(X) = Div(X)/{principal divisors}.
Example: X = Pnk , a prime divisorP
is Y = V (f
P) and deg(Y ) = deg(f ).
There is an ismorphism C`(Pn ) → Z by
ni [Yi ] 7→
ni deg(Yi ).
1. U ⊆ X open implies there exists a surjective group hom C`(X) → C`(U )
by [Y ] 7→ [Y ∩ U ] if Y ∩ U 6= ∅ and 0 otherwise. So if Z = X \ U and
codim(Z; X) ≥ 2, then C`(X) ' C`(U ). If Z ⊆ X is a prime divisor, then
have Z → C`(X) → C`(U ) → 0.
2. A is a Nötherian domain. Then A is a UFD iff U = Spec A is normal and
C`(U ) = 0.
3. π : X × Am → X gives an isomorphism π ∗ : C`(X) → C`(X × Am ) by
[Y ] 7→ [Y × Am ].
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Example: X = Pn × Pm , p : X → Pn and q : X → Pm , then p∗ : C`(Pn ) →
C`(X) is injective, by [Y ] 7→ [Y × Pm ]. Let H ⊆ Pm be a hyperplane, U =
X \ Pn × H = Pn × Am . Then we have an isomorphism C`(Pn ) → C`(X) →
C`(Pn × An ). Similarly q ∗ : C`(Pm ) → C`(X) is injective. In fact, the image of
q ∗ is Z[Pn × H].
So we get 0 → C`(Pm ) = Z → C`(X) → C`(Pn × Am ) = Z → 0, but the last
map has a section, so we get C`(X) = Z[Pn × H] ⊕ Z[H 0 × Pm ].
Carier Divisors
Let X be any scheme. For U ⊆ X open, let S(U ) ⊆ OX (U ) be the set of
non-zerodivisors. We define K to be the sheafification of pre − K , which has
pre − K (U ) = S(U )−1 OX (U ). This is a sheaf of rings, OX -module, OX → K .
Example: if X is an integral scheme, S(U ) = OX (U )\{0}. Then for U affine,
Γ(U, pre − K ) = K(OX (U )) = k(U ) and Γ(U, K ) = k(X) for any nonempty
U ⊆ X open.
Note that OX ⊆ pre − K is a sub-presheaf, so pre − K is a decent presheaf.
Recall that a decent presheaf satisfies the first sheaf axiom, and OX ⊂ pre−K ⊂
K.
Define K ∗ (U ) to be the set of invertible elements of K (U ). This is a sheaf
∗
⊆ K ∗.
of abelian groups. So OX
Definition 6.42 (Cartier Divisor). A Cartier Divisor is an element D ∈ Γ(X, K ∗ /O ∗ ).
It is principleif it is in the image of Γ(X, K ∗ ) → Γ(X, K ∗ /O ∗ ).
NOTE: This is not surjective! There is a sheafification involved.
Convention: We use additive notation.
Let D be a Cartier Divisor. Then there exists an open cover X = ∪Ui
and fi ∈ Γ(Ui , K ∗ ) such that D|Ui is the image of fi and on Ui ∩ Uj , fi /fj ∈
∗
Γ(Ui ∩ Uj , OX
).
Whenever {fi } satisfy this condition, they define a Cartier divisor D.
Definition 6.43. We define CaC`(X) = {cartier divisors}/{principal cartier
divisors}. Write D ∼ D0 iff D − D0 = 0 ∈ CaC`(X).
Assume that X is a *-scheme. Then D is a Cartier, Y ⊆ X a prime divisor,
we write vY (D) = vY (fi ) where Y ∩ Ui 6= ∅ and D|Ui =image of
P di . This gives
∗
) → Div(X) by D 7→ Y vY (D)[Y ].
us a group homomorphism Γ(X, K ∗ /OX
This induces a map CaC`(X) → C`(X), and is injective when X is normal.
Definition 6.44 (Locally Factorial). X is locally factorial if OX,P is a UFD
for all P ∈ X.
Proposition 6.42. X is locally factorial *-scheme, then CaC`(X) ' C`(X).
Definition 6.45 (Picard Group). The set of all isomorphism classes of invertible OX -modules under tensor product.
∗
Let D ∈ Γ(X, K ∗ /OX
) cartier. Then D|Ui is the image of fi ∈ Γ(U, K ∗ ).
Definition 6.46. L (D) ⊆ K is the sub OX -module such that L (D)|Ui is
generated by fi−1 .
85
L (D) is an invertible OX -mod as OUi → L (D)|Ui given by multiplication
by fi−1 .
Proposition 6.43.
∗
1. Γ(X, K ∗ /OX
) ↔ {invertible subsheafs of L }.
2. L (D1 ) ⊗ L (D2 )−1 ' L (D1 − D2 ).
3. D1 ∼ D2 iff L (D1 ) ' L (D2 ).
Proof.
1. Let L ⊆ K be any invertible OX -module. There exists an open
cover X = ∪Ui such that L |Ui ' OUi , so Γ(Ui , L ) ' Γ(Ui , OUi ). Define fi to be the section corresponding to 1 ∈ Γ(Ui , OUi ) by the chosen
∗
isomorphism. fi /fj ∈ Γ(Ui ∩ Uj , OX
) ⇒ {fi−1 } define a Cartier divisor D.
2. L (D1 ) ⊗ L (D2 )−1 → L (D1 − D2 ) by h1 ⊗ h2 7→ h1 h2 .
3. Assume L (D1 ) ' L (D2 ) as OX -modules. By the last part, K ⊂ L (D1 −
D2 ) ' OX . Let 1 ∈ Γ(X, OX ) which corresponds to f ∈ Γ(X, K ). f ∈
Γ(X, K ∗ ), so D1 − D2 =image of f −1 .
Corollary 6.44. If X is any scheme, then the map CaC`(X) → Pic(X) is
injective.
Proposition 6.45. If X is integral, then CaC`(X) ' Pic(X).
Proof. Must show that any invertible L is a submodule of K .
Let L |U ' OU . Then 1 ∈ Γ(U, OU ) corresponds to f ∈ Γ(U, L ). Assume
that ∅ =
6 V ⊆ X open, h ∈ L (V ), then U ∩ V 6= ∅. h ⊗ f −1 ∈ Γ(U ∩ V, L ⊗
−1
L ) = Γ(U ∩ V, OX ) ⊆ Γ(U ∩ V, K ) = Γ(V, K ).
Define an OX -hom L → K by h 7→ h ⊗ f −1 . This is injective, as fp
generator for Lp for all p ∈ U .
Corollary 6.46. X Nötherian, Integral, Separated, Locally Factorial implies
C`(X) = CaC`(X) = Pic(X)
Corollary 6.47. Pic(Pnk ) = {O(m)|m ∈ Z}.
Proof. X = Pnk . We know that C`(X) = Z[H] ' Pic(X).
Must show that [H] ↔ O(1). H = V (h) where h ∈ k[x0 , . . . , xn ] is a linear
form. Set fi = h/xi ∈ OX (D+ (xi )). fi /fj = xj /xi is a unit on D+ (xi xj ), so
{fi } defines a Cartier Divisor. D ∈ CaCl(X) thar corresponds to [H] since
vY (D) = 0 if Y 6= H and 1 if Y = H. So we get OX (1) → L (D) an isomorphism, by s 7→ s/h.
Definition 6.47 (Effective Divisor). A Cartier Divisor D is effective if ∃ an
open cover X = ∪Ui and nonzerodivisors fi ∈ OX (Ui ) such that D|Ui is the
image of fi .
86
D is effective gives us a closed subscheme of X by ID generated by fi on
U . This gives us a correspondence between effective Cartier Divisors and closed
subschemes that are locally generated by one nonzerodivisor.
Note: If X P
is a *-scheme and D an effectve Cartier Divisor, then vY (D) =
vY (fi ) ≥ 0 so Y vY (D)[Y ] is effective. If X is a locally factorial *-scheme,
then effective Cartier divisors correspond to effective Weil divisors.
Commutative Algebra Fact: If A is a normal Nötherian Domain, then A =
∩ht(P )=1 AP ⊆ K(A). So fi ∈ k(Ui ) gives fi ∈ OUi (Ui ) ⇐⇒ (fi ) ∈ Div(Ui )
effective.
Proposition 6.48. D ⊆ X an effective Cartier divisor implies that ID =
L (−D) ⊆ K .
This is because OD ⊆ OX ⊆ K is locally generated by fi .
Let ϕ : X → Y , f ∈ OY (Y ) then ϕ−1 (Yf ) = Xϕ] f .
In fact, we can do this for any line bundle L . Let L be an OY -module. Then
ϕ∗ = ϕ−1 L ⊗ϕ−1 OY OX . For s ∈ Γ(Y, L ) define ϕ∗ s = ϕ−1 s ⊗ 1 ∈ Γ(X, ϕ∗ L ).
/ mP (ϕ∗ L )P } = ϕ−1 (Ys ).
Xϕ∗ s = {P ∈ X|(ϕ∗ s)P ∈
n
Morphisms to P .
If ϕ : X → PnA is a morphism, then
1. X is a scheme over A
2. L = ϕ∗ O(1) is an invertible OX -module.
3. Generated by si = ϕ∗ xi ∈ Γ(X, L ) for 0 ≤ i ≤ n where O(1) is generated
by x0 , . . . , xn ∈ Γ(Pn , O(1)).
Claim: 1+2+3 determines a unique ϕ : X → PnA .
Note: s−1
∈ Γ(Xsi , L −1 ). sj /si = sj ⊗ s−1
∈ Γ(Xsi , L ⊗ L −1 ) =
i
i
Γ(Xsi , OX ).
We must have Xsi = ϕ−1 (D+ (xi )). This defines ϕ : Xsi → D+ (xi ) by Aalgebra homomorphism A[x0 /xi , . . . , xn /xi ] → Γ(Xsi , OX ) by xj /xi 7→ sj /si .
Theorem 6.49. Let X be a scheme over A. Then there is a correspondence
{ϕ : X → PnA over A} to {(L , s0 , . . . , sn )|L generated by s0 , . . . , sn }/ '.
Example: if k is a field, then Pnk = Ank ∪ Pn−1
, An = D+ (x0 ) and P ∈ Pn .
k
n
n
n−1
Then P = 0 ∈ A . So π : A \ {0} → P
, id : Pn−1 → Pn−1 defines the
n
n−1
projection from a point P \ {P } → P
.
Let x1 , . . . , xn ∈ Γ(Pn , O(1)). These generate O(1)|Pn \{P } . This defines the
projection from P .
Note: Assume ϕ : X → PnA is given by si = ϕ∗ xi ∈ L = ϕ∗ O(1). ϕ is a
closed immersion iff ϕ : Xsi → D+ (xi ) is a closed immersion for all i iff Xsi
affine and A[x0 /xi , . . . , xn /xi ] → OX (Xsi ) is surjective.
Recall the definition of an ample line bundle.
Example: If X is projective over a Nötherian ring, L = O(1) is very ample.
If X is affine, and L is any invertible sheaf.
87
Proposition 6.50. X a Nötherian scheme, L an invertible OX -module. Then
TFAE
1. L ample
2. L ⊗m ample for all m ≥ 1
3. L ⊗m ample for some m ≥ 1.
Proof. 1 ⇒ 2 ⇒ 3: trivial
Assume that L ⊗m is ample, F a coherent OX -module. F ⊗ L ⊗k is coherent. For 0 ≤ k ≤ m − 1 choose nk > 0 such that (F ⊗ L ⊗k ) ⊗ (L ⊗m )⊗n is
generatedby global sections for all n ≥ nk .
N = max{k + nmk } implies that F ⊗ L ⊗n is generated by global sections
for all n ≥ N .
Theorem 6.51. X is of finite type over Nötherian A. Then L is ample iff
L ⊗m isvery amples relative to Spec A for some m > 0.
Pnk : O(m) is ample iff m ≥ 1. If m < 0 then Γ(Pn , O(m)) = 0. O(0) gen by
global sections.
Remark: ϕ : X → Y a morphism, (X Nötherian OR ϕ separated and
quasi-compact) Let Z = ϕ(X) ⊆ Y . IZ = ker(ϕ] : OY → ϕ∗ OX ). IZ is
quasicoherent implies that (Z, OX /IZ ) ⊆ Y is a closed subscheme called the
scheme-theoretic image of X.
Exercise: X is reduced implies that Z = ϕ(X) is reduced.
Example: Y = Spec k[x, y]/(xy, y 2 ), X = D(x) = Spec k[x, x−1 ] ⊆ Y . j :
X ⊆ Y . So j(X) = Yred = A1 .
f
g
Application: X → Y → Z an immersion. Then X ⊆ gf (X) → Z is an
immersion.
Exercise: Check This! Hint: Assume f is closed and g is open.
Lemma 6.52. X Nötherian Scheme, U ⊆ X open, F a coherent OU -module.
Then there exists a coherent OX -module F 0 such that F 0 |U ' F .
Note: i : U → X, so i∗ F is quasicoherent has (i∗ F )|U ' F .
Proof. If not, let U ⊆ X be a maximal open set such that the lemma is false,
and WLOG, F is a counterexample.
Let Y ⊆ X be open affine, Y 6⊆ U . Then j : U ∩ Y ⊆ Y . G = j∗ (F |U ∩V )
is a quasicoherent OY -module, and G |U ∩Y = F |U ∩Y . Let M = Γ(Y, G ) =
Γ(U ∩ Y, F ) be a module over A = Γ(Y, OX ).
G = M̃ . Now, U ∩ Y = Yf1 ∪ . . . ∪ Yfn , fi ∈ A. Then M̃fi = G |Yfi = F |Yfi is
coherent. Thus Mfi is a finitely generated Afi -module. Choose m1 , . . . , mN ∈
M generated Mfi for all i. Set M 0 = (m1 , . . . , mN ) ⊆ M . G 0 = M̃ 0 is a coherent
OY -module. G 0 ⊆ G and G 0 |U ∩Y = G |U ∩Y = F |U ∩Y .
Set X 0 = U ∪ Y . Define an OX 0 -module F 0 by F 0 (V ) = {(a, b)|a ∈ F (U ∩
V ), b ∈ G 0 (Y ∩ V ) with a|U ∩Y ∩V = b|U ∩Y ∩V ∈ F (U ∩ Y ∩ V )}.
F 0 |U = F , and F 0 |Y = G 0 , so F 0 is a coherent OX 0 -module. The contradicts that F is a maximal counterexample.
88
Theorem 6.53. X is a scheme of finite type over a Nötherian ring A. L
an invertible OX -module. Then L is ample iff L ⊗m is very ample relative to
Spec A for some m > 0.
Proof. Assume that L ⊗m is very ample over A. Then there exists an immersion
X ⊆ X̄ → PnA where the first is open and the second is closed, such that
L ⊗m ' OX (1).
Let F be coherent OX -module. We know that OX̄ (1) is ample, so the
lemma implies that there exists a coherent OX̄ -module F¯ such that F¯ |X = F .
F¯ ⊗ OX̄ (N ) generated by global sections for all N >> 0, so F ⊗ OX (N ) is
generated by global sections for N >> 0. Thus OX (1) = L ⊗m is ample, so L
is ample.
Now we assume that L is ample. Let P ∈ X. There exists an open affine
neighborhood P ∈ U ⊆ X such that L |U ' OU . Y = (X \ U )red → X is a
closed subscheme, IY ⊆ OX is coherent implies htat IY ⊗ L ⊗n is gneerated
by global sections for some n. So there exists s ∈ Γ(X, IY ⊗ L ⊗n ) such that
sP ∈
/ mP (IY ⊗ L ⊗n )P .
So IY ⊆ OX , so I ⊗ L ⊗n ⊆ L ⊗n , thus s ∈ Γ(X, L ⊗n ).
P ∈ Xs ⊆ U , so (IY )P = OX,P , (IY )Q ⊆ mQ for all Q ∈ Y , so L |Y ' OU .
s|U corresponds to f ∈ Γ(U, OX ).
Xs = Us = Uf is affine. Therefore for all P ∈ X, there exists n > 0
and s ∈ Γ(X, L ⊗n ) such that P ∈ Xs and Xs affine. As X is Nötherian,
Q
X = Xs1 ∪. . .∪Xsk , where s1 , . . . , sk ∈ Γ(X, L ⊗n ) and Xsi are affine. n = ni ,
n/n
so we can replace si with si i ∈ Γ(L ⊗n ).
WLOG, ni = n for all i. We replace L with L ⊗n . So WLOG, there exist
s1 , . . . , sk ∈ Γ(X, L ) such that Xsi is affine and X = Xs1 ∪ . . . ∪ Xsk .
Set Bi = Γ(Xsi , OX ). As X is of finite type over Spec A, we know that Bi is
a finitely gneerated A-algebra generated by bi1 , . . . , biN ∈ Γ(Xsi , OX ). So there
exists n > 0 and cij ∈ Γ(X < OX ⊗ L ⊗n ) such that cij |Xsi = bij sni .
Now, L ⊗n is an invertible OX -module generated by the global setions
n
{si , cij }.
k(N +1)−1
= Proj A[xij : 1 ≤ i ≤
We define a morphism over A, ϕ : X → PA
∗
⊗n
∗
n, 0 ≤ j ≤ N ]. Then ϕ O(1) = L , ϕ (xij ) = cij and ϕ∗ (xi0 ) = sni . Note that
Xsi = ϕ−1 (D+ (xi0 )) → D+ (xi0 ) is a closed immersion, so Bi ← O(D+ (xi0 )) is
surjecitve, mapping xij /xi0 to cij /sni = bij .
k(N +1)−1
Thus, ϕ : X → ∪ki=1 D+ (xi0 ) ⊆ PA
is an immersion, so L ⊗n '
∗
ϕ O(1) is very ample.
Remark: Y a scheme, X ⊆ Y an integral closed subscheme. D ⊆ Y is an
effective Cartier divisor. Assume that X 6⊆ D, then D|X = X ∩ D ⊆ X is an
effective Cartier divisor, and L (D)|X = i∗ L (D) = L (D ∩ X). So ID |X =
ID∩X .
Example: P2k = Proj k[x, y, z]. X = V (zy 2 − x3 + xz 2 ) ⊆ P2 , and P0 = (0 :
1 : 0) corresponds to (x, z) ⊆ k[x, y, z]. L (P0 ) is ample, but not very ample.
89
Claim: OX (1) ' L (3P0 ). Let L = V (z) ⊆ P2 , this is just the line at infinity.
C`(P2 ) = Z[L] and OP2 (1) = L ([L]). So OX (1) = OP2 (1)|X = L ([L])|X =
L (L ∩ X).
Show that L ∩ X = 3P0 ∈ Div(X). Notice that X = Xy ∪ Xz IL∩X ⊆ OX
is generated by z/y on Xy and 1 on Xz .
OX,P0 = k[x/y, z/y](x/y,z/y) /(z/y − (x/y)3 + (x/y)(z/y)3 ) and mP0 = (x/y).
Then vP0 (L ∩ X) = vP0 (z/y) = 3 as z/y = (x/y)3 times a unit, therefore
L ∩ X = 3P0 ∈ Div(X) so L (P0 ) is ample.
Claim: L (P0 ) is not very ample. Otherise there would exist s ∈ Γ(X, L (P0 ))
such that sP0 ∈
/ mP0 L (P0 )P0 . (s)0 ⊆ X is an effective cartier divisor. If
L (P0 )|U = OU then s|U corresponds to f ∈ Γ(U, OU ), and (s)0 ∩U = V (f ) ⊆ U
is a closed subscheme. (s)0 ∼ P0 so deg((s)0 ) = deg P0 = 1, so (s0 ) = Q ∈
Div(X), Q 6= P0 . So P0 ∼ Q and P0 6= Q on a nonsingular projective curve X,
so X is rational and X ' P1 , contradiction.
Let f : Y → X be an affine morphism, A = f∗ OY is a quasi-coherent sheaf
of OX algebras. f ] : OX → f∗ OY = A .
If U ⊆ X open affine, then V = f −1 (U ) ⊆ Y is open affine, and A (U ) =
OY (f −1 (U )) = OY (V ), thus, f : V → U given by OX (U ) → A (U ).
Let X be any scheme, A a quasi-coherent OX -algebra. Want: affine f :
Y → X such that A = f∗ OY .
Functor of Points
Let X be a scheme.
Definition 6.48 (Functor of Points). We define the functor of points to be
a contravariant functor FX :schemes to sets with FX (Y ) = homsch (Y, X). If
h : Y → Y 0 is a morphism, then F (h) = h∗ : FX (Y 0 ) → FX (Y ) by g 7→ gh.
Example: X a variety over k = k̄. Then FX/k (Y ) = homk (Y, X), then
FX/k (Spec k) = {the set of points of X}.
Proposition 6.54. The set of morphisms ϕ : X → X 0 are in correspondence
with the natural transformations α : FX → FX 0 .
Proof. Given ϕ : X → X 0 , then αY : FX (Y ) → FX 0 (Y ) can be defined by
g 7→ ϕg.
Given α : FX → FX 0 , αX : FX (X) → FX 0 (X) set ϕ = αX (id) : X → X 0 .
Let Y be a scheme, and g ∈ FX (Y ). As we have a natural transformation,
we know that αY g ∗ = g ∗ αX . Thus, if we take id ∈ FX (X), it is mapped to
g ∈ FX (Y ) then to αY (g) in FX (Y ). But also it is mapped to ϕ in FX 0 (X) and
then to ϕg in FX (Y ), so they must be equal, and so αY (g) = ϕg.
Corollary 6.55. X ' X 0 ⇐⇒ FX ' FX 0 .
Example: Let X ×S Y be the fibered product over f : X → S and g : Y → S.
Then X ×S Y is uniquely determined by FX×S Y (Z) = {(p, q)|p : Z → X, q :
Z → Y, f p = gq} = FX (Z) ×FS (Z) FY (Z).
X is a scheme, A a quasi-coherent OX -algebra, γ : OX → A .
90
We define FA :schemes→sets by FA (Y ) = {(f, f˜)|f : Y → X, F̃ : A →
f˜
γ
f∗ OY an OX -alg hom, and f ] : OX → A → f∗ OY } on objects and if h : Y → Y 0
is a morphism, then we define h∗ : FA (Y 0 ) → FA (Y ) by (f, f˜) 7→ (f g, f∗ (h] )◦f˜).
Definition 6.49 (SpecA ). SpecA is the unique scheme represented by FA if
it exists. FSpecA = FA .
Note: FSpecA (SpecA ) = FA (SpecA ) so id maps to (π, π̃) the natural proγ
π̃
jection π : SpecA → X, π ] : OX → A → π∗ O with O = OSpecA .
Example: f : Y → X an affine morphism, A = f∗ OY , γ = f ] : OX → A ,
then SpecA = Y
Let Z be any scheme, g : Z → Y a morphism, g ] : OY → g∗ OZ , so f∗ (g ] ) :
f∗ OY = A → (gf )∗ OZ define FY (Z) → FA (Z) by g 7→ (gf : Z → X, f∗ (g ] )).
f]
ϕ̃
Assume (ϕ, ϕ̃) ∈ FA (Z). ϕ : Z → X and ϕ] : OX → A → ϕ∗ OZ .
Let U ⊆ X be open affine, V = f −1 (U ) ⊆ Y affine.
ϕ̃ : A (U ) = OY (V ) → ϕ∗ OZ (U ) = IZ (ϕ−1 (U )) defines a morphism
−1
ϕ (U ) → V .
Proposition 6.56. Let X be a scheme, A be a quasi-coherent OX -algebra.
Then SpecA exists, π : SpecA → X is affine, and π̃ : A → π∗ O is an
isomorphism.
Proof. If U ⊆ X open affine, then SpecA |U = Spec(A (U )), because π :
˜ ) = A (U
˜ ) = A |U .
Spec A (U ) → X has π∗ OSpecA = π∗ A (U
−1
0
Assume U ⊆ U open subset, U affine, then SpecA |U 0 = πU
(U 0 ) We
πU
0
have Spec A |U 0 = Spec A |U ×U 0 U ⊆ Spec A |U → U ⊇ U , so we glue
{Spec A (U )|U ⊆ X affine } together to Spec A . U1 , U2 ⊆ X open affine, and
−1
−1
Spec A (U1 ) ⊇ πU
(U1 ∩U2 ) = Spec A |U1 ∩U2 = πU
(U1 ∩U2 ) ⊆ Spec A (U2 ).
1
2
Let X be a Nötherian Scheme, S = ⊕d≥0 Sd a graded quasi-coherent OY algebra. For U ⊆ X open affine, πU : Proj S (U ) → U . If U 0 ⊆ U a smaller
open affine, then S (U ) → S (U 0 ) define Proj S (U 0 ) → Proj S (U ). So we have
a fiber square
π..................
..
Proj S (U ) ................U
U
.
.........
...
..
...
...
...
...
...
...
..
inc
.
.........
...
..
...
...
...
...
...
...
..
inc
π........0........... 0
.. U
Proj S (U 0 ) .............U
S is quasi coherent implies that S (U 0 ) = S (U ) ⊗OX (U ) OX (U 0 ).
Define FS (Y ) = homsch (Y, Proj S ) = {(f, gU )|f : Y → X, gU : f −1 (U ) →
Proj S (U ) compatible}.
Proposition 6.57. There exists a unique scheme Proj S representing FS .
π
U
Proof. If U ⊆ X open affine, then Proj(S|U ) = Proj(S(U )) →
U . If U 0 ⊆ U
−1
0
any open subset, then Proj(S|U 0 ) = πU (U ), now glue.
91
Example: If X is a scheme, and S = OX [τ0 , . . . , τn ] then Proj(S) = PnX =
P × X.
Example: P1 = Proj(k[x, y]), S = ⊕d≥0 Sd . Let a ∈ Z, then Sd = OP1 ⊕
O(a) ⊕ O(2a) ⊕ . . . ⊕ O(da). So if f ∈ Sd , f 0 ∈ Sd0 , then f ∈ O(ia) and
f 0 ∈ O(ja), so f f 0 ∈ O((i + j)a).
So we get π : Proj S → P1 with π −1 (D+ (x)) = Proj S(D+ (x)) = Proj k[y/x][s, xa t] =
1
A × P1 , and π −1 (D+ (y)) = Proj(k[x/y][s, y a t]) = A1 × P1 . Set λ = y/x, then
this is k[λ−1 ][s, xa λa t], so we glue along the (A1 \ 0) × P1 ’s vis (λ, (s : t)) 7→
(λ−1 , (s : λa t)).
π
Then F−a = Proj(S) → P1 is the Hirzebruch Surface.
X is a scheme, S = ⊕d≥0 Sd graded OX -algebra. Then ProjS is the unique
scheme such that hom(Y, ProjS ) = FS (Y ).
Note: Have OU (1) on π −1 (U ), compatible: U 0 ⊆ U a smaller open affine,
then Proj S (U 0 ) ⊆ Proj S (U ), so OU 0 (1) =pullback of OU (1). Glue to get
O(1) on ProjS : Γ(V, O(1)) = {(σU )|σU ∈ Γ(V ∩ π −1 (U )), OU (1) which are
compatible.
Remark: S = ⊕d≥0 Sd is a graded ring, u ∈ S0 a unit. Then define θd : Sd →
Sd by θd (s) = ud s. This gives an isomorphism of graded S0 -algebras θ : S → S.
Note: h ∈ S+ homogeneous implies that θ = id : S(h) → S(h) , so it induces
id : Proj S → Proj S.
n
Definition 6.50. Let S = ⊕d≥0 Sd graded OX -algebra. L an invertible OX module. Then S ∗ L = S 0 = ⊕d≥0 Sd0 where Sd0 = Sd ⊗ L ⊗d . Let π : P =
ProjS → X and π 0 : P 0 = ProjS 0 → X.
Lemma 6.58. We have an natural isomorphism ϕ : P 0 → P over X and
OP 0 (1) = ϕ∗ OP (1) ⊗ (π 0 )∗ L .
Proof. Let U ⊂ X open affine, with OU → L |U an isomorphism with 1 corresponding to f ∈ γ(U, L ). This defines an isomorphism S (U ) → S 0 (U ), with
S (U )d → S 0 (U )d = S (U )d ⊗ Γ(U, L ⊗d ) by s 7→ s ⊗ f d .
This defines an isomorphism ϕ : Proj S 0 (U ) → Proj S (U ).
Remark implies that ϕ is independent of f ∈ Γ(U, L )∗ . So we can glue to
an isomorphism ϕ : P 0 → P .
The sections of OP 0 (1) correspond to elements in Γ(U, S10 ) = Γ(U, S1 ) ⊗
Γ(U, L ) correspond to sections of OP (1) ⊗ π ∗ L .
Definition 6.51. Assume X is Nötherian. S = ⊕d≥0 Sd satisfies (+) if
1. S0 = OX
2. S1 is a coherent OX -module
3. S is generated (locally) by S1 .
Note: If U ⊆ X is a small open affine, then ∃S1⊗d |U → Sd |U → 0 implies
that Sd is coherent.
92
Lemma 6.59. Assume that S satisfies (+). Let π : P = Proj S → X. Then
π is prokper and if there exists an ample invertible OX -module L on X, then
π is projective, and OP (1) ⊗ L ⊗n is very ample relative to X for some n > 0.
Proof. Let U ⊂ X open affine such that S (U ) generated by S1 (U ) as A-algebra,
with A = OX (U ). Then S1 -coherent implies that S1 (U ) is a finitely gneerated
A-module, so ∃ a graded A-algebra homomorphism A[x0 , . . . , xN ] → S (U )
close
π
surjective. Thus, Proj S (U ) → PN
U → U . Thus π|U is projective, so it is
proper. Thus, π : Proj S → X is proper.
If L is ample, then S1 ⊗ L ⊗n is generated by global sections. As X is
Nötherian, S1 ⊗ L ⊗n is coherent, so it is generated by finitely many global sections. Thus, there exists a surjection of graded OX -algberas OX [T0 , . . . , TN ] →
S ∗ (L ⊗n ), and so P 0 = Proj(S ∗ L ⊗n ) → Proj OX [Ti ] = PnX is closed, and so
OP 0 (1) = ϕ6 ∗ (OP (1) ⊗ π ∗ (L ⊗n )) is very ample.
Definition 6.52 (Tensor Algebra). Let A be a ring and M be an A-module.
Then T d M = M ⊗ . . . ⊗ M , d-times. T (M ) = ⊕d≥0 T d M is called the tensor
algebra. S(M ) = ⊕d≥0 S d (M ) = T (M )/(x ⊗ y − y ⊗ x) is the symmetric algbera.
If M = A⊕r , then S(M ) = A[T1 , . . . , Tr ].
If X is a Nötherian scheme, E a locally free OX -module of rank r and
E ∨ = H om(E , OX ) the dual sheaf is also locally free of rank r. Then S(E ∨ ) =
[U 7→ S(Γ(U, E ∨ ))+ (sheafification) is a graded OX -algebra.
S(E ∨ )0 = OX , and S(E ∨ )1 = E ∨ , ... this satisfies (+). Set π : Y =
Spec S(E ∨ ) → X.
If U = Spec A ⊆ X open, E |U ' OU⊕r , then S(E ∨ )|U ' OU [T1 , . . . , Tr ], so
−1
π (U ) = Spec OU [T1 , . . . , Tr ] = U × Ar . Thus π : Y → X affine bundle (in
fact, a vector bundle), so assume f : U → π −1 (U ) ⊆ Y is a section, (πf = idU ),
then
f]
Γ(U, E ∨ ) → Γ(U, S(E ∨ )) → Γ(U, OX )
gives an OX -homomorphism f : E ∨ → OX over U , ie, a section f ∈ Γ(U, E ).
THus E is the sheaf of sections of π : Spec S(E ∨ ) → X.
Definition 6.53. P(E ) = ProjS(E ∨ ).
π : P(E ) → X, OE (1).
r−1
If E |U ' OU⊗r , then π −1 (U ) = U × Pr−1 = PU
.
Example: L an invertible OX -module, π : P(L ) → X an isomorphism,
OL (1) = π ∗ (L −1 ).
Proposition 6.60.
0 for m < 0.
1. If rank(E ) ≥ 2, then π∗ OE = S m (E ∨ ) for m ≥ 0 and
2. Have surjection π ∗ E ∨ → OE (1).
Proof. Have global S m (E ∨ ) → π∗ OE (m), and in U ⊆ X open affine, then
f ∈ Γ(U, S m (E ∨ )) gives a section of OE (m) over π −1 (U ) = Proj Γ(U, S(E ∨ ))
93
isomorphism locally when E |U ' OU⊗r , π −1 (U ) = Pr−1
U , already computed
Γ(Pr−1
,
O(m)).
U
For the secton part, we have an OX -homomorphism E ∨ → π∗ OE (1) which
gives an OP(E ) -homomorphism π ∗ E ∨ → OE (1), and can check that it is surjecitve locally.
Universal Property
FP(E ) (Y ) = {(f, L , θ)|f : Y → X, L an invertible OY -module, and θ :
f ∗ E ∨ → L surjective}. Left to reader.
Exercise: P2k , k = k̄. Then P2 = {L ⊂ k 3 | dim L = 3} = {E2 ⊆ k 3 | dim E2 =
2}.
0 → E → OP⊕3
→ OP2 (1) → 0 with E locally free of rank 2, then P(E ) =
2
F `(k 3 ) = {(E1 , E2 )|E1 ⊂ E2 ⊂ k 3 , dim(Ei ) = i}.
⊕3
π : P(E ) → P2 . Flag of locally free OX -module, OE (−1) ⊆ π ∗ E ⊆ OP(E
) cor3
respond to flags of vector bundles B1 ⊆ B2 ⊆ P(E ) × k , Bi = {((E1 , E2 ), ~v )|~v ∈
Ei }.
7
Schemes II
Differentials
Let R be a ring, S a commutative R-algebra and M an S-module.
Definition 7.1 (R-derivation). D : S → M is an R-derivation if D(f g) =
f D(g) + gD(f ), D(f + g) = D(f ) + D(g) for all f, g ∈ S and D(f ) = 0 if f ∈ R
(iff D is R-linear)
Universal Derivation: Let F be the free S-module with basis {df |f ∈ S},
and F 0 ⊂ F the submodule generated by d(f g)−f dg −gdf, d(f +g)−df −dg and
df for all f ∈ R. Then defime ΩS/R = F/F 0 and d = dS = dS/R : S → ΩS/R by
d(F ) = df + F 0 . This satisfies the universal property that if D : S → M is any
R-derivation, then ∃!S-homomorphism D̃ : ΩS/R → M such that D = D̃ ◦ dS/R .
P (x1 , . . . , xn ) ∈ R[x1 , . . . , xn ] and f1 , . . . , fn ∈ S. Then D(P (f1 , . . . , fn )) =
PnNote:
∂P
(f
i=1 ∂xi 1 , . . . , fn )D(fi ).
Proposition 7.1. S = R[x1 , . . . , xn ], then ΩS/R ' S ⊕n = ⊕ni=1 S · dxi and
d : S → S ⊕n is given by df = (df /dx1 , . . . , df /dxn ).
Proposition 7.2. R → S → T ring homomorphisms, then ΩS/R ⊗S T →
ΩT /R → ΩT /S → 0 is exact as T -modules.
δ
Proposition 7.3. If S is an R-algebra and T = S/I, then I/I 2 → ΩS/R ⊗S T →
ΩT /R → 0 wotj δ taking f + I 2 to df ⊗ 1.
Proposition 7.4. Let R0 and S be R-algebras and S 0 = S⊗R R0 . Then ΩS 0 /R0 =
ΩS/R ⊗S S 0
Proposition 7.5. U ⊆ S a multiplicative subset, then U −1 ΩS/R = ΩU −1 S/R .
94
Corollary 7.6. If S is a localization of a finitely generated R-algebra, then
ΩS/R is a finitely generated S-module.
Proof. S = U −1 S 0 , S 0 = R[s1 , . . . , sn ]. Then ΩS/R is generated by ds1 , . . . , dsn
as an S 0 -module, so ΩS/R = U −1 ΩS 0 /R is generated by ds1 , . . . , dsn as an Smodule.
Sheaves of Differentials
Let X be a topological space, R, S sheaves of rings, and R → S a sheaf
homomorphism.
Define pre − Ω = pre − ΩS /R = [U 7→ ΩS(U )/R(U ) ]. If V ⊆ U is open, then
we get a diagram S (U ) → S (V ) → pre − Ω(V ), but also S (U ) → pre − Ω(U ),
which goes to pre − Ω(V ) by restriction, and this is an R(U )-derivation. Then
pre − Ω is a presheaf.
Define ΩS /R = (pre − Ω)+ .
Let ϕ : X → Y be a morphism of schemes, ϕ] : ϕ−1 OY → OX .
Define ΩX/Y = ΩOX /ϕ−1 OY , and if we have X → Spec(k) is a scheme over
k, then ΩX = ΩX/k = ΩX/ Spec(k) . We call this the relative cotangent sheaf and
the cotangent sheaf.
Proposition 7.7. If ϕ : X → Y is a morphism of affine schemes, X = Spec(S)
˜ .
and Y = Spec(R), then ΩX/Y ' ΩS/R
Corollary 7.8. ΩX/Y is always quasi-coherent. If ϕ : X → Y is locally of
finite type, then ΩX/Y is coherent.
Proof. P ∈ X, take open affine neighborhoods ϕ(P ) ∈ V ⊆ Y adn P ∈ U ⊆
ϕ−1 (V ) ⊆ X. ΩX/Y |U = ΩU/V is quasicoherent. If ϕ is lcoally of finite type,
then we can take U, V such that OX (U ) finitely generated OY (V )-algebra.
Proposition. 7.9. Take a fiber square:
X ....................................................... Y
g
.
.........
....
...
.
0 .....
...
...
...
....
.
.
.........
....
...
...
...
...
...
...
....
.
g
f .
X 0 ....................................................... Y 0
Then g 0∗ ΩX/Y = ΩX 0 /Y 0
f
g
Proposition 7.10. If X → Y → Z morphisms then f ∗ ΩY /Z → ΩX/Z →
ΩX/Y → 0 is exact.
2
Proposition 7.11. If g : Y → Z, X ⊆ Y a closed subscheme, then IX /IX
→
ΩY /Z ⊗ OX → ΩX/Z → 0 is exact.
Theorem 7.12. Let Y be a scheme. X = PnY = PnZ × Y . Then ∃ an exact
sequence 0 → ΩX/Y → OX (−1)⊕n+1 → OX → 0.
95
X
........................................................
...
...
...
...
...
...
...
...
...
..
..........
.
PnZ
...
...
...
...
...
...
...
...
...
..
..........
.
.
Y ............................................... Spec Z
Proof. WLOG, Y = Spec(Z) and X = Proj(S) for S = Z[x0 , . . . , xn ]. Set
E = S(−1)⊕n+1 . We have a map E → S by ei 7→ xi , it has kernel M , so we
get an exact sequence 0 → M → E → S. This gives us an exact sequence
0 → M̃ → O(−1)⊕n+1 → OX → 0
Pn ∂f
xi = df .
Notice: f ∈ Q(x0 , . . . , xn ) homogeneous of degree d, then i=0 ∂x
i
Pn ∂f
Define d : OX → Ẽ by d(f ) = i=0 ∂xi ei . Note that d(OX ) ⊆ M̃ , ie d : OX →
M̃ is a derivation. This induces d˜ : ΩX/Z → M̃ , we will check that this is an
isomorphism locally on D+ (xi ) = Spec Z[x0 /xi , . . . , xn /xi ].
Enough to check that Γ(D+ (xi ), M̃ ) is a free S(xi ) -module with basis {d(xj /xi )|j 6=
i}.
0 → M(xi ) → E(xi ) → S(xi ) → 0 which takes ej → xj but there’s a map from
x
S(xi ) → E(xi ) taking 1 → ei /xi . And therefore, d(xj /xi ) = x1i ej − x2j ei .
i
Singular Varieties
Let k = k̄
Definition 7.2 (Nonsingular Variety). A variety X over k is nonsingular at
P ∈ X if OX,P is a regular local ring. X is nonsingular if all points are nonsingular.
Theorem 7.13. X is an irreducible separated scheme of finite type over k = k̄.
Set n = dim(X). Then ΩX/k is locally free of rank n iff X is a nonsingular
variety.
Recall that F ⊂ K is a field extension, and each a ∈ K has a minimal
polynomial pa (T ) ∈ F [T ] such that pa (a) = 0 ∈ K. K is separable over F if
pa (T ) does not have multiple roots for all a ∈ K.
Exercise: If K/F is separable, then ΩK/L = 0. (0 = dK (pa (a)) = p0a (a)dK (a)
with p0a (a) 6= 0)
Corollary 7.14. X is a variety over k implies that a dense open subset of X
is nonsingular.
Proof. K = k(X).
FACT: k is a perfect field implies that any finitely generated extenseion k ⊂
K is separaly generated. IE, there exists a transcendence basis x1 , . . . , xn ∈ K
suchj that k(x1 , . . . , xn ) ⊆ K is separable.
K is separable over F = k(x1 , . . . , xn ) for n = dim(X). Let S = k[x1 , . . . , xn ],
F = S0 ⇒ ΩF/k = (ΩS/k )0 = F ⊕n .
α
So ΩF/k ⊗F K → ΩK/k → ΩK/F = 0.
96
As any k-derivation D : F → K can be extended to D : K → K we can
conclude that α is injective.
Thus, ΩK/k ' K n . Spec(R) ⊆ X open. K = R0 ⇒ (ΩR/k )0 = ΩK/k = K n .
Thus ∃0 6= f ∈ R : (ΩR/k )f = ΩRf /k = (Rf )n .
'
Lemma 7.15. If (R, m) is a local ring, k = R/m, and k ⊆ R. Then δ : m/m2 →
ΩR/k ⊗R k is an isomorphism.
Theorem 7.16. X nonsing variety over k = k̄ and Y ⊆ X irreducible closed
subscheme, then Y is nonsingular iff ΩY /k is locally free and 0 → IY /IY2 →
ΩX ⊗ OY → ΩY → 0 is exact.
Proof. Assume the latter conditions, set q = rank(ΩY ), n = rank(ΩX ) =
dim(X). It is enough to show that q = dim(Y ). The second condition causes
IY /IY2 to be locally free of rank n − q, and by Nakayama, IY is locally generated by n − q elements.
The Principle Ideal Theorem says that dim(Y ) ≥ q. Let P ∈ Y , mP ⊆
OY,P , then the lemma says that mP /m2P = ΩY,P ⊗ OY,P /mP = ΩY,P ⊗ k. So
dimk (mP /m2P ) = q. Thus dim(Y ) = dim OY,P ≤ q, so dim(Y ) = q, so Y is
nonsingular.
Assume that Y is nonsingular. ΩY /k is then locally free of rank q = dim(Y ).
ϕ
We know that IY /IY2 → ΩX ⊗ OY → ΩY → 0 is exact, so all that remains is
2
showing that δ : IY /IY → ΩX ⊗ OY is injective.
Let P ∈ Y be any closed point. Localize at P . Then ker ϕP is a free OY,P module of rank r = n−q, so there exist x1 , . . . , xr ∈ IY,P such that dx1 , . . . , dxr
form a basis for ker ϕP . Let I 0 ⊆ IY be a subideal generated by x1 , . . . , xr and
Y 0 = Z(I 0 ) ⊆ X.
Thus, dx1 , . . . , dxr must generate a free subsheaf of rank r in ΩX ⊗ OY 0 . So
δ
we get that I 0 /I 02 → ΩX ⊗ OY 0 → ΩY 0 → 0 and δ 0 must be injective.
So the first part tells us that Y 0 is a nonsingular variety and dim(Y 0 ) =
rank ΩY 0 = n − r = q, and Y ⊆ Y 0 ⊆ X closed subschemes, Y, Y 0 both have
dimension q, so Y 0 = Y and δ 0 = δ.
Exercise: (R, m) is a regular local ring, f ∈ m. Then R/(f ) is regular local
iff f ∈
/ m2 . Use: Minimum number of gens of m = dimk m/m2 , k = R/m.
Theorem 7.17 (Bertini’s Theorem). Let X ⊆ Pnk a closed irreducible nonsingular subvariety, k algebraically closed. Then there exists a hyperplane H ⊆ Pn
such that X 6⊂ H and X ∩ H is nonsingular.
Proof. V = Γ(Pn , O(1)) = kx0 ⊕ . . . ⊕ kxn . Then points of P(V ) correspond to
hyperplanes by f 7→ Z(f ). Let P ∈ X be a closed point, BP = {f ∈ P(V )|X ⊆
Z(f ) or P ∈ X ∩ Z(f ) is a singular point}.
Will show: ∪P ∈X BP ( P(V ) is proper closed.
Set r = dim(X). Claim: BP ⊆ P(V ) is a linear subspace of dimension
n − r − 1. Check that f0 ∈ V such that P ∈
/ Z(f0 ) ⊆ Pn . For f ∈ V , we have
97
f /f0 ∈ OX,P . Then OX∩Z(f ),P = OX,P /(f /f0 ). f /f0 ∈ mP ⇐⇒ P ∈ Z(f ),
f /f0 = 0 ∈ OX,P ⇐⇒ X ⊆ Z(f ), f /f0 ∈ mP /m2P ⇐⇒ f ∈ BP .
Note: ϕP : V → OX,P /m2P by f 7→ f /f0 is surjective because k = k̄ so mP
is generated by linear forms.
dimk V = n + 1, dimk (OX,P /m2P ) = r + 1. So dim ker ϕP = n − r, so
BP = P(ker ϕP ) so dim BP = n − r − 1.
Now we define B = {(P, f ) ∈ X × P(V )|f ∈ BP }
Check: This is actually a closed subvariety of X × P(V ). π1 : B → X is a
surjective morphism and π1−1 (P ) = BP has dim BP = n − r − 1 and dim(X) = r
and so dim B = dim(X) + dim BP = n − 1 and ∪P ∈X BP = π2 (B) ⊆ P(V ).
Then π2 : X × P(V ) → P(V ) is a proper map, so π2 (B) is closed of dimension
at most n − 1.
8
Cohomology
Let R be a ring.
Definition 8.1 (Injective). An R-modules I is injective if for every R-module
M and submodule M 0 and R-homomorphism ϕ : M 0 → I, there exists an extension to M .
Definition 8.2 (Divisible Abelian Group). A Z-module T is divisible if for all
n ∈ Z the map T → T by multiplication by n is surjective.
Examples: Q, Q/Z.
Lemma 8.1. Any divisible Z-module is injective.
Proof. M 0 ⊂ M , ϕ0 : M 0 → T . Consider {(N, ψ)|M 0 ⊂ N ⊂ M, ψ : N → T
extends ϕ0 }.
Zorn’s Lemma says that there is a maximal element (N, ψ).
Claim: N = M . Else take x ∈ M \ N . 0 → N → (N, x) → Z/mZ → 0.
Either m = 0 in which case (N, x) = N ⊕ Zx and we can extend ϕ(x) = 0 or
m 6= 0 then mn ∈ N . So there exists y ∈ T with ϕ0 (mx) = my, so ψ(x) = y.
Definition 8.3. If M is a Z-module, M ˆ = homZ (M, Q/Z).
Exercise: M → M ˆˆ, x 7→ [f 7→ f (x)] is injective.
Corollary 8.2. Every Z-module M is contained in a divisible module.
Proof. Take F free, F → M ˆ → 0. So M → M ˆˆ → F ˆ =direct product of
Q/Zs, with all the maps inclusions.
Note: T a Z-module, M an R-module, then homZ (R, T ) is an R-module.
homZ (M, T ) ' homR (M, homZ (R, T )) by f 7→ [x 7→ [r 7→ f (rx)]].
Lemma 8.3. T a divisible Z-module implies that I = homZ (R, T ) is an injective
R-module.
98
Proof. M 0 ⊆ M are R-modules. Want:
surj
homR (M, I) ............................................ homR (M 0 , I)
.
..........
...
..
...
...
...
...
...
....
.
'
.
..........
...
..
...
...
...
...
...
....
.
'
surj
homZ (M, T ) ............................................ homZ (M 0 , T )
Which is true.
Theorem 8.4. Every R-module M is a submodule of an injective R-module.
Proof. Lemma implies that there exists f : M → T injective Z-hom when T is
divisible. We define M → I = homZ (R, T ) by x 7→ fx with fx (r) = f (xr)
Note that M ⊆ I: this is because f was injective to begin with.
Definition 8.4 (Abelian Category). A category C is abelian if hom(A, B) is
an abelian group for all A, B ∈ obj(C ), hom(B, C) × hom(A, B) → hom(A, C)
is a group homomorphism, finite direct products exist, kernels exist, cokernels
exist, etc. See Weibel.
For any I an object of C , we have a contravariant functor hom(−, I) : C →
Ab. It is left exact, ie 0 → A0 → A → A00 → 0 exact implies 0 → hom(A00 , I) →
hom(A, I) → hom(A0 , I) exact.
Definition 8.5 (Enough Injectives). I is injective if hom(−, I) is exact.
A category C has enough injectives if every object is a subobject of an injective object.
Corollary 8.5. If (X, OX ) is a ringed space, then the category of OX -modules
has enough injectives.
Proof. Let F be an OX -module. For p Q
∈ X, there exists an inclusion FP → IP
an injective OX,P -module. Set I(U ) = p∈U IP .
Claim: I is Q
an injective OX -module. For any OX -module, G , we have
homOX (G , I) = p∈X homOX homOX,P (GP , IP ).
So if G 0 ⊆ G submodule, then hom(G , I) → hom(G 0 , I ) is surjective.
Corollary 8.6. Let X be a topological space, then the category Ab(X) of sheaves
of abelian groups on X has enough injectives.
Proof. Set OX to be the constant sheaf Z, the sheaf of locally constant functions
to Z.
Injective Resolution
Let M be an object of C , then if C has enough injectives, there exists an
exacct sequence 0 → M → I 0 → I 1 → . . . with I j injective, that is, M has an
injective resolution.
Let F : C → D be a left exact functor. Then we have a complex F (I · ) =
0 → F (I 0 ) → F (I 1 ) → . . ..
99
Definition 8.6. Rj F (M ) = H j (F (I · )).
Example: Left Exact implies that 0 → F (M ) → F (I 0 ) → . . .. We throw
away M so that H 0 (I · ) = R0 F (M ) = F (M ).
Definition 8.7. Let X be a topological space, F a sheaf of abelian groups, and
Γ : Ab(X) → Ab by Γ(F ) = Γ(X, F ) is left exact. Then set H j (X, F ) =
Rj Γ(F ).
Let A· , B · be complexes in C , a morphism f : A· → B · is a collection of
fn : An → B n such that the appropriate squares all commute.
This induces a map H(f ) : H n (A· ) → H n (B · ).
f, g : A· → B · are homotopic (f ∼ g) iff there exists morphisms hn : An →
n−1
n
B
with fn − gn = dn−1
B hn + hn+1 dA . This implies that H(f ) = H(g) :
n
·
n
·
H (A ) → H (B ).
Lemma 8.7. Consider two chain complexes A· and I · , 0 → M → A· and
0 → M 0 → I · an injective resolution, with ϕ : M → M 0 . Then there exists a
morphism f : (M → A· ) → (M 0 → I · ) and any other is homotopic to f .
Proof. See Weibel.
Note: (f1 − g1 − d0B h1 )d0A = d0B (f0 − g0 − h1 d0A ) = d0B d−1
B h0 = 0.
Corollary 8.8. Rj F (M ) is independent of the choice of injective resolution.
Corollary 8.9. A morphism ϕ : M → M 0 induces a unique ϕ∗ : Rj F (M ) →
Rj F (M 0 )
Lemma 8.10 (Horseshoe Lemma). Let 0 → M 0 → M → M 00 → 0 be a short
exact sequence in a category with enough injectives. Then there exists...
Theorem 8.11. Let C be abelian with enough injectives and F : C → C 0 a left
exact functor.
1. Rn F : C → C 0 is additive, ie, Rn F (M ⊕ M 0 ) = Rn F (M ) ⊕ Rn F (M 0 )
2. R0 F (M ) = F (M ) and F n F (I) = 0 if I is injective and n ≥ 1.
3. 0 → M 0 → M → M 00 → 0 short exact gives a long exact sequence 0 →
F (M 0 ) → F (M ) → F (M 00 ) → R1 F (M 0 ) → . . .
4. The δ morphisms from Rn F (M 00 ) → Rn+1 F (M 0 ) are natural.
Definition 8.8 (F-acyclic). J ∈ obC is F -acyclic if Rn F (J) = 0 for all n ≥ 1.
Example: J injective implies that J is F -acyclic for all left exact functors
F.
Lemma 8.12. 0 → Y 0 → Y 1 → . . . exact with Y i F -acyclic implies that
0 → F (Y 0 ) → F (Y 1 ) → . . . is exact.
100
Theorem 8.13. 0 → M → J 0 → . . . a resolution by F -acyclic objects J n .
Then Rn F (M ) = H n (F (J · ))
Definition 8.9 (δ-functor). If C and C 0 are abelian categories, then a δ-functor
from C → C 0 is a sequence of functors T = (T n : C → C 0 )n≥0 together
with a morphism δ n : T n (M 00 ) → T n+1 (M 0 ) for each short exactr sequence
0 → M 0 → M → M 00 → 0 such that
1. Long exact sequence
2. Naturality.
Example: T n = Rn F .
Definition 8.10 (Universal δ-functor). A δ-functor T is universal if, given any
δ functor U and natural transformation f0 : T 0 → U 0 , then there are unique
fi : T i → U i such that the appropriate diagram commutes.
Definition 8.11 (Erasable). An additive functor F : C → C 0 is erasable if for
all M ∈ ob(C ) there exists a monomorphism P : M → J such that F (P ) = 0 :
F (M ) → F (J).
Example: Rn F is erasable for n ≥ 1, M → I a monomorphism and I
injective, then Rn F (M ) → Rn F (I) = 0.
Theorem 8.14. Let T = (T n ) be a δ-functor. If every T n is erasable for all
n ≥ 1 then T is a universal δ-functor.
Proof. Assume that U = (U n ) is a δ-functor with f0 : T0 → U0 a natural
transformation. Let M ∈ ob(C ).
p
q
0 → M → J → X → 0 such that T 1 (p) = 0.
Then we take the long exact sequences, horizontally, and f0 vertically. We
0
◦f0 . Then ∃!f1 (M ) : T 1 (M ) → U 1 (M ). Must
use that ker δT0 = Im(q∗ ) ⊆ ker δU
check that f1 is a natural transformation.
Check: f1 commutes with δ.
To get f2 , we use 0 → M → J → X → 0 such that T 2 (p) = 0. Proceed by
induction using the long exact sequence.
Definition 8.12. Let X be a topological space, and F a sheaf of abelian groups
(hereafter referred to as an abelian sheaf ). Let Γ = Γ(X, −). Then H n (X, F )
is defined to be Rn Γ(F ).
Definition 8.13 (Flasque). A sheaf F is flasque if all the restriction maps are
surjective.
Lemma 8.15. (X, OX ) a ringed space. F injective as an OX -module. Then
F is flasque.
Proof. For U ⊆ X open, Set OU = j! (OX |U ) by j : U → X the inclusion and
for W ⊆ X open, OU (W ) = OX (W ) if W ⊆ U and 0 else.
Note, hom(OU , F ) = F (U ). If V ⊆ U , then 0 → OV → OU implies that
hom(OU , F ) → hom(OV , F ) is surjective.
101
Lemma 8.16. Assume that 0 → F 0 → F → F 00 → 0 exact.
1. F 0 is flaque implies that 0 → F 0 (X) → F (X) → F 00 (X) → 0 is exact
2. F 0 and F flasque implies that F 00 is flasque.
Proof. Let γ ∈ Γ(X, F 0 ). Consider {(U, β)|U ⊆ X open and β ∈ Γ(U, F ) with
β 7→ γ|U ∈ F 00 (U )}.
Then (U 0 , β 0 ) ≤ (U, β) iff U 0 ⊆ U and β 0 = β|U 0 . Zorn’s lemma gives us a
maximal element (U, β).
Claim: U = X. Assume not, then there exists (V, σ) such that V 6⊆ U and
σ ∈ Γ(V, F ), σ 7→ γ|V . Then β|U ∩V − σ|U ∩V ∈ F 0 (U ∩ V ). Then flasque
implies that ∃α ∈ F 0 (V ) with α|U ∩V = β − σ. Set β1 = σ + α ∈ F (V ). Then
β|U ∩V = β1 |U ∩V , so glue β, β1 to β̃ ∈ Γ(U ∪ V, F ), then (U ∪ V, β̃) ≥ (U, β),
contradiction.
Proposition 8.17. If F is flasque, then H n (X, F ) = 0 for all n ≥ 1.
Proof. 0 → F → I → G → 0 is a short exact sequence, with I injective and
G flasque. So we get a long exact sequence 0 → F (X) → I(X) → G (X) →
H 1 (X, F ) → . . .. But H i (X, I) = 0, so we get H 1 (X, F ) = 0, H 2 (X, F ) =
H 1 (X, G ) = 0, etc.
Corollary 8.18. If 0 → M → F 0 → . . . is a flasque resolution, then H n (X, M ) =
H n (Γ(X, F ∗ )).
Proposition 8.19. (X, OX ) a ringed space. The right derived functos fo Γ :
M od(X) → M od(OX (X)) are given as M 7→ H n (X, M ).
Proof. Injective resolution in M od(X) 0 → M → I 0 → . . ., we know that I n is
flasque for all n, so the corollary implies this.
Definition 8.14 (Direct Limit Sheaf). lim Fα = [U 7→ lim Fα (U )]+ .
−→
−→
Lemma 8.20. X Nötherian implies that Γ(U, lim Fα ) = lim Γ(U, Fα ).
−→
−→
Proof. Check sheaf axioms using finite open covers.
Lemma 8.21. X Nötherian and Fα flasque, then lim Fα is flasque.
−→
Proof. V ⊆ U ⊆ X open, Fα (U ) → Fα (V ) surjective implies that lim Fα (U ) →
−→
lim Fα (V ) is surjective.
−→
Proposition 8.22. X Nötherian, {Fα } directed system, then H n (X, lim Fα ) =
−→
lim H n (X, Fα ).
−→
Proof. Let I = indA (Ab(X)) be the category of all directed systems of abelian
sheaves.
T n : I → Ab by T n ({Fα }) = lim H n (X, Fα ) and U n : I → Ab by
−→
n
U ({Fα }) = H n (X, lim Fα ).
−→
Note that lim is exact, so T n and U n form δ-functors.
−→
102
And T 0 ({Fα }) = lim Γ(X, Fα ) = Γ(X, lim Fα ) = U 0 ({Fα }).
−→
−→
By theorem from before, ETS that T n and U n are universal δ-functors,
because they agree on the 0th term. Q
Given F ∈ Ab(X), set F˜ (U ) = p∈U Fp , the sheaf of discontinuous sections.
0 → F → F˜ exact and F˜ is flasque.
{Fα } ∈ ob(I), and {F˜α } ∈ ob(I), and we have 0 → {Fα } → {F˜α }.
So now sett that T n ({F˜α }) = lim H n (X, F˜α ) = 0 and simialry for U n .
−→
Lemma 8.23. Y ⊆ X a closed subspace, j : Y → X the inclusion, and F an
abelian sheaf on Y . Then H n (Y, F ) = H n (X, j∗ F ).
Proof. 0 → F → I 0 → I 1 → . . . an injective resolution on Y . I n is flasque, so
j∗ (I n ) is flasque. So 0 → j∗ F → j∗ I 0 → . . . is a flasque resolution (exactness
follows from Y being closed).
So H n (Y, F ) = H n (Γ(Y, I ∗ )) = H n (Γ(X, j∗ I ∗ )) = H n (X, j∗ F )
Remark: Y ⊆ X closed, U = X \Y . Then F an abelian sheaf on X, we have
j : Y → X and i : U → X inclusions. Then we can set FU = i! (F |U ) = 0 unless
V ⊆ U , and if V ⊆ U , then F (V ) and also FY = j∗ (F |Y ) where FY (U ) = 0
if V ⊆ U and equals j −1 F (Y ∩ V ) if V 6⊆ U . Then FY,P = FP if P ∈ Y and
zero else, as FU = Fp iff p ∈ U and is 0 otherwise.
So we have 0 → FU → F → FY → 0 is exact.
Theorem 8.24. Let X be a Nötherian topological space, and F an abelian
sheaf on X. Then H n (X, F ) = 0 for all n > dim(X).
Proof. Step 1: Reduce to X irreducible: If X is reducible, let Y be a maximal
component in X. Take U = X \Y , then we have the above short exact sequence.
This gives us a long exact sequence on cohomology, H n (X, FU ) → H n (X, F ) →
H n (X, FY ) = H n (Y, F |Y ). If the theorem is true for irreducible spaces, then
the last one is zero, and so we have isomorphisms H n (X, FU ) → H n (X, F ), so
by induction, it is enough to show that the theorem holds when X is irreducible.
Step 2: dim X = 0, X irreducible implies that X has open sets ∅, X and
nothing else. Then Ab(X) = Ab, so Γ : Ab(X) → Ab is the identity, and so it is
exact, so H n (X, F ) = 0 for n ≥ 1.
Step 3: Assume that X is irreducible of dimension > 0. If σ ∈ F (U ), then
get ZU → F by 1 7→ σ. Let α = {σ1 , . . . , σm } with σi ∈ F`
(Ui ). Define
Fα = Im(⊕ZUi → F ) the subsheaf generated by α. Set B = U ⊆X F (U ).
A = {α ⊆ B|α finite}. A is a directed poset, so {Fα } is a directed system.
lim Fα = F . And ten, as H n (X, F ) = H n (X, lim Fα ) = lim H n (X, Fα ), it
−→
−→
−→
suffices to prove this for F = Fα finitely generated.
If α0 ⊆ α, then we have 0 → Fα0 → Fα → G → 0, with G generated by the
number of sections in α \ α0 . So WLOG, F is generated by one section. So we
have 0 → K → ZU → F → 0 exact. Thus, it is enough to show vanishing for
F ⊆ ZU a subsheaf.
103
For p ∈ U , Fp ⊆ ZU,p = Z. Define d = min{e ∈ Z+ |Fp = eZ for some p}.
Choose p such that Fp = dZ, d = σp , σ ∈ F (V ) ⊆ ZU (V ), p ∈ V . Must have
F |V = dZU |V = dZV .
·d
So we have 0 → ZV → F → F /ZV → 0. Supp(F /ZV ) ⊆ U \ V , so
dim U \ V < dim(X). So induction on dimension gives us that H n (X, F /ZV ) =
0 for n > dim(X).
Set Y = X \ U , 0 → ZU → ZX → ZY → 0 is ses, dim Y < dim X, so
H n (X, ZY ) = 0 for n ≥ dim(X). H n−1 (ZY ) → H n (ZU ) → H n (ZX ). So
WLOG, F = ZX .
If X is irreducible, then ZX is flasque. So H n (X, ZX ) = 0 for all n ≥ 1.
NOTE: Above, FU is not a sheaf, to correct things, sheafify
Lemma 8.25. Z ⊆ X closed, G a sheaf on Z, then H n (Z, G ) = H n (X, j∗ G ).
Note: Supp(FU ) = U ⊆ Ū , then FU = j∗ (FU |Ū ), so H n (X, FU ) =
H (Ū , FU |Ū ).
Next: If X = Spec(A) and F is a quasi-coherent OX -module, then H n (X, F ) =
0 for n ≥ 1.
Exercise: If M is an A-module, assume homA (A, M ) → homA (I, M ) is
surjective for all ideals I ⊆ A. Then M is injective. Hint: N 0 ⊆ N , ϕ : N 0 → M .
Let M 0 ⊆ M an A-module, I ⊆ A an ideal, I p (M 0 ∩I n M ) ⊆ M 0 ∩I n+p M for
all p, n ≥ 0. The Artin-Rees lemma says that if A is Nötherian and M finitey
generated, then for some n > 0 we have equality for all p ≥ 0.
n
Definition 8.15. ΓI (M ) = {m ∈ M |I n m = 0 for some n > 0}.
Lemma 8.26. A Nötherian, M injective implies that ΓI (M ) is injective.
Proof. Let J ⊆ A be an ideal. ϕ : J → ΓI (M ) an A-homomorphism. If a ∈ J,
then I n ϕ(a) = 0.
J finitely generated implies that ∃p > 0 such that ϕ(I p J) = 0. By ArtinRees, there exists n > 0 such that I p (J ∩ I n A) = J ∩ I p+n A.
So we have ϕ(J ∩ I n+p ) = 0. So the map from J → A gives a map from
J/J ∩ I n+p → A/I n+p . As we have J/J ∩ I n+p → ΓI (M ) ⊆ M , and M
injective, we get a map A/I n+p → M . The image is contained in ΓI (M ), as
I n+p ψ(A/I n+p ) = 0, and so the inclusion really extends to a map to ΓI (M ).
Lemma 8.27. M an injective A-module, A Nötherian, f ∈ A. Then θ : M →
Mf is surjective.
Proof. Ann(f ) ⊆ Ann(f 2 ) ⊆ . . . ⊆ Ann(f r ) = Ann(f r+1 ) = . . .. Let x ∈ Mf ,
then x = θ(y)/f n for some y ∈ M . So we have a ses 0 → Ann(f r ) → A →
(f n+r ) → 0. Then A → M by f r y and (f n+r ) ⊆ A, so we get a map (f n+r ) →
M by f n+r 7→ f r y. This map can be extended to A, call the image of 1 in this
map z. Then f n+r z = f r y, then θ(z) = θ(y)/f n = x.
F a sheaf on X, σ ∈ F (U ). Supp(σ) = {p ∈ U |σp 6= 0 ∈ Fp } ⊆ U relatively
closed. For Z ⊆ X closed, ΓZ (U, F ) = {σ ∈ F (U )| Supp(σ) ⊆ Z}
104
Definition 8.16. Sheaf HZ0 (F ) by Γ(U, HZ0 (F )) = ΓZ (U, F ).
Let j : X \ Z → X, then 0 → HZ0 (F ) → F → j∗ (F |X\Z ), which becomes
short exact if F is flasque.
Assume X = Spec(A) and F = M̃ . For m ∈ M = Γ(X, M̃ ), then
Supp(m) = V (Ann(m)), p ∈ Supp(m) ⇐⇒ m/1 6= 0 ∈ Mp ⇐⇒ Ann(m) ⊆
P.
Lemma 8.28. X = Spec(A), Nötherian, Z = V (I) ⊆ X. Then HZ0 (M̃ ) =
Γ^
I (M )
Proof. 0 → HZ0 (F ) → F → j∗ (F |U \Z ) tells us that HZ0 is the kernel of a map
of quasi-coherent sheaves.
0
Enough to show that Γ(X, H
pZ (M̃ )) = ΓI√(M ), mn∈ ΓZ (X, M̃ ) ⇐⇒ Supp(m) ⊆
Z iff V (Ann(m)) ⊆ V (I), iff Ann(m) ⊇ I iff I ⊆ Ann(m) (by Nötherian
property), and this is iff I n m = 0, iff m ∈ ΓI (M ).
Proposition 8.29. A a Nötherian ring, X = Spec(A). Then M an injective
A-module implies that M̃ is a flasque OX -module.
Proof. Nötherian induction on Y = Supp M̃ . On Y = {point}, clear.
Let U ⊆ X open. Show Γ(X, M̃ ) → Γ(U, M̃ ) surjective. WLOG, Y ∩ U 6= ∅.
Choose f ∈ A such that Xf ⊆ U and Xf ∩ Y 6= ∅. Set Z = V (f ) = X \ Xf . Set
I = (f ) ⊆ A.
˜ )) ⊆ Y ∩ Z. (f n m = 0 for
We know that ΓI (M ) is injective, and Supp(ΓI (M
all m ∈ ΓI (M ), so f ∈
/ P implies ΓI (M )P = 0)
Induction implies that Γ^
I (M ) is flasque, and M = Γ(X, M̃ ) → Γ(U, M̃ ) →
˜ )) = ΓZ (X, M̃ ) → ΓZ (U, M̃ ) = Γ(U, ΓI (M
˜ )).
Γ(Xf , M̃ ) = Mf gives us Γ(X, ΓI (M
Recall that M → Mf is surjective.
Let σ ∈ Γ(U, M̃ ). Then ∃τ 0 ∈ Γ(X, M̃ ) with the same image in Γ(Xf , M̃ ).
Set τ = τ 0 |U . Then (σ − τ )|Xf = 0, so σ − τ ∈ ΓZ (U, M̃ ).
So ∃α0 ∈ ΓZ (X, M̃ ) with α0 7→ σ − τ . Therefore, α0 + τ 0 ∈ Γ(X, M̃ ) maps to
σ ∈ Γ(U, M̃ ).
Theorem 8.30. X affine Nötherian scheme, F a quasi-coherent OX -module,
then H n (X, F ) = 0 for all n > 0.
Proof. X = Spec(A), F = M̃ . A resolution of M by injective A modules gives
us a resolution of F by flasque OX -modules.
The global section functor gives back the original sequence, so H n (X, M̃ ) =
n ·
H (I ) which is M if n = 0 and 0 otherwise.
Corollary 8.31. X Nötherian scheme, F a quasi-coherent OX -module, then
there exists a monomorphism 0 → F → G where G is a flasque quasicoherent
OX -module.
105
Proof. X = U1 ∪ U2 ∪ . . . ∪ Un where Ui = Spec Ai where Ai is Nötherian.
F |Ui = M̃i , Mi an Ai -module. There exists Mi ⊆ Ii for some injective Ii .
FUi → I˜i injective, so F → j∗ (I˜i ) is injective over Ui . As Ii is injective, I˜i
is flasque, and j∗ (I˜i ) is flasque.
0 → F → ⊕ni=1 j∗ (I˜i ) is flasque and quasicoherent.
Exercise: X be a scheme, f1 , . . . , fr ∈ Γ(X, OX ), then if Xfi is affine for all
i, and (f1 , . . . , fr ) = (1) ⊆ Γ(X, OX ), then X is affine.
The ideas is that A = Γ(X, OX ), and we have ϕ : X → Spec(A). Then show
that Γ(Xfi , OX ) = Afi , and so we get isomorphisms ϕ : Xfi → Spec(Afi ). So
X = ∪Xfi and so ϕ is a global isomorphism.
Theorem 8.32 (Serre’s Criterion). Let X be a Nötherian scheme. TFAE
1. X is affine
2. H n (X, F ) = 0 for all quasicoherent OX -modules F and n > 0.
3. H 1 (X, I ) = 0 for all coherent sheaves of ideals I ⊆ OX .
Proof. 1 implies 2 implies 3 are done already.
Assume 3. Let P ∈ X be a closed point. Let P ∈ U be an open affine
neighborhood of P . Let Y = X \ U . We get 0 → IY ∪P → IY → k(P ) → 0,
Γ(X, IY ) → Γ(X, k(P )) → H 1 (X, IY ∪P ) = 0.
∃f ∈ Γ(X, IY ) ⊆ A = Γ(X, OX ) such that Y ⊆ V (f ) and f 7→ 1 ∈ k(P ),
so P ∈ Xf ⊆ U , Xf = Uf is affine. Choose f1 , . . . , fr ∈ A such that X =
Xf1 ∪ . . . ∪ Xfr and Xfi is affine. It remains to show that (f1 , . . . , fr ) = (1).
r
is a coherent subsheaf implies that H 1 (X, F ) = 0. For
Claim: F ⊆ OX
r−1
r
→ OX → 0 giving
r = 1 this is three. For r > 1, we have 0 → OX
→ OX
r−1
1
us 0 → F ∩ OX → F → I → 0, and so H (I ) = 0 and by induction,
r−1
H 1 (X, F ∩ OX
) = 0, so the claim is proved.
P
r
gi fi . This map is surjective, and
Take OX → OX → 0 by (g1 , . . . , gr ) 7→
so we take the kernel and call it F , and get a short exact sequence.
r
) → Γ(X, OX ) → H 1 (X, F ) = 0 exact so (f1 , . . . , fr ) = (1).
Γ(X, OX
Cech Cohomology
Let X be a topological space and X = ∪i∈I Ui . U = (Ui )i∈I , for i0 , . . . , ip ∈ I,
set Ui0 ,...,ip = Ui0 ∩ . . . ∩ Uip . Let F be an abelian sheaf on X.
Definition 8.17 (Alternating Function). A function α : I p+1 → {sections of
F } is called alternating if α(i0 , . . . , ip ) ∈ F (Ui0 ,...,ip ) and if α(i0 , . . . , it , it+1 , . . . , ip ) =
−α(i0 , . . . , it+1 , it , . . . , tp ) and if α(i1 , i1 , i2 , . . . , ip ) = 0.
Definition 8.18 (Cech Complex). C p (U, F ) = {alternating α : I p+1 → secPp+1
tions of F } and maps d : C p → C p+1 by dα(i0 , . . . , ip ) = k=0 (−1)k α(i0 , . . . , îk , . . . , ip+1 )|Ui0 ,...,ip+1 .
Check: d2 α = 0.
Then we have Ȟ p (U, F ) = H p (C ∗ (U, F )).
106
Lemma 8.33. Ȟ 0 (U, F ) = Γ(X, F ).
Q
Proof. C 0 (U, F ) = i∈I F (Ui ). If α ∈ C 0 (U, F ), then dα(i, j) = α(j)|Ui,j −
α(i)|Ui,j , so dα = 0 iff the sections are all compatible which is true iff ∃σ ∈ F (X)
such that σ|Ui = α(i).
Therefore, Ȟ 0 (U, F ) = ker(C 0 → C 1 ) = F (X).
Example: X = P1 = Proj k[x, y]. F = ΩP1 = Ω. U = {U, V } with U =
D+ (x) = Spec k[t], t = y/x and V = D+ (y) = k[t−1 ]. U ∩ V = Spec k[t, t−1 ].
Then we denote C p = C p (U, Ω).
C 0 = Γ(U, Ω) ⊕ Γ(V, Ω) = k[t]dt ⊕ k[t−1 ]d(t−1 ), C 1 = Γ(U ∩ V, Ω) =
k[t, t−1 ]dt. C 2 = 0.
d : C 0 → C 1 by dt 7→ −dt and d(t−1 ) 7→ −t−2 dt. f (t)dt ⊕ g(t−1 )d(t−1 ) ∈
ker d iff −f (t) − t−2 g(t−1 ) = 0 iff f = g = 0, so Ȟ 0 (U, Ω) =P
Ω(X) = 0.
Im(d) = {(−f (t) − t−2 g(t−1 ))dt} ⊆ k[t, t−1 ]dt. This is { ai ti |a−1 = 0}, so
Ȟ 1 (U, Ω) ' k generated by t−1 dt.
Let X = ∪i∈I Ui , and U = (Ui )i∈I . Then we have defined C p (U, F ).
Definition 8.19. Define an abelian sheaf C p (U, F ) = C p given by Γ(V, C p (U, F )) =
C p (U, j∗ (F |V )) where j is the inclusion V → X.
d
0 → F → C 0 → C 1 → . . .. σ ∈ F (V ), then σ ∈ C 0 (U, j∗ F |V ) and
σ(i) = σ|Ui ∩V .
Lemma 8.34. This is an exact sequence of sheaves.
Proof. Exact at p = 0. 0 → Γ(X, j∗ (F |V )) → C 0 → C 1 → . . .. Let p ≥ 1.
x ∈ X, choose k ∈ I with x ∈ Uk . Let αx ∈ Cxp . Then α ∈ C p (V ) for x ∈ V .
WLOG, V ⊆ Uk .
Define hα ∈ C p−1 by hα(i0 , . . . , ip−1 ) = α(k, i0 , . . . , ip−1 ) ∈ F (Uk,i0 ,...,ip−1 ∩
V ) = F (Ui0 ,...,ip ∩ V ).
Now define h : Cxp → Cxp−1 by αx 7→ (hα)x . Check that (dh + hd)α = α, so
dh + hd = id − 0 so id ∼ 0 and so H p (Cx∗ ) = 0, and so C ∗ is exact.
Proposition 8.35. If F is flasque, then Ȟ p (U, F ) = 0 for p > 0.
Proof. Let C p = C p (U, F ). This is flasque for all p. So 0 → F → C 0 →
C 1 → . . . is an exact sequence of flasque sheaves. Taking the global section
functo we get C p , but it is still exact because it is a flasque sequence. So
Ȟ p (U, F ) = H p (C ∗ ) = 0.
Note: F is an abelian sheaf, 0 → F → I 0 → . . . an inejctive resolution,
then we get a map Ȟ p (U, F ) → H p (X, F ).
Theorem 8.36. If X is Nötherian separated scheme, and U = (Ui )i∈I is an
open affine covering, and F is quasi-coherent OX -module, then Ȟ p (U, F ) =
H p (X, F ).
107
Proof. 0 → F → G → R → 0 with G flasque and quasicoherent, then R is quasicoherent. If X is seprated, then Ui0 ,...,ip is affine, so we get 0 → F (Ui0 ,...,ip ) →
G (Ui0 ,...,ip ) → R(Ui0 ,...,ip ) → 0.
So we get the long exact sequence 0 → Ȟ 0 (F ) → Ȟ 0 (G ) → Ȟ 0 (R) →
Ȟ 1 (F ) → 0 and an exact sequence H 0 (F ) → H 0 (G ) → H 0 (R) → H 1 (F ) → 0,
and so taking the vertical maps, for p = 1, this proves it.
We can proceed by induction obtaining 0 → Ȟ p (R) → Ȟ p+1 (F ) → 0 and
0 → H p (R) → H p (F ) → 0 and looking at the vertical maps, whcih must be
isomorphisms.
Let A be a Nötherian ring, S = A[x0 , . . . , xr ], and X = Proj S = PrA .
Then let F be any OX module. Recall that Γ∗ (F ) = ⊕n∈Z Γ(X, F (n)).
Note: f ∈ Sm = Γ(X, O(m)) gives f : F (n) → F (n + m) by σ 7→ σ ⊗ f .
Thus, f : H p (X, F (n)) → H p (X, F (n + m)), so Γ∗ (F ) is a graded Smodule.
Theorem 8.37.
1. S → Γ∗ (OX ) is an isomorphism of graded S-modules.
2. H p (X, OX (n)) = 0 for 0 < p < r any n.
3. H r (X, OX (−r − 1)) ' A
4. Perfect Pairing: H 0 (X, OX (n))×H r (X, OX (−n−r−1)) → H r (X, OX (−r−
1)) = A.
Proof. F = ⊕n∈Z OX (n) is quasicoherent. Then H p (X, F ) = ⊕n∈Z H p (X, OX (n)).
Let U = {U0 , . . . , Ur } with Ui = D+ (xi ) ⊆ X. Then Ui0 ,...,ip = D+ (xi0 . . . xip ),
so we look at Γ(Ui0 ,...,ip , OX (n)) = S(n)(xi0 ...xip ) , and so Γ(Ui0 ,...,ip , F ) =
Sxi0 ...xip .
C ∗ (U, F ): ⊕i0 Sxi0 → ⊕i0 <i1 Sxi0 xi1 → . . . → ⊕rk=0 Sx0 ...x̂k ...xr → Sx0 ...xr
1. ker(first map) = ∩Sxi = S.
Q
Q
2. C p (U, F )xr = i0 <...<ip Sxi0 ...xip xir = i0 <...<ip ⊕n∈Z Γ(Ui0 ,...,ip ∩Ur , OX (n)).
Q
This equals Γ(Ui0 ,...,ip ∩ Ur , F ) = C p (U 0 , F |Ur ) where U 0 = {Ui ∩ Ur }
is an affine open covering of Ur . Thus, Ȟ p (U, F )xr = Ȟ p (U 0 , F |Ur ) =
H p (Ur , F |Ur ) = 0. So every element of H p (X, F ) is killed by xN
r . It is
xr
p
p
enough to show that H (X, F ) → H (X, F ) is an isomorphism. WLOG,
r ≥ 2.
H = V (xr ) ⊆ X, H ' Pr−1
A . We have 0 → OX (−1) → OX → OH → 0.
As F is locally free, tensoring with it is flat, so we get 0 → F (−1) → F →
FH → 0, and FH = ⊕n∈Z OH (n), with the first map being multiplication
by xr . By induction on r, we get that H p (X, FH ) = H p (H, FH ) = 0 for
0 < p < r − 1.
So we get a long exact sequence H 0 (X, F ) → H 0 (X, FH ) → H 1 (X, F (−1)) →
H 1 (X, F ) → H 1 (X, FH ), so we get that H 1 (X, F ) ' H 1 (X, F (−1)).
108
For 1 < p < r − 1, we have H p−1 (X, FH ) = 0 → H p (X, F (−1)) '
H p (X, F ) → H p (X, FH ) = 0.
x
0
So we now have H r−2 (X, FH ) → H r−1 (X, F (−1)) →r H r−1 (X, F ) →
δ
x
H r−1 (X, FH ) → H r (X, F (−1)) →r H r (X, F ) → 0. Note that if r = 2,
we don’t get H r−2 (X, FH ) = 0, but the first map is zero.
`
r−1 −1
xr |`i < 0} and H r−1 (X, FH )
Ann(xr ) ⊆ H r (X, F (−1)) has basis {x`00 . . . xr−1
`r−1
`0
has basis {x0 . . . xr−1 |`i < 0}. T his tells us that δ is injective, and so
the map before it is zero and we have an isomorphism H r−1 (X, F (−1)) =
H r−1 (X, F ).
3. Image(last map)= spanA {x`00 . . . x`rr |`i ∈ Z with some `i ≥ 0}, so we get
H r (X, F ) = spanA {x0`0 . . . x`rr |`i < 0∀i}. And so H r (X, OX (−r − 1)) =
A(x0 x1 . . . xr )−1 ' A.
P
4. H r (X, OX (−n−r −1)) = spanA {x`00 . . . x`rr | `i = −n−r −1, `i < 0}. So
look at H 0 (X, OX (n)) × H r (X, OX (−n − r − 1)) → H r (X, OX (−r − 1)).
`0
`0 +m0
mr
`r
0
map (xm
. . . xr`r +mr .
0 . . . xr ) × (x0 . . . xr ) 7→ x0
The right hand side 6= 0 ∈ H r (X, OX (−r − 1)) iff mi + `i = −1 for all i,
so `i = −mi − 1 for all i.
Theorem 8.38. X a projective scheme over Spec A, A Nötherian and OX (1) is
a very ample invertible sheaf relative to Spec(A). If F is a coherent OX -module,
then
1. H p (X, F ) is a finitely generated A-module for all p ≥ 0.
2. H p (X, F (n)) = 0 for all p > 0 for n >> 0.
Proof. Let f : X → Pr , OX (1) = f ∗ OPr (1). f∗ F is coherent on Pr . f∗ (F (n)) =
f∗ (f ∗ OPr (n) ⊗ F ) = OPr (n) ⊗ f∗ F = (f∗ F )(n).
H p (X, F (n)) = H p (Pr , f∗ F (n)) = H p (Pr , (f∗ F )(n)).
WLOG, X = Pr . Note: Theorem is true for F = OPr (n).
The theorem is true for p > r, as all the cohomology vanishes. We proceed
by decreasing induction on p. There exists E = ⊕OPr (qi ) with finitely many
terms with 0 → R → E → F → 0 exact, so 0 → R(n) → E (n) → F (n) → 0
exact.
Thus, H p (Pr , E (n)) → H p (Pr , F (n)) → H p+1 (Pr , R(n)). The outer modules are finitely generated by the previous theorem and inductions. As A is
Nötherian, the middle module is finitely generated.
If the outer modules are zero, then H p (Pr , F (n)) = 0 for all n >> 0,
p > 0.
Application: if X is a nonsingular curve, defined genus to be g = dimk Γ(X, ΩX ) <
∞. With Serre duality (to be shown), we see that g = dimk H 1 (X, OX )
109
Euler Characteristic
Let X be a projective scheme over k. Let F be a coherent OX -module and
dim(X) = r.
Definition
8.20 (Euler Characteristic). The Euler characteristic of F is χ(F ) =
P
p
p
(−1)
dim
k H (X, F ).
p≥0
Then 0 → F 0 → F → F 00 → 0 gives a long exact sequence 0 → H 0 (X, F 0 ) →
H 0 (X, F ) → . . . → H r (X, F ) → H r (X, F 00 ) → 0 an exact sequence of k-vector
spaces.
SoP
χ(F ) = χ(F 0 )+χ(F 00 ). In general, if we have 0 → F 0 → . . . → F ` = 0,
then (−1)i χ(F i ) = 0.
Proposition 8.39. X a proper scheme over Nötherian Spec(A). L an invertible OX -module. Then L is ample iff ∀ coherent OX -modules F , we have
H p (X, F ⊗ L n ) = 0 for all p > 0, n >> 0.
Proof. ⇒: L ample, then L ⊗m is very ample relative to Spec A for some
m > 0. Thus, there exists an immersion X ⊆ X 0 ⊆ PrA with L ⊗m = OX (1).
X is proper, so X = X 0 is closed in PrA . Choose n0 > 0 such that H p (X, F ⊗
L i ⊗ OX (n)) = 0 for all p > 0 and for 0 ≤ i < m and n ≥ n0 . Thus,
H p (X, F ⊗ L n ) = 0 for n ≥ m + mn0 .
⇐: F a coherent OX -module. Show: F ⊗ L n generated by global sections
for all n >> 0. Let p ∈ X be a closed point. Then we get 0 → Ip → OX →
k(P ) → 0 ses, so we have Ip ⊗ F → F → F ⊗ k(p) → 0, and to get a short
exact sequence, we take 0 → Ip F → F → F ⊗ k(P ) → 0.
So then we have 0 → Ip F ⊗ L n → F ⊗ L n → F ⊗ L n ⊗ k(p) → 0. For
n ≥ n0 = n0 (p), we get Γ(X, F ⊗L n ) → Γ(X, F ⊗L n ⊗k(P )) → H 1 (X, Ip F ⊗
L n ) = 0, so the second term is (F ⊗ L n )p ⊗ k(p). By Nakayama, (F ⊗ L n )p
is generated by global sections from Γ(X, F ⊗ L n ). p ∈ U = Spec(B) ⊆ X, B
Nötherian.
So M = Γ(U, F ⊗ L n ) is a finitely generated B-module. M 0 ⊆ M is the
submodule generated by the imeage of Γ(X, F ⊗ L n ). Mp = Fp = Mp0 . Then
there exists f ∈ B \ p suc that Mf0 = Mf .
Set Up,n = D(f ) ⊆ U , then p ∈ Up,n and (F ⊗ L n )|Up,n is generated by
global sections.
Special case: may choose n1 and open Vp 3 p such that L n1 |Vp is generated
by global sections. Set Up = Vp ∩ Up,n0 ∩ . . . ∩ Up,n0 +n1 −1 . Then if n ≥ n0 (p),
0
we have F ⊗ L n = (F ⊗ L n ) ⊗ (L n1 )m where n0 ≤ n0 < n0 + n1 with m ≥ 0.
Then (F ⊗ L n )|Up is generated by global sections from Γ(X, F ⊗ L n ).
So X = Up1 ∪ Up2 ∪ . . . ∪ Up` with n ≥ max(n0 (p1 ), . . . , n0 (p` )), and we get
F ⊗ L n generated by global sections.
Applications: A Nötherian, X = PrA = Proj(S) for S = A[x0 , . . . , xr ]. Let
M be a finitely generated graded S-module. We have maps Mn → Γ(Pr , M̃ (n)),
a homomorphism of graded S-modules ϕ : M → Γ∗ (M̃ ) = ⊕n∈Z Γ(M̃ (n)).
Claim: Mn ' Γ(Pr , M̃ (n)) for all n >> 0
110
Injective: N = ker ϕ ⊆ M . Show that Nn = 0 for n >> 0. As N is finitely
generated, it is enough to show that ∀m ∈ N and 0 ≤ i ≤ r, there exists p > 0
so that xpi m = 0. This is true because m = 0 in Γ(D+ (xi ), M̃ (n)) = M (n)(xi ) ⊆
Mx i .
Surjective: 0 → F 0 → F → M → 0, F = ⊕S(qi ) finite sum. So then
0 → F̃ 0 (n) → F̃ (n) → M̃ (n) → 0, and so for n >> 0, we get Γ(Pr , F̃ (n)) →
Γ(Pr , M̃ (n)) → H 1 (Pr , F̃ 0 (n)) = 0, so we have a surjection, but this is really an
onto map Fn → Mn , and must also give a surejction Mn → Γ(Pr , M̃ (n)).
X is projective over a field, k. X ⊆ Prk . Then there is OX (1) very ample.
F is a coherent OX -module.
Definition 8.21 (Hilbert Polynomial). We define the hilbert polynomial to be
PF (n) = χ(F (n)). In particular, PX (n) = POX (n).
Claim: PF (n) ∈ Q[n].
Proof. Nötherian induction on Y = Supp F . Y = ∅ ⇒ F = 0 ⇒ PF (n) = 0.
f·
If Y 6= ∅, take f ∈ Γ(X, OX (1)) such that Y ∩ Xf 6= 0. 0 → R → F (−1) →
F → M → 0 is an exact sequence of coherent OX -modules.
Taking the Euler characteristics at n, we get PF (n) − PF (n − 1) = PM (n) −
PR (n). The closures of Supp R and Supp M ⊆ Y ∩ V (f ) ( Y . By Nötherian
induction, the functions on the right hand side are polynomials in Q[n], and
therefore PF (n) ∈ Q[n].
Note Pr = Proj S, S = k[x0 , . . . , xr ], M is a finitely generated graded Smodule. If n >> 0, then Mn = Γ(Pr , M̃ (n)) = χ(M̃ (n)).
Note: PF (n) depends on OX (1) on OX (1), but PF (0) = χ(F ). Assume
dim X = r. Then the arithmetic genus if pa (X) = (−1)r (χ(OX ) − 1) =
(−1)r (PX (0) − 1), so if X is a connected curve, then pa (X) = 1 − χ(OX ) =
1 − dimk H 0 (X, OX ) + dimk (H 1 (X, OX )). So then, as X is projective, this is
dimk H 1 (X, OX ).
Ext Groups and Sheaves
(X, OX ) a ringed space, F , G are OX -modules. Remember that hom(F , −)
is a functor from M od(X) → Ab is left exact.
Definition 8.22 (Ext). Extp (F , −) = Rp hom(F , −) from M od(X) → Ab.
E xtp (F , −) = Rp H om(F , −).
Proposition 8.40.
1. E xt0 (OX , G ) = G , E xtp (OX , G ) = 0 for p > 0.
2. Extp (OX , G ) = H p (X, G )
Lemma 8.41. I ∈ M od(X) injective, U ⊆ X open, then I |U ∈ M od(U )
injective.
Proof. j : U → X inclusion.
111
j∗ F ...................................................... j! G
...
...
...
...
...
...
...
...
...
..
.........
.
...
...
...
...
...
...
...
...
...
..
.........
.
............................
j∗ (I |U ) ................⊂
I
Proposition 8.42. U ⊆ X open. Then E xtpX (F , G )|U = E xtpU (F |U , G |U ).
Note: 0 → G 0 → G → G 00 → 0 gives a long exact sequence on Ext(F , −)
Proposition 8.43. 0 → F 0 → F → F 00 → 0 gives a long exact sequence
0 → hom(F 00 , G ) → hom(F , G ) → hom(F 00 , G ) → Ext1 (F 00 , G ) → . . ..
Proof. 0 → G → I 0 → . . .. Then as hom(−, I P ) is exact and contravariant,
we get 0 → hom(F 00 , I ∗ ) → hom(F , I ∗ ) → hom(F 0 , I ∗ ) → 0, and so we get
a long exact sequence in the first variable.
Definition 8.23 (Locally Free Resolution). A locally free resolution of F is a
resolution . . . → E2 → E1 → E0 → F → 0 exact with Ep locally free of finite
rank.
Proposition 8.44. If E∗ → F → 0 is locally free resolution, then E xtp (F , G ) =
H p (H om(E∗ , G )).
Proof. Both sides are δ-functors. RHS: 0 → G 0 → G → G 00 → 0, Ep is locally
free implies that H om(Ep , −) is exact, and so we get 0 → H om(E∗ , G 00 ) →
H om(E∗ , G ) → H om(E∗ , G 0 ) → 0, so get long exact sequence.
Agree for p = 0: E1 → E0 → F → 0, so 0 → H om(F , G ) → H om(E0 , G ) →
H om(E1 , G ) → . . . is exact, and so H 0 (H om(E∗ , G )) = H om(F , H ) =
E xt0 (F , G ).
Universality: Both sides vanish when G is injective.
Note: E is locally free OX -module of finite rank, then E ∨ = H om(E , OX )is
locally free.
Exercise: H om(F , G ) ⊗ E ' H om(F , E ⊗ G ) ' H om(F ⊗ E ∨ , G ) by
ϕ ⊗ e 7→ [f 7→ e ⊗ ϕ(f )] and ψ 7→ [f ⊗ e∨ 7→ (e∨ ⊗ 1)(ψ(f ))]
Special Case: E = H om(O, E ) = H om(E ∨ , OX ) = E ∨∨ .
Lemma 8.45. I is an injective OX -module, E is a locally free OX -module,
then I ⊗ E is injective.
Proof. hom(−, I ⊗ E ) = hom(− ⊗ E ∨ , I ) is exact.
Proposition 8.46. E locally free of finite rank, then Extp (F , E ⊗G ) = Extp (F ⊗
E ∨ , G ) and E xtp (F , G ) ⊗ E = E xtp (F , E ⊗ G ) = E xtp (F ⊗ E ∨ , G ).
Proof. Everything is a δ-functor. They are the same for p = 0.
They vanish for G injective.
112
Note: If A is a ring, M an A-module, then ExtpA (M, −) = Rp homA (M, −).
({pt}, A) is a ringed space, and modules on it and A-mod are the same thing.
If E is locally free of finite rank, then H om(E , F )x = homOX,x (Ex , Fx ).
Proposition 8.47. X a Nötherian Scheme, F a coherent OX -module and G
any OX -module. Then E xtp (F , G )x = ExtpOX,x (Fx , Gx ) for all x ∈ X, p ≥ 0.
Proof. WLOG, X is affine. There exists a locally free resolution E∗ → F → 0,
so there exists a free resolution of OX,x -modules E∗,x → Fx → 0.
E xtp (F , G )x = H p (H om(E∗ , G ))x = H p (H om(E∗ , G )x ) = H p (hom(E∗,x , Gx )) =
ExtpOX,x (Fx , Gx ).
Proposition 8.48. X a projective scheme over Nötherian Spec(A) with OX (1)
is very ample relative to A. F , G coherent OX -modules, p ≥ 0. Then Extp (F , G (n)) =
Γ(X, E xtp (F , G (n))) for all n >> 0.
Proof. p = 0: hom(F , G (n)) = Γ(X, H om(F , G (n))) is true for all n.
F is lcoally free of finite rank. Extp (F , G (n)) = Extp (OX , F ∨ ⊗ G (n)) =
p
H (X, F ∨ ⊗ G (n)) = 0. E xtp (F , G (n)) = E xtp (OX , F ∨ ⊗ G (n)) = 0 for all
p > 0 for all n, so it is ture for all n >> 0.
F coherent: 0 → R → E → F → 0 where E is locally free. 0 →
H om(F , G (n)) → H om(E , G (n)) → H om(R, G (n)) → E xt1 (F , G (n)) →
E xt1 (E , G (n)) = 0 and we can take the twisting outside.
So for n >> 0, we have 0 → hom(F , G (n)) → hom(E , G (n)) → hom(R, G (n)) →
Γ(X, E xt1 (F , G (n))) → 0.
We also get 0 → hom(F , G (n)) → hom(E , G (n)) → hom(R, G (n)) →
Ext1 (F , G (n)) → Ext1 (E , G (n)) = 0 for n >> 0. As the first three are the
same, the last one must be.
For p ≥ 2, we have long exact 0 → Extp−1 (R, G (n)) → Extp (F , G (n)) → 0
for n >> 0. Also, for all n, 0 → E xtp−1 (R, G (n)) → E xtp (F , G (n)) → 0.
By induction, we get vertical maps, which prove the theorem.
Vp
Let A be a ring, M an A-module,
M = M ⊗ . . . ⊗ M/ < . . . x ⊗ V
x . . . >.
n
0
→
M
→
N
→
R
→
0
a
ses
of
free
modules
of
ranks
m,
n,
r,
then
N '
Vm
Vr
M ⊗A
R is a natural isomorphism.
This
is
important
for
sheafification.
Vp
Vp
Let F be an OX -module,
F = (U 7→
F (U ))+ .
Then if 0 → M → N → R V
→ 0 a short
of locally free
Vm exactVsequence
n
r
OX -modules of ranks m, n, r, then
N '
M⊗
R.
Let X be a nonsingular variety over a field k. Then ΩX is locally free of
rank n.
Definition
8.24 (Canonical Sheaf of X). The canonical sheaf on X is ωX =
Vn
ΩX .
V
On Pnk , we have 0 → ΩPn → O(−1)⊕n+1 → OPn → 0. So ωPn = ΩPn =
V1
Vn+1
Vn
ΩP n ⊗
OPn =
(O(−1)⊕n+1 ) = O(−1)⊗n+1 = O(−n − 1).
Theorem 8.49 (Serre Duality, Version I). X = Pnk , k a field.
113
1. H n (X, ωX ) ' k.
2. F coherent OX -module implies that hom(F , ωX )×H n (X, F ) → H n (X, ωX ) =
k is a perfect pairing.
3. Extp (F , ωX ) ' H n−p (X, F )∗ , a natural isomorphism of functors in F
from Coherent Sheaves on X to k-modules.
(If V is a k-vector space, V ∗ = homk (V, H n (X, ωX )))
Proof.
1. H n (X, ωX ) = H n (X, O(−n − 1)) ' k.
2. Let F = O(q). Then hom(F , ωX ) × H n (X, F ) = Γ(X, O(−n − 1 − q)) ×
H n (X, O(q)) → H n (X, O(−n − 1)) = k is a perfect pairing. So now
consider the case where F is coherent. Then ∃E1 → E0 → F → 0 exact,
Ei = ⊕O(qij ) a finite sum. Note that H n (X, −) is right exact (end of the
long exact sequence). So we get H n (X, E1 ) → H n (X, E0 ) → H n (X, F ) →
0, which gives us 0 → H n (X, F )∗ → H n (X, E0 )∗ → H n (X, E1 )∗ . We also
have 0 → hom(F , ωX ) → hom(E0 , ωX ) → hom(E1 , ωX ), with the last two
on each being equal. Thus, hom(F , ωX ) = H n (X, F )∗ .
3. Both sides are contravariant δ-functors from Coh(X) → M od(k), ie δfunctors Coh(X)op → M od(k). They agree for p = 0. It remains to show
that they are universal: we will show that both sides are erasable. That is,
for all coherent F , there exists an epi u : E → F such that both functors
vanish on u when p > 0.
So there is an epi E → F when E is a finite ⊕O(qi ). WLOG, qi <
0 as O(q − 1)⊕n+1 → O(q). So Extp (E , ωX ) = ⊕ Extp (O(qi ), O(−n −
1)) = H p (X, OX (−n − 1 − qi )) = 0 for p > 0. Also, H n−p (X, E ) =
⊕H n−p (X, O(qi )) = 0 for p > 0.
Definition 8.25 (Dualizing Sheaf). Let X be a proper scheme of dimension n
◦
together
over a field k. A dualizing sheaf for X is a coherent OX -module ωX
n
◦
with a k-linear trace map t : H (X, ωX ) → k such that forall coherent F we
t
◦
◦
get perfect pairings hom(F , ωX
) × H n (X, F ) → H n (X, ωX
) → k.
Proposition 8.50. Let (ω, t) and (ω 0 , t0 ) be dualizing sheaves for X. Then there
exists a unique isomorphism ϕ : ω → ω 0 such that t = t0 ◦ ϕ∗ : H n (X, ω) →
H n (X, ω 0 ) → k.
t0
Proof. The perfect pairing requirement gives hom(ω, ω 0 )×H n (X, ω) → H n (X, ω 0 ) →
k, and so t : H n (X, ω) → k is given by some ϕ ∈ hom(ω, ω 0 ). That is, there
exists a unique ϕ : ω → ω 0 with t = t0 ◦ ϕ∗ . By symmetry, there exists a unique
ψ : ω 0 → ω such that t0 = t ◦ ψ∗ . ψϕ : ω → ω satisfies t = t ◦ (ψϕ)∗ , and the
identity does this, and so ψϕ = id and similarly for ϕψ.
Assume that X is projective over k, choose a closed embedding X ⊆ PN
k .
Set r = codim(X, PN ).
114
∗
Definition 8.26. ωX
= E xtrp (OX ωPN ).
Lemma 8.51. E xtiP (OX , ωP ) = 0 for i < r.
Proof. E xtiP (OX , ωP ) is coherent as an OX -module. Thus, E xtiP (OX , ωP )(q) is
generated by global sections for all q >> 0.
Γ(P, E xtiP (OX (−q), ωP )) ' H N −i (P, OX (−q)) = ExtiP (OX (−q), ωP ). This
is zero if N − i > dim X iff i < r.
Note: Assume 0 → M 0 → M → M 00 → 0 ses. Then M 0 injective implies
split exact, so hom(M, M 0 ) → hom(M 0 , M 0 ) surjective. If M 0 , M injective, then
M 00 injective. And 0 → M 0 → M 1 → . . . → M r → 0 with M i injective for
i < r, then M r injective.
◦
Lemma 8.52. There is a natural isomorphism homX (−, ωX
) ' ExtrP (−, ωP )
of functors from M od(X) → M od(k).
Proof. 0 → ωP → I ∗ an injective resolution in M od(P ). Set J m = H omP (OX , I m ) ⊆
I m the subsheaf of sections killed by the ideal sheaf of X.
Note: F an OX -module, ϕ : F → I m an OP -hom, then Im(F ) ⊆ J m .
Thus, J m is an injective OX -module. So homX (−, J m ) = homP (−, I m )
is exact.
H i (J ∗ ) = E xtiP (OX , ωP ) = 0 for i < r. So we have exact sequence 0 →
0
J → . . . → J r−1 → J r , and we get J r−1 → J1r → 0 with J1r =
Im d(dr−1 ) ⊆ J r . The Note implies that J1r is injective, so we get 0 →
J1r → J r → J2r → 0 with J r = J1r ⊕ J2r , so all are injective.
So the exact complex can be called J1∗ and J2r → J r+1 → . . . can be
called J2∗ .
◦
ωX
= E xtrP (OX , ωP ) = H r (J ∗ ) = ker(J2r → J r+1 ), so we get 0 →
◦
◦
)→
ωX → J2r → J r+1 which gives, for F an OX -module, 0 → homX (F , ωX
r
r
r
r+1
homX (F , J2 ) → homX (F , J
), so ExtP (F , ωP ) = H (homP (F , I ∗ )) =
◦
).
H r (homX (F , J ∗ )) = H r (homX (F , J2∗ )) = homX (F , ωX
◦
is a dualizing sheaf for X.
Proposition 8.53. ωX
◦
)×
Proof. F is any coherent OX -module. n = N − r = dim(X). homX (F , ωX
r
n
N −r
N
H (X, F ) = ExtP (F , ωP ) × H
(P, F ) → H (P, ωP ) ' k is a perfect pairing, by Serre Duality for PN .
◦
◦
◦
◦
Take F = ωX
. Then homX (ωX
, ωX
)×H N (X, ωX
) → H N (P, ωP ) = k is per◦
◦
◦
) → H N (P, ωP ),
fect pairing. id ∈ homX (ωX , ωX ) corresponds to t : H n (X, ωX
the trace map. Because everything has been natural, we can use t to factor the
perfect pairing.
Let X be a proper scheme over k of dimension n.
Definition 8.27 (Dualizing Sheaf). A dualizing sheaf for X is a coherent OX ◦
◦
module ωX
with a k-linear trace map t : H n (X, ωX
) → k such that for all
◦
coherent OX -mod F , we have a perfect pairing homX (F , ωX
) × H n (F ) →
t
◦
H n (X, ωX
) → k.
115
VN
Thus, it must be unique. ωPN =
ΩPN is the dualizing sheaf for PN . IF
N
◦
X ⊆ P is a closed subscheme of codimension r, then ωX
= E xtrPN (OX , ωPN )
dualizing for X.
◦
Notice: A choice of trace map t gives a natural isomorphism θ◦ : homX (−, ωX
)→
n
∗
op
H (X, −) of functors Coh(X) → M od(k).
◦
Corollary 8.54. ∃ natural transformations θi : ExtiX (−, ωX
) → H n−i (X, −)∗ .
Proof. We must show that both sides are δ-functors that agree for i = 0 and
◦
that ExtiX (−, ωX
) is erasable in Coh(X)op for i > 0.
Choose OX (1) very ample. F coherent implies that there exists a surjection
◦
◦ ⊕N
O(−q)⊕N → F for q >> 0. ExtiX (O(−q)⊕N , ωX
) = ExtiX (OX (−q), ωX
)
=
i
◦
⊕N
i
◦
⊕N
Ext (OX , ωX (q))
= H (X, ωX (q))
= 0 for q >> 0, so done.
Definition 8.28 (Regular Sequence). Let A be a local ring, m ⊆ A maximal
ideal, and f1 , . . . , fr ∈ m. Then f1 , . . . , fr is a regular seqeunce if each fi is a
nonzero divisor on A/(f1 , . . . , fi−1 ).
Definition 8.29 (Cohen-Macaulay Ring). A is Cohen-Macaulay (CM) if ∃ a
regular sequence f1 , . . . , fr ∈ m such that dim(A) = r.
Facts:
1. A regular local ring is CM
2. If A is CM then f1 , . . . , fr ∈ m is a regular sequence iff dim A/(f1 , . . . , fr ) =
dim A − r.
3. A is CM and f1 , . . . , fr ∈ m a regular sequence, then A/(f1 , . . . , fr ) is CM.
4. A CM, f1 , . . . , fr ∈ m a regular sequence, then I = (f1 , . . . , fr ) ⊆ A
implies that (A/I)[t1 , . . . , tr ] → grI (A) = ⊕n≥0 I n /I n+1 by ti 7→ f¯i ∈ I/I 2
is an isomorphism, so I/I 2 ' (A/I)⊕r .
Koszul Complex
Vp ⊕r
A a ring, f1 , . . . , fr ∈ A and A⊕r has basis e1 , . . . , er . Define Kp =
(A )
for 0 ≤ p ≤ r. Basis {ei1 ∧ . . . ∧ eip }.
1 → K0 by ei 7→ fi . dp : Kp → Kp−1 by dp (ei1 ∧ . . . ∧ eip ) =
Pp d1 : Kj−1
fij eij ∧ . . . ∧ êij ∧ . . . ∧ eip .
j=1 (−1)
So K∗ = K∗ (f1 , . . . , fr , A) is this complex.
Proposition 8.55. A a local ring, f1 , . . . , fr a regular sequence for A. Then
0 → K∗ → A/(f1 , . . . , fr ) → 0 is exact.
Definition 8.30 (Cohen-Macaulay). A scheme X is Cohen-Macaulay if OX,P
is CM for all P ∈ X.
Definition 8.31 (Local Complete Intersection). If Y is a nonsingular variety
over a field k and X ⊆ Y is a closed subscheme of codim r, and IX ⊆ OY
the ideal sheaf of X. Then X is a local complete intersection in Y if IX,P is
generated by r elements, ∀P ∈ X.
116
Note: X ⊆ Y a local complete intersection implies that X is Cohen-Macaulay
and IX /IX2 is a locally free OX -module of rank r, the ”Conormal Sheaf” of X
in Y .
Definition 8.32 (Normal Sheaf). NX/Y = (IX /IX2 )∨ is the Normal Sheaf of
X in Y .
Recall that X ⊆ Y nonsingular gives 0 → I /I 2 → ΩY ⊗ OX → ΩX → 0 is
exact, so dualizing we get 0 → TX → TY ⊗ OX → NX/Y → 0
N
◦
Theorem
Vr 8.56. X ⊆ Pk local complete intersection of codim r, then ωX '
ωPN ⊗
NX/Pn .
Proof. Let U = Spec(A) ⊆ PN open affine such that I(X ∩ U ) ⊆ A is generated
by f1 , . . . , fr ∈ A. Then let P ∈ U ∩ X. m = I(P ) ⊆ A is a maximal ideal.
Then Am = OPN ,P is regular local, so CM. dim Am /(f1 , . . . , fr ) = dim OX,P =
dim X = N − r. So f1 , . . . , fr is a regular sequence.
So we look at 0 → K∗ (f1 , . . . , fr , Am ) → Am /(f1 , . . . , fr ) → 0. Replace U
with some Uh , so WLOG, 0 → K∗ (f1 , . . . , fr ; A) → A/(f1 , . . . , fr ) → 0 is exact.
Thus, 0 → K̃∗ → OX∩U → 0 is locally free resolution on U .
◦
|U = E xtrPN (OX , ωPN ) = E xtrU (OX∩U , ωU ) = H r (H omU (K̃∗ , ωU )).
Therefore, ωX
⊕R
K̃∗ , ωU ) gives the sequence . . . → ωU
→ ωU → 0 by (σ1 , . . . , σr ) 7→
P H om(
j−1
(−1) fj σj .
◦
|U ' ωU /(f1 , . . . , fr )ωU = ωU ⊗ OX .
Thus, ωX
2
f1 ∧ f2 ∧ . . . ∧ fr is a basis of
Note:
f
,
.
1 . . , fr form a basis of IX /IX , then V
Vr
r
−1
2
(IX /IX ). Thus (f1 ∧ . . . ∧ fr ) is a basis of
(NX/Pn ).
◦
We now define γ : ωX |U = ωU /(f1 , . . . , fr )ωU → ωU ⊗ NX/PN by taking
σ̄ 7→ σ ⊗ (f1 ∧ . . . ∧ fr )−1 .
Claim: γ is independent of choices, and so defines a global isomorphism
◦
→ ωPn ⊗ NX/PN .
ωX
Let
g1 , . . . , gr ∈ I(X ∩ U ) ⊆ A be another set of generators.
P
P Write giV=p
cij fj with cij ∈ A. Define a map ϕ : A⊕r → A⊕r by ϕ(ei ) = cij ej . As
is a functor, we get a map on exterior powers.
So we get 0 → K∗ (g1 , . . . , gr ; A) → A/I → 0 and 0 → K∗ (f1 , . . . , fr ; A) →
A/I → 0, with vertical maps given by exterior powers of ϕ and the identity
from A/I
Vr to itself, with everything commuting.
As
ϕ = det(cij ) : Kr (g) → Kr (f ) and so det(cij ) : H omU (K̃r (f ), ωU ) →
H omU (K̃r (g), ωU ) gives a map ωU → ωU .
det(cij )
Tus we getVmap ωU (f1 , . . . , fr )ωU → ωU /(g1 , . . . , gr )ωU and maps from
r
each to ωU ⊗
NX/PN giving a commutative diagram.
◦
Corollary
8.57. If X is a nonsingular projective variety then ωX
= ωX =
Vn
ΩX for n = dim X.
Proof. X ⊆ PN
subvariety. So 0 → IX /IX2 → ΩPN ⊗ OX → ΩX → 0.
k closed
VN
Vr
Thus, ωPN ⊗ OX =
IX /IX2 ⊗ ωX .
P P N ⊗ OX ) =
V(Ω
r
NX/PN
Thus, ωX = ωPn ⊗
117
9
Curves
Let k = k̄ be an algebraically closed field.
Definition 9.1 (Curve). A curve is a complete connected non-singular variety
of dimension 1 over k.
Fact: All curves P
are projective.
P
Div(X) = {D = p∈X mp P } and deg(D) = mp .
P
L an invertible OX -module, s ∈ Γ(U, L ), U ⊆ X, then (s) = P ∈X vP (s)P
gives Pic(X) ' C`(X) by L 7→ (s), s ∈ Γ(U, L ). The inverse map
P takes D to
OX (D) by Γ(U, OX (D)) = {f ∈ k(X)|vP (f ) ≥ −mp } where D =
mP P
Fact: s ∈ k(X)∗ , then s ∈ Γ(U, OX ), deg(s) = 0. So deg : C`(X) → Z is
well-defined.
ωX = ΩX is the canonical sheaf.
Serre Duality: H i (X, L −1 ⊗ ωX ) = ExtiX (L , ωX ) = H 1−i (X, L )∗ .
Definition 9.2 (Genus). The geometric genus is pg (X) = dimk H 0 (X, ωX ).
The arithmetic genus is pa (X) = (−1)dim X (χ(OX ) − 1).
When X is a curve, pa (X) = dimk H 1 (X, OX ). By Serre Duality, we have:
Proposition 9.1. For a curve X, pa (X) = pg (X).
Proof. Set i = 0 then Serre Duality says that H 0 (X, L −1 ⊗ ωX ) = H 1 (X, L )∗ .
Then if L = OX , we get H 0 (X, ωX ) = H 1 (X, OX )∗ .
As ωX ∈ Pic(X), it corresponds to a divisor K = KX ∈ C`(X) the Canonical
Divisor, with ωX = OX (KX ).
If D ∈ Div(X), we can define `(D) = dimk H 0 (X, OX (D)). So, for example,
`(KX ) = g, the genus of X.
Lemma 9.2.
1. `(D) 6= 0 ⇒ deg(D) ≥ 0.
2. If `(D) 6= 0 and deg(D) = 0, then D ∼ 0 (ie, D = 0 in C`(X))
P
Proof.
1. ∃0 6= s ∈ Γ(X, OX (D)), and (s) = P ∈X vP (s)P and so deg(D) =
deg(s) ≥ 0.
2. If deg(D) = 0, then (s) = 0, so then D ∼ 0.
Theorem 9.3 (Riemann-Roch). Let D ∈ Div(X). Then
`(D) − `(K − D) = deg(D) + 1 − g
Proof. `(K − D) = dimk H 0 (X, OX (D)−1 ⊗ ωX ) = dimk H 1 (X, OX (D)). So
`(D) − `(K − D) = dimk H 0 (X, OX (D)) − dim H 1 (X, OX (D)) = χ(OX (D)).
If D = 0, then `(0) − `(K) = deg(0) + 1 − g, that is, 1 − g = 1 − g.
ETS that RR holds for D iff RR true for D + P (P a point).
118
Let k(P ) = OX /IP and note that IP = OX (−P ). So we get 0 →
OX (−P ) → OX → k(P ) → 0 ses.
Tensor with OX (D + P ) and get 0 → OX (D) → OX (D + P ) → k(P ) → 0
ses. As Euler Characteristic is additive, χ(OX (D + P )) = χ(OX (D)) + 1.
Examples:
1. g − 1 = `(K) − `(K − K) = deg(K) + 1 − g, so deg(K) = 2g − 2.
2. deg(D) > 2g − 2, then `(K − D) = 0, so `(D) = deg(D) + 1 − g.
3. Assume g(X) = 0. Then deg(K) = −2. Let P 6= Q ∈ X, D = P − Q ∈
Div(X). `(D) = `(D)−`(K−D) = 0+1−0 = 1. So then D ∼ 0, so P ∼ Q.
Thus, X = P1 from last semester because f ∈ k(X) with (f ) = P − Q
gives a map f : X → P1 , which turns out to be an isomorphism.
Fact: A morphism f : X → Y of curves is finite or constant.
Definition 9.3 (Degree of a Map). deg(f ) = [k(X) : k(Y )].
If P ∈ X and Q = f (P ) ∈ Y with Y a nonsingular curve, then mQ = (t) ⊆
OY,Q and f ∗ (t) ∈ OX,P .
Define eP = vP (f ∗ (t)).
This number is called the ramification index of f at P . If eP = 1, then f is
unramified at P . If eP > 1, then f is ramified at P . In this case, Q is a branch
point.
A ramification point is called tame if char(k) = 0 or if char(k) 6 |eP . If it is
not tame, then it is wild.
P
Recall: f ∗ : Div(Y ) → Div(X) by f ∗ (Q) = P ∈f −1 (Q) eP P . This map in
fact gives a map f ∗ : C`(Y ) → C`(X). This is becuase D ∈ Div(Y ), then
f ∗ (OY (D)) ' OX (f ∗ D)
Definition 9.4 (Separable Map). f : X → Y is separable if it is finite and
f ∗ : k(Y ) ⊆ k(X) is a separable extension of fields.
Proposition 9.4. f : X → Y is a separable map of curves, then 0 → f ∗ ΩY →
ΩX → ΩX/Y → 0 is ses.
Proof. We already know everything except that the map f ∗ ΩY → ΩX is injective. Note that these are both invertible OX -modules on an irreducible variety.
ETS that the map is nonzero.
Let P0 ∈ X be the generic point. It is enough to show that (ΩX/Y )P0 = 0.
This stalk is equal to Ωk[X]/k[Y ] localized at 0. Taking modules of differentials commutes with localization, do we get Ωk(X)/k(Y ) = 0 as the extension is
separable.
Let f : X → Y be finite morphism of curves. P ∈ X has image Q = f (P ) ∈
Y . mP = (u) ⊆ OX,P and similarly mQ = (t) ⊆ OY,Q . Then du ∈ ΩX,P is a
generator, and similarly, dt ∈ ΩY,Q is a generator. f ∗ (dt) ∈ (f ∗ ΩY )P ⊆ ΩX,P .
So ∃h ∈ OX,P with f ∗ (dt) = hdu. We will set the notation dt/du = h. This
just means that f ∗ (dt) = dt/du · du ∈ Ω(X, P ).
119
Proposition 9.5. Let f : X → Y be separable map of curves.
1. supp(ΩX/Y ) = {P ∈ X|eP > 1}, which is a finite set.
2. lengthOX,P (ΩX/Y,P ) = vP (dt/du).
3. If f is tamely ramified at P , then length(ΩX/Y,P ) = ep − 1. If f is wildly
ramified, then length(ΩX/Y,P ) > ep − 1.
Proof.
1. ΩX/Y,P = 0 iff ΩX,P generated by f ∗ (dt) iff f ∗ t generates mP ⊆
OX,P iff eP = 1.
2. (ΩX/Y )P ' Ω(X, P )/f ∗ ΩY,Q = ΩX,P /(f ∗ (dt)) ' OX,P /(dt/du).
3. Set e = eP and t = f ∗ (t). Then t = aue with a ∈ OX,P a unit. Then
dt = daue + aeue−1 du. If the ramification is tame, then e 6= 0 ∈ OX,P , so,
as da = hdu for some h, vP (dt/du) = e − 1.
If wild, then e = 0 ∈ OX,P , so dt = ue da, so vP (dt/du) = vP (ue h) >≥
e > e − 1.
Assume from now: f : X → Y separable finite morphism of curves.
P
Definition 9.5 (Ramification Divisor). R = P ∈X length(ΩX/Y,P )P ∈ Div(X).
Proposition 9.6. KX ∼ f ∗ KY + R.
Proof. s ∈ Γ(V, ΩY ), then KY = (s) ∈ Div(Y ). f ∗ (s) ∈ Γ(f −1 (V ), f ∗ ΩY ) ⊆
Γ(f −1 (V ), ΩX ). KX = (f ∗ s) ∈ Div(X).
Let (KX ; P ) denote the coefficient of P in KX . Then (KX ; P ) = lengthOX,P (ΩX,P /(f ∗ s)) =
length(f ∗ ΩY,P /(f ∗ s)) + length(ΩX/Y,P ) = (f ∗ KY ; P ) + (R; P ).
Theorem 9.7 (Hurwitz Theorem). 2g(X) − 2 = n(2g(Y ) − 2) + deg(R) where
n is deg(f ).
Example: f : X → Y any finite morphism of curves, then g(X) ≥ g(Y ).
Why? Because g(X) = g(Y ) + (n − 1)(g(Y ) − 1) + 21 deg(R).
The ramification divisor has even degree.
Theorem 9.8 (Lyroth’s Theorem). If k ⊆ L ⊆ k(t) and L is a field, then k = L
or L = k(u) for some u ∈ L.
Proof. k 6= L, then k has transcendence degree 1, so let X = CL . Then L → k(t)
gives a map f : P1 → X a finite map of curves. As g(P1 ) ≥ g(X) and g(P1 ) = 0,
g(X) = 0, so X ' P1 . Thus, L ' k(u) for some u ∈ L.
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