1
Affine Varieties and the Nullstellensatz
All rings in this document mean “Commutative rings with a multiplicative unit 1”.
Theorem 1.1. Let A be a ring. The following are equivalent:
1. Every ideal I ⊂ A is finitely generated; that is, for every ideal I ⊂ A, there exists f1 , ..., fk ∈
I so that I = (f1 , ..., fk ).
2. Every ascending chain
I1 ⊂ I2 ⊂ ⋯Ik ⊂ ⋯
of ideals of A terminates; that is the chain is eventually constant. This is known as the
ascending chain condition, or a.c.c.
3. Every nonempty set of ideals A has a maximal element.
If these conditions hold, the ring A is said to be Noetherian.
Proposition 1.2. Let A be a ring.
1. Suppose that A is noetherian and I ⊂ A an ideal; then the quotient ring B = A/I is
noetherian.
2. Let A be a noetherian integral domain and A ⊂ K its field of fractions. Let 0 ∈/ S ⊂ A be
a subset and set
a
B = A[S −1 ] = { ∈ K∣a ∈ A and b = 1 or a product of elements in S}
b
Then B is again Noetherian.
Theorem 1.3 (Hilbert Basis Theorem). For a ring A,
A Noetherian ⇒ A[X] Noetherian
Proof . Let J ⊂ A[X] be any ideal; I prove that J is finitely generated. Define the ideal of
leading terms of degree n in J to be
Jn = {a ∈ A∣∃f = aX n + bn−1 X n−1 + ⋯ + b0 ∈ J}
Then Jn is obviously an ideal of A and as x ∈ A[X], this implies Jn ⊂ Jn+1 . Hence using the
a.c.c, there exists an N so that JN = JN +1 = ⋯.
We now build a set of generators for J. For i ≤ N , let ai1 , ⋯, aim(i) be generators of Ji and, as
in the definition of Ji , for each of the aik , let fik = aik X i + ⋯ ∈ J be an element of degree i and
leading term aik .
I claim that the set
{fik ∣i ∈ {0, ..., N }, k ∈ {1, ..., m(i)}}
just constructed generates J: for given h ∈ J, suppose deg g = m. Then the leading term of g is
bX m with b ∈ Jm . Therefore write b = ∑k cm′ k am′ k (where m′ = m if m ≤ N otherwise m′ = N ).
Then consider
′
g1 = g − X m−m ∑ cm′ k fm′ k
k
by construction, the term of degree m is zero so that deg g1 ≤ deg g − 1. By induction, I can
therefore write out g as a combination of the fik , so that these generate J.
1
Corollary 1.4. If k is a field, then a finitely generated k-algebra is Noetherian.
Proof . Any finitely generated k-algebra is a ring of the form A = k[a1 , ..., an ]; every such ring
is isomorphic to a quotient of the polynomial ring A ≅ k[X1 , ..., Xn ]/I. A field is noetherian,
then apply Theorem 1.3 inductively and Proposition 1.2 to see that A is Noetherian.
Definition 1.5. Let k be any field and a = k[X1 , ..., Xn ]. Write An = k n for the n-dimensional
Affine Space over k; given a polynomial f (X1 , ..., Xn ) ∈ A and a point P = (a1 , ..., an ) ∈ An , the
element f (a1 , ..., an ) ∈ k is thought of as ‘evaluating the function f at P ’. Define a correspondence taking an ideal J ⊂ A to V (J), the zero set in An of all f ∈ J.
A subset X ⊂ An is called an algebraic subset if X = V (I) for some I. Note that by Corollary
1.4, I is finitely generated and V (I) is simply the zero set of the generators.
That is, an algebraic set is the locus of points satisfying a finite number of polynomial equations.
If I = (f ) is principal, we just write V (f ) = V (I) or V ∶ (f = 0).
Proposition 1.6. The correspondence V satisfies the following properties:
1. V (0) = An , V (A) = ∅
2. I ⊂ J ⇒ V (I) ⊃ V (J)
3. V (I1 ∩ I2 ) = V (I1 ) ∪ V (I2 )
4. V (∑λ∈Λ Iλ ) = ⋂λ∈Λ V (Iλ )
Hence the algebraic subsets of An form the closed sets of a topology on An called the Zariski
Topology.
Definition 1.7. For any subset X ⊂ An , let I(X) be the ideal of functions vanishing on it.
Proposition 1.8.
1. X ⊂ Y ⇒ I(X) ⊃ I(Y )
2. For any subset X ⊂ An , we have X ⊂ V (I(X)) with equality if and only if X is an algebraic
subset.
3. For J ⊂ A, we have J ⊂ I(V (J)); the inclusion may well be strict.
The main two ways in which the inclusion in 1.8 part 3 can be strict. Firstly, if the field
is not algebraically closed; consider x2 + 1 in R. Secondly, for any f ∈ k[X1 , ..., Xn ] and a ≥ 2
note that f a defines the same locus of zeroes as f so V (f a ) = V (f ), but f ∈/ f a ; think of
V (X 2 ) = V (X) in R2 .
Definition 1.9. An algebraic set X ⊂ An is irreducible if there does not exists a decomposition
X = X1 ∪ X2 of X as a union of two strict algebraic subsets.
Proposition 1.10.
then
1. Let X ⊂ An be an algebraic subset and I(X) the corresponding ideal;
X is irreducible ⇔ I(X) is prime
2. Any algebraic set X has a (unique) expression as a union of irreducible subsets.
The way to prove the second part is to notice that any non-empty set of irreducible subsets
of An has a minimal element.
2
Definition 1.11. If I is an ideal of A, then define
√
rad I = I = {f ∈ A∣f n ∈ I for some n}
Note that rad I is an ideal, and call an ideal I radical if rad I = I. It is easy to see that prime
ideals are radical.
Theorem 1.12 (Hilbert’s Zeroes Theorem). Let k be an algebraically closed field. Then
1. Every maximal ideal of k[X1 , ..., Xn ] is of the form mP = (X1 − a1 , ..., Xn − an ) for some
point P = (a1 , ..., an ) ∈ k n ; that is, it’s the ideal I(P ) of all functions vanishing at P .
2. Let J ⊂ A be an ideal with J ≠ (1). Then V (J) ≠ ∅.
3. For any J ⊂ A,
I(V (J)) = rad J
Proof . We will use the following hard fact without proof:
Lemma 1.13. For an infinite field k, and A = k[a1 , ..., an ] a finitely generated k-algebra, then
if A is a field, A is an algebraic field extension of k.
1. Let m ⊂ k[X1 , ..., Xn ] be a maximal ideal; then K = k[X1 , ..., Xn ]/m is a finitely generated
field extension of k. As k is algebraically closed, we have k = K so the composite of the
natural maps
φ ∶ k ↪ k[X1 , ..., Xn ] → k[X1 , ..., Xn ]/m = K
is an isomorphism.
Now for each i, Xi ∈ k[X1 , ..., Xn ] maps to some element bi ∈ K; taking ai = φ−1 (bi )
gives Xi − ai ∈ ker{k[X1 , ..., Xn ] → K} = m. Hence there exist a1 , ..., an ∈ k so that
(X1 −a1 , ..., Xn −an ) ⊂ m. But the LHS is a clearly a maximal ideal so (X1 −a1 , ..., Xn −an ) =
m.
2. If J ≠ A = k[X1 , ..., Xn ] then there exists a maximal ideal m of A such that J ⊂ m. By
part 1, m = (X1 − a1 , ..., Xn − an ); then J ⊂ m means that f (P ) = 0 for all f ∈ J where
P = (a1 , ..., an ). Therefore P ∈ V (J).
3. Let J ⊂ k[X1 , .., Xn ] be any ideal and f ∈ k[X1 , ..., Xn ]. Introduce another variable Y and
consider the ideal
J1 = (J, f Y − 1) ⊂ k[X1 , ..., Xn , Y ]
Then a point Q = (a1 , ..., an , b) ∈ V (J1 ) ⊂ An+1 such that P = (a1 , ..., an ) ∈ V (J) and
f (P ) ≠ 0 and b = f (P )−1 .
Now suppose that f (P ) = 0 for all P ∈ V (J); then by the above, V (J1 ) = ∅. From part 2,
we see 1 ∈ J1 and so there exists an expression
1 = ∑ gi fi + g0 (f Y − 1) ∈ k[X1 , ..., Xn , Y ]
i
with fi ∈ J and g0 , gi ∈ k[X1 , ..., Xn , Y ].
In the RHS, Y can appear in each of the gi . Suppose N is the highest power of Y appearing
in any of the gi and g0 . Then multiplying both sides through by f N , we get
f N = ∑ Gi (X1 , ..., Xn , f Y )fi + G0 (X1 , ..., Xn , f Y )(f Y − 1)
i
3
Reduce modulo (f Y − 1) to get
f N = ∑ hi (X1 , ..., Xn )fi ∈
i
k[X1 , ..., Xn , Y ]
(f Y − 1)
now simply replace Y with f −1 Then f N ∈ J for some N .
Corollary 1.14. The correspondences V and I induce bijections between radical ideals and
algebraic subsets, and between prime ideals and irreducible algebraic subsets.
Theorem 1.15 (Noether Normalisation). Let k be an infinite field and A = k[a1 , ..., an ] be a
finitely generated k-algebra. Then there exist m ≤ n and y1 , ..., ym ∈ A such that
1. y1 , ..., ym are algebraically independent over k
2. A is a finite k[y1 , ..., ym ]-algebra; that is every element of A satisfies a monic equation
over k[y1 , ..., ym ].
Proof . Let I be the kernel of the natural surjection
k[X1 , ..., Xn ] → k[a1 , ..., an ] = A
′
so f becomes a monic
suppose that 0 ≠ f ∈ I. We replace X1 , ..., Xn−1 by special X1′ , ..., Xn−1
′
′
′
equation for an over A = k[a1 , ..., an−1 ].
Write a′1 = a1 − α1 an , ..., a′n−1 = an−1 − αn−1 an , where αi ∈ k are to be determined.
Then 0 = f (a′1 + α1 an , ..., a′n−1 + αn−1 an , an ) since f ∈ I.
Proposition 1.16. There exists a choice of α1 , ..., αn−1 ∈ k so the polynomial f (X1′ +
′
α1 Xn , ..., Xn−1
+ αn−1 Xn , Xn ) is monic in Xn .
Proof . Let d = deg f . Write F = Fd + G where Fd is the homogeneous part of degree d
and G is the remainder. Then
′
+ αn−1 Xn , Xn )
f (X1 , ..., Xn−1 , Xn ) = f (X1′ + α1 Xn , ..., Xn−1
= Fd (α1 , ..., αn−1 , 1)Xnd + (terms of lower degree)
Now Fd (α1 ..., αn−1 , 1) ≠ 0 for some choice of α1 , ..., αn−1 , since otherwise Fd would be
constantly zero. As Fd (α1 , ..., αn−1 , 1) ∈ k, we are done.
If I = 0, nothing to prove, otherwise pick 0 ≠ f ∈ I. Let α1 , ..., αn−1 be as in the proposition
above. Then f gives a monic relation satisfied by an with coefficients in A′ = k[a′1 , ..., a′n−1 ] ⊂ A.
By inductive assumption, there exist y1 , ..., yn ∈ A′ so that
1. y1 , ..., ym are algebraically independent.
2. A′ is a finite k[y1 , ..., ym ]-algebra.
Then A = A′ [an ] is finite over A′ so A is finite over k[y1 , ..., ym ], proving the theorem.
4
2
Functions on Varieties
Definition 2.1. For an algebraic set V ⊂ An , define the coordinate ring by
k[V ] =
k[X1 , ..., Xn ]
I(V )
; that is, f ∈ k[V ] is the restriction of some polynomial on An .
Definition 2.2. A map f ∶ V → W between algebraic sets V, W is a polynomial map if there
exist m polynomials F1 , ..., Fm ∈ k[X1 , ..., Xn ] so that
f (P ) = (F1 (P ), ..., Fm (P )) ∈ Am
for all P ∈ V .
A polynomial map is an isomorphism if there exists an inverse polynomial map.
Theorem 2.3. Let V ⊂ An and W ⊂ Am be algebraic sets.
1. A polynomial map f ∶ V → W induces a ring homomorphism f ∗ ∶ k[W ] → k[V ], defined
by composition of functions.
2. Conversely, any k-algebra homomorphism Φ ∶ k[W ] → k[V ] is of the form Φ = f ∗ for a
uniquely defined polynomial map f ∶ V → W .
3. If f ∶ V → W and g ∶ W → U are polynomial maps then the two ring homomorphisms
(g ○ f )∗ = f ∗ ○ g ∗ ∶ k[U ] → k[V ] coincide.
Proof .
1. Note f ∗ (g) is a polynomial map V → k, and f ∗ (a) = a for all a ∈ k and f ∗ is
clearly a ring homomorphism defined pointwise.
2. For i = 1, ..., m, let yi ∈ k[W ] be the ith coordinate function on W , so that
k[W ] = k[y1 , ..., ym ] = k[Y1 , ..., Ym ]/I(W )
Now Φ ∶ k[W ] → k[V ] is given, so I can define fi ∈ k[V ] by fi = Φ(yi ) and consider
the map f ∶ V → Am defined by f (P ) = f1 (P ), ..., fm (P ). This is a polynomial map
since fi ∈ k[V ]. Furthermore, f (V ) ⊂ W ; suppose G ∈ I(W ) ⊂ k[Y1 , ..., Ym ]. Then
G(y1 , ..., ym ) = 0 ∈ k[W ]. Therefore Φ(G(y1 , ..., ym )) = 0 ∈ k[V ]; but Φ is a k-algebra
homomorphism, so that
k[V ] ∋ 0 = Φ(G(y1 , ..., ym )) = G(Φ(y1 ), ..., Φ(ym )) = G(f1 , ..., fm )
The fi are functions on V , for P ∈ V , and for every G ∈ I(W ), the coordinates (f1 (P ), ..., fm (P ))
of f (P ) satisfy G(f1 (P ), ..., fm (P )) = 0. Since W is the subset of Am defined by the vanishing of G ∈ I(W ), it follows that f (P ) ∈ W . This proves that f given above is a
polynomial map f ∶ V → W . Since f ∗ , Φ coincide on the generators yi , shows that the
map f is uniquely defined by Φ.
3. Composition of functions is associative.
Corollary 2.4. A polynomial map f ∶ V → W is an isomorphism if and only if f ∗ ∶ k[W ] → k[V ]
is an isomorphism.
5
Definition 2.5. An Affine Variety is an irreducible algebraic subset V ⊂ An , defined up to
isomorphism.
Equivalently, an affine variety over a field k is a set V , together with a ring k[V ] of k-valued
functions f ∶ V → k such that
1. k[V ] is a finitely generated k-algebra.
2. For some choice x1 , ..., xn of generators of k[V ] over k, the map
V → An sending P → x1 (P ), ..., xn (P )
embeds V as an irreducible algebraic subset.
Definition 2.6. The function field k(V ) of V is the field of fractions of k[V ]. An element
f ∈ k(V ) is a rational function on V .
We say that f is regular at a point P ∈ V if there exists an expression f = g/h with g, h ∈ k[V ]
and h(P ) ≠ 0.
Definition 2.7. Write
dom f = {P ∈ V ∣f is regular at P }
for the domain of definition of f and
OV,P = {f ∈ k(V )∣f is regular at P }
to be the local ring of V at P .
Theorem 2.8.
1. dom f is open and dense in the Zariski topology.
2. If the field k is algebraically closed, then dom f = V ⇔ f ∈ k[V ]; that is, a polynomial
function is a regular rational function.
3. Furthermore, for any h ∈ k[V ], let
Vh = V ∖ V (h) = {P ∈ V ∣h(P ) ≠ 0}
then
dom f ⊃ Vh ⇔ f ∈ k[V ][h−1 ]
Proof . Define the ideal of denominators of f ∈ k(V ) by
Df = {h ∈ k[V ]∣hf ∈ k[V ]} ⊂ k[V ]
= {h ∈ k[V ]∣∃ an expression f =
g
with g ∈ k[V ]} ∪ {0}
h
Obviously, Df is an ideal of k[V ]. Then,
V ∖ dom f = {P ∈ V ∣h(P ) = 0 for all h ∈ Df } = V (Df )
so that V ∖ dom f is an algebraic set of V ; hence dom f = V ∖ V (Df ) is the complement of a
closed set, so open in the Zariski topology. It is obvious that dom f is non-empty, hence dense.
Now, using part 2 of the Nullstellensatz,
dom f = V ⇔ V (Df ) = ∅ ⇔ 1 ∈ Df i.e. f ∈ k[V ]
Finally,
dom f ⊃ Vh ⇔ h vanishes on V (Df )
and using part 3 of the Nullstellensatz, this happens if and only if hn ∈ Df for some n; that is
f = g/hn ∈ k[V ][h−1 ]
6
Definition 2.9. A rational map f ∶ V ⇢ An is a partially defined map given by rational
functions f1 , ..., fn ; that is,
f (P ) = f1 (P ), ..., fn (P ) for all P ∈ ⋂ dom fi
i
By definition, dom f = ⋂i dom fi .
F is said to be regular at P ∈ V if and only if P ∈ dom f .
A rational map V ⇢ W between two affine varieties V ⊂ An and W is defined to be a rational
map f ∶⇢ Am so that f (dom f ) ⊂ W .
Definition 2.10. The composition g ○ f of two rational maps f and g is only defined on
dom f ∩ f −1 (dom g). It is possible that this is empty.
In terms of algebra, suppose f is given by f1 , ..., fm ∈ k(V ) so that
f ∶ V ⇢ W ⊂ Am , defined by P ↦ f1 (P ), ..., fm (P )
for P ∈ ⋂ dom fi ; any g ∈ k[W ] is of the form g = G mod I(W ) for some G ∈ k[Y1 , ..., Ym ] and
g ○ f = G(f1 , ..., fm ) is well defined in k(V ). So exactly as in Theorem 2.3, there is a k-algebra
homomorphism f ∗ ∶ k[W ] → k(V ) corresponding to f . However, if h ∈ k[W ] is in the kernel
of f ∗ , then no meaning can be attached to f ∗ (g/h) so that f ∗ cannot be extended to a field
homomorphism k(W ) → k(V ).
Definition 2.11. A rational map f ∶ V ⇢ W is dominant if f (dom f ) is dense in W for the
Zariski topology.
Geometrically, this means f −1 (dom g) ⊂ dom f is a dense open set for any rational map g ∶ W ⇢
U so that g ○ f is defined on a dense open set of V .
Algebraically,
f is dominant ⇔ f ∗ ∶ k[W ] → k(V ) is injective.
For given g ∈ k[W ],
g ∈ ker f ∗ ⇔ f (dom f ) ⊂ V (g)
that is, f ∗ is not injective if and only if f (dom f ) is contained in a strict algebraic subset of W .
Theorem 2.12.
1. A dominant rational map f ∶ V ⇢ W defines a field homomorphism
∗
f ∶ k(W ) → k(V )
2. Conversely, a k-homomorphism Φ ∶ k(W ) → k(V ) comes from a uniquely defined dominant
rational map f ∶ V ⇢ W .
3. If f and g are dominant then (g ○ f )∗ = f ∗ ○ g ∗
The proof is only a minor modification of Theorem 2.3.
Definition 2.13. Let V, W be affine varieties and U ⊂ V be an open subset. A morphism
f ∶ U → W is a rational map f ∶ V ⇢ W such that U ⊂ dom f , so that f is regular at every
P ∈ U.
If U1 ⊂ V and U2 ⊂ W are open, then a morphism f ∶ U1 → U2 is just a morphism f ∶ U1 → W
such that f (U1 ) ⊂ U2 . An isomorphism is a morphism with an inverse morphism.
Note that by Theorem 2.8 part 2
{morphisms f ∶ V → W } = {polynomial maps f ∶ V → W }
where the LHS consists of rational objects, the RHS is more concrete in terms of polynomials.
7
Definition 2.14. For an affine variety V and f ∈ k[V ], write Vf for the open set Vf = V ∖V (f ).
The Vf are called the standard open sets of V .
Proposition 2.15. Vf is isomorphic to an affine variety, and
k[Vf ] = k[V ][f −1 ]
Proof. Let J = I(V ) ⊂ k[X1 , ..., Xn ] and choose F ∈ k[X1 , ..., Xn ] such that f = F mod I(V ).
Now define I = (J, Y F − 1) ⊂ k[X1 , ..., Xn , Y ] and let
V (I) = W ⊂ An+1
Now the maps (X1 , ..., Xn , Y ) → (X1 , ..., Xn ) and (X1 , ..., Xn ) → (X1 , .., Xn , F (X1 , ..., Xn )−1 )
are inverse morphisms between W and Vf
3
Plane Conics
Definition 3.1. A conic in R2 is a plane curve given by a quadratic expression
q(x, y) = ax2 + bxy + cy 2 + dx + ey + f = 0
Lemma 3.2. All conics in R2 are one of the following form:
1. Ellipse;
x2
a2
+
y2
b2
=1
2. Parabola; y = mx2
3. Hyperbola;
x2
a2
−
y2
b2
=1
4. Single point; x2 + y 2 = 0
5. Emptyset; x2 + y 2 = −1,
x2 = −1,
0=1
6. Line; x = 0
7. Line Pair; xy = 0
8. Parallel Lines; x(x − 1) = 0
9. Double Line; x2 = 0
10. Whole plane; 0 = 0
The first three cases are known as the “non-degenerate” conics.
Definition 3.3. A conic C ⊂ P2 is the curve given by C ∶ (Q(X, Y, Z) = 0), where Q is a
homogeneous quadratic expression.
Definition 3.4. A line in P2 is by definition given by L ∶ (aX + bY + cZ = 0) and
L passes through (X, Y, 0) ⇔ aX + bY = 0
8
Theorem 3.5. Let k be any field of characteristic not 2. Then, using the bijection between
homogeneous quadratic polynomials and symmetric bilinear forms on k 3 , given by:
⎛a b d ⎞
aX + 2bXY + cY + 2dXZ + 2eY Z + f Z ↔ ⎜ b c e ⎟
⎝d e f ⎠
2
2
2
with the Gram-Schmidt Orthogonalisation, saying that any quadratic form Q can be expressed
as a sum of squares in some choice of coordinates.
We have, in a suitable system of coordinates, any conic in P2 is one of the following:
1. Non-degenerate conic: X 2 + Y 2 − Z 2 = 0
2. Emptyset: X 2 + Y 2 + Z 2 = 0
3. Line Pair: X 2 − Y 2 = 0
4. One Point: X 2 + Y 2 = 0
5. Double Line: X 2 = 0
6. Whole of P2 : 0 = 0
Corollary 3.6. Any non-empty, non-degenerate conic of P2 is projectively equivalent to the
curve XZ = Y 2 ; the curve parameterised by (U 2 ∶ U V ∶ V 2 ).
Definition 3.7. Let F (U, V ) be a nonzero homogeneous polynomial of degree d in U, V with
coefficients in a fixed field k. F has an associated inhomogeneous polynomial in 1 variable
f (U ) = F (U, 1) and clearly for α ∈ k we have
f (α) = 0 ⇔ (u − α)∣f (u) ⇔ (U − αV )∣F (U, V ) ⇔ F (α, 1) = 0
Define the multiplicity of a zero of F on P1 to be
1. The multiplicity of f at the corresponding α ∈ k
2. d − deg f if (1, 0) is the zero.
So the multiplicity of zero of F at a point (α, 1) is the greatest power of (U − αV ) dividing F .
Proposition 3.8. Let F (U, V ) be a non-zero form of degree d in U, V . Then F has at most d
zeroes in P1 ; furthermore if k is algebraically closed then F has exactly d zeroes on P1 provided
these are counted with multiplicities as defined above.
Theorem 3.9. Let L ⊂ P2 be a line (respectively C ⊂ P2 a non-degenerate conic), and let
D ⊂ P2 be a curve defined by D ∶ (Gd (X, Y, Z) = 0) where G is a form of degree d in X, Y, Z.
Assume that L ⊂/ D (respectively C ⊂/ D); then
#{L ∩ D} ≤ d
(respectively #{C ∩ D} ≤ 2d)
Proof . A line L ⊂ P2 is given by an equation λ = 0, with λ a linear form; assume it is
parameterised in the form
X = a(U, V ), Y = b(U, V ), Z = c(U, V )
where a, b, c are linear forms in U and V .
Similarly, a nondegenerate conic can be parameterised by a, b, c each quadratic forms in U, V .
9
The intersection of L (respectively C) with D is given by finding the values of the ratios (U ∶ V )
such that
F (U, V ) = Gd (a(U, V ), b(U, V ), c(U, V )) = 0
But F is a form of degree d (respectively 2d in U, V , so the result follows by Proposition 3.8.
Corollary 3.10. If P1 , ..., P5 ∈ P2 are distinct points such that no four are collinear, then there
exists at most one conic through P1 , .., P5 .
Proof . Suppose by contradiction that C1 and C2 are conics with C1 ≠ C2 and C1 ∩ C2 contains
P1 , ..., P5 .
C1 is non-empty, so that if it’s non-degenerate, by Corollary 3.6 it is projectively equivalent to
the parameterised curve
C1 = {(U 2 , U V, V 2 )∣(U, V ) ∈ P1 }
By Theorem 3.9, we have C1 ⊂ C2 . Now if Q2 is the equation of C2 then Q2 (U 2 , U V, V 2 ) ≡ 0 for
all (U, V ) ∈ P1 and hence Q2 is a multiple of XZ − Y 2 ; hence C1 = C2 which is a contradiction.
Now suppose that C1 is degenerate. By the classification of conics, either it is a line pair or a
line and there is no other option than C1 = L0 ∪ L1 , C2 = L0 ∪ L2 , so C1 ∩ C2 = L0 ∪ (L1 ∩ L2 );
hence 4 out of the five points P1 , .., P5 lie on L0 , a contradiction.
Let S2 be the set of all quadratic forms on R3 . For Q ∈ S2 , write Q = aX 2 + 2bXY + ⋯ + f Z 2 .
Then for a fixed P0 = (X0 , Y0 , Z0 ), the set of conics passing through this point is
S2 (P0 ) = {Q ∈ S2 ∣Q(P0 ) = 0} ≅ R5
which is a 5-dimensional hyperplane. Similarly, for P1 , ..., Pn ∈ P2 , define
S2 (P1 , ..., Pn ) = {Q ∈ S2 ∣Q(Pi ) = 0∀i}
Proposition 3.11. dim S2 (P1 , ..., Pn ) ≥ 6 − n
Corollary 3.12. If n ≤ 5 and no 4 of P1 , ..., Pn are collinear, then
dim S2 (P1 , ..., Pn ) = 6 − n
Proof . Corollary 3.10 implies that if n = 5 then dim S2 (P1 , ..., P5 ) ≤ 1 which gives the result
in this case. If n ≤ 4 add the points Pn+1 , ..., P5 while preserving the condition that no 4 of the
points are collinear, and since each point imposes at most one linear condition this gives
1 = dim S2 (P1 , ..., P5 ) ≥ dim S2 (P1 , ..., Pn ) − (5 − n)
According to the previous corollary, the set of conics passing through 4 points is a 2dimensional vector space, so choose a basis Q1 , Q2 representing conics C1 , C2 so that C1 ∩ C2 =
{P1 , ..., P4 }.
Definition 3.13. A pencil of conics is a family of the form
C(λ,µ) ∶ (λQ1 + µQ2 = 0)
where the variables (λ ∶ µ) ∈ P1 .
Clearly, C(λ,µ) is degenerate if and only if det(λQ1 + µQ2 ) = 0.
Proposition 3.14. Suppose C(λ,µ) is a pencil of conics of P2 with at least one non-degenerate
conic. Then the pencil has at most 3 degenerate conics. If k = R then the pencil has at least
one degenerate conic.
Proof . The value det(λQ1 + µQ2 ) is a homogeneous cubic in λ and µ, and hence has at most
3 roots, each corresponding to a degenerate conic. If k = R then it must have at least one
root.
10
4
Projective Varieties
Definition 4.1. A polynomial f ∈ k[X0 , ..., Xn ] is homogeneous of degree d if all monomials
have degree d.
Definition 4.2. An ideal I ⊂ k[X0 , ..., Xn ] is homogeneous if for all f ∈ I, the homogeneous
decomoposition f = f0 + f1 + ⋯ + fN of f satisfies fi ∈ I for all i.
This is equivalent to saying that I is generated by finitely many homogeneous polynomials.
As before, we have the V, I correspondence; we need homogeneity of ideals and polynomials
to make sure everything is well-define on Pn .
Definition 4.3. As before, a subset of the form V (I) ⊂ Pn is an algebraic subset of Pn
Theorem 4.4 (Projective Nullstellensatz). Assume that k is algebraically closed. Then
1. V (J) = ∅ ⇔ rad J ⊃ (X0 , ..., Xn )
2. If V (J) ≠ ∅ then I(V (J)) = rad J
Corollary 4.5. I and V determine inverse bijections between
homogeneous radical ideals
{
} ↔ {algebraic subsets X ⊂ Pn }
J ⊂ k[X0 , ..., Xn ] with J ≠ k[X0 , ..., Xn ]
homogeneous prime ideals
{
} ↔ {irreducible algebraic subsets X ⊂ Pn }
J ⊂ k[X0 , ..., Xn ] with J ≠ k[X0 , ..., Xn ]
Proof . Let π ∶ An+1 ∖ 0 → Pn be the defining map of Pn . For a homogeneous ideal J write
V a (J) ⊂ An+1 for the affine algebraic set defined by J. Then, since J is homogeneous V a (J)
has the property
(α0 , ..., αn ) ∈ V a (J) ⇔ (λα0 , ..., λαn ) ∈ V a (J)
and V (J) = V a (J) ∖ 0/ ∼⊂ Pn . Hence
V (J) = ∅ ⇔ V a (J) ⊂ {0} ⇔ I(V (J)) = rad J ⊃ (X0 , ⋯, Xn )
where the last implication uses the affine Nullstellensatz.
Definition 4.6. A rational function on V is a partially defined function f ∶ V ⇢ k given by
f (P ) = g(P )/h(P ) where g, h ∈ k[X0 , ..., Xn ] are homogeneous polynomials of the same degree
d.
Note that g/h and g ′ /h′ define the same rational function on V if and only if h′ g − g ′ h ∈ I(V )
so that the set of all rational functions is the function field of V .
k(V ) = {g/h∣g, h ∈ k[X0 , ..., Xn ] homogeneous of same degree , h ∈/ I(V )}/ ∼
where ∼ is the equivalence relation g/h ∼ g ′ /h′ ⇔ h′ g − g ′ h ∈ I(V ).
We say f is regular at P if there exists an expression f = g/h with g, h homogeneous polynomials
of the same degree, such that h(P ) ≠ 0. Write
dom f = {P ∈ V ∣f is regular at P }
and
OV,P = {f ∈ k(V )∣f is regular at R}
Clearly, dom f ⊂ V is a dense Zariski open set in V and OV,P is a subring.
Similarly, we define rational maps f ∶ V ⇢ Pm by m rational functions in k(V ). We usually
assume one of them is 1.
11
Definition 4.7. A rational map f ∶ V ⇢ Pm is regular at P ∈ V if there exists an expression
f = (f0 , ..., fm ) such that
1. each of f0 , ..., fm is regular at P
2. at least one of the fi (P ) ≠ 0
If U ⊂ V is an open subset of a projective variety V then a morphism f ∶ U → W is a rational
map V ⇢ W such that dom f ⊃ U ; that is, a morphism is a rational map that is everywhere
regular on U .
Definition 4.8. Let V and W be varieties, then a rational map V ⇢ W is birational if it has
a rational inverse.
Proposition 4.9. The following are equivalent:
1. f is a birational map.
2. f is dominant and f ∗ ∶ k(W ) → k(V ) is an isomorphism.
3. There exist open sets V0 ⊂ V and W0 ⊂ W such that f restricted to V0 is an isomorphism
f ∶ V0 → W0 .
Corollary 4.10. For a variety V , the following are equivalent:
1. The function field k(V ) is a purely transcendental extension of k; that is k(V ) ≅ k(t1 , ..., tn )
for some n.
2. There exists a dense open set V0 ⊂ V which is isomorphic to a dense open subset U0 ⊂ An ;
that is, V can be parameterised by n independent variables.
A variety satisfying these conditions is called rational.
5
Tangent Space, Non-Singularity and Dimension
Definition 5.1. For a hypersurface V = V (f ), P ∈ V ⊂ An is a non-singular point of V if
∂f
∂Xi (P ) ≠ 0 for some i; otherwise P is a singular point of V .
Note that the derivatives appearing here are formal; no calculus is here.
Proposition 5.2. For an irreducible hypersurface V = V (f ), Vnonsing = {P ∈ V ∣P is nonsingular}
is a dense open set of V in the Zariski Topology.
Proof . Note that the set of singular points is the set
Vsing = V (f,
∂f
) ⊂ An
∂Xi
which is closed by the definition of the Zariski topology. Moreover, suppose Vsing = V . Then all
partial derivatives of f must vanish on V , so must be divisible by f ; but the partial derivatives
of f have degree strictly smaller than f so all derivatives must be zero and hence f is constant
(over R or C). Over a general field, this implies that f is an inseparable polynomial in Xi . If
this happens then f is a pth power, meaning that f is not irreducible.
12
Definition 5.3. Let V ⊂ An be a subvariety, with V ∋ P = (a1 , ..., an ). For any f ∈ k[X1 , ..., Xn ]
write
∂f
(1)
(P )(Xi − ai )
fP = ∑
i ∂Xi
This is an affine linear polynomial; the first order part of f at P . Define the tangent space to
V at P by
(1)
TP V = ⋂ (fP = 0) ⊂ An
f ∈I(V )
Definition 5.4. There exists an integer r and a dense open subset V0 ⊂ V such that
dim TP Y = r for P ∈ V0 , and dim TP V ≥ r for all P ∈ V
Define r to be the dimension of V and write dim V = r. Say that P ∈ V is non-singular if
dim TP V = r and singular if dim TP V > r.
Definition 5.5. If k ⊂ K is a field extension, the transcendence degree of K over k is the
maximum number of elements of K algebraically independent over k. It is denoted tr degk K.
Let P = (a1 , ..., an ) ∈ V ⊂ An . Take new coordinates Xi′ = Xi − ai to bring P to the origin
and so assume that P = (0, ..., 0).
Write mP = ideal of P ∈ k[V ] and
MP = ideal (X1 , ..., Xn ) ⊂ k[X1 , ..., Xn ]
Then, mP = MP /I(V ) ⊂ k[V ].
Theorem 5.6. In the above notation,
1. There is a natural isomorphism of vector spaces
mP
(TP V )∗ = 2
mP
2. If f ∈ k[V ] is such that f (P ) ≠ 0 and Vf ⊂ V is the standard affine open set, the natural
map
TP (Vf ) → TP V
is an isomorphism.
Proof .
1. Write (k n )∗ for the vector space of linear forms on k n ; this is the vector space
(1)
with basis X1 , ..., Xn . Since P = (0, ..., 0), for any f ∈ k[X1 , ..., Xn ], the linear part fP is
naturally an eleent of (k n )∗ ; define a map d ∶ MP → (k n )∗ by taking f ∈ MP into df = fP .
d is surjective, since the Xi ∈ MP go into the natural basis of (k n )∗ ; also ker d = MP2 , since
(1)
fP = 0 ⇔ f starts with quadratic terms in X1 , ..., Xn ⇔ f ∈ MP2
(1)
Hence
map
MP
MP2
≅ (k n )∗ . Now, use the restriction map (k n )∗ → (TP V )∗ . Composing defines a
D ∶ MP → (k n )∗ → (TP V )∗
The composite D is surjective since each factor is. Now
f ∈ ker D ⇔ fP ∣TP V = 0
(1)
⇔ fP = ∑ ai {gi }P for some gi ∈ I(V )
(1)
(1)
i
⇔ f − ∑ ai gi ∈ MP2
i
⇔ f ∈ MP2 + I(V )
13
2. It is left as an exercise...
Corollary 5.7. TP V only depends on a neighbourhood of P ∈ V up to isomorphism. More
precisely, if P ∈ V0 ⊂ V and Q ∈ W0 ⊂ W are open subsets of affine varieties and φ ∶ V0 → W0
is an isomorphism taking P into Q, there is a natural isomorphism TP V0 → TQ W0 ; hence
dim TP V0 = dim TQ W0 .
Proof . By passing to a smaller neighbourhood of P in V , I can assume V0 is isomorphic to an
affine variety, then so is W0 and φ induces isomorphisms k[W0 ] ≅ k[V0 ] taking mP → mQ .
Thanks to this corollary, we can define tangent spaces and dimension for projective varieties;
a point P in a projective variety is non-singular if it is non-singular is some affine piece; the
corollary ensures this fact is independent of the choice of affine piece.
6
The 27 Lines on a Cubic Surface
In this section, S ⊂ P3 will be a non-singular cubic surface, given by a homogeneous cubic
f = f (X, Y, Z, T ). Consider the lines l of P3 lying on S.
Proposition 6.1.
1. There exists at most 3 lines of S through any point P ∈ S; if there are
2 or 3, they must be coplanar.
2. Every plane Π ⊂ P3 intersects S in one of the following:
i An irreducible cubic
ii A conic plus a line
iii 3 distinct lines
Proof .
1. If l ⊂ S then l = TP l ⊂ TP S so that all lines of S through P are contained in the
plane TP S; there are at most 3 of them by part 2.
2. I have to prove that a multiple line is impossible. If Π ∶ (T = 0) and l ∶ (Z = 0) ⊂ Π then
to say that l is a multiple line of S ∩ Π means that f is of the form
f = Z 2 A(X, Y, Z, T ) + T B(X, Y, Z, T )
with A a linear form, B a quadratic form. Then S ∶ (f = 0) is singular at a point where
Z = T = B = 0; this is a non-empty set since it is the set of roots of B on the line Z = T = 0.
Theorem 6.2. There exists at least one line on S.
Proposition 6.3. Given a line l ⊂ S, there exist exactly 5 pairs (li , li′ ) of lines of S meeting l,
in such a way that
1. for each i, l ∪ li ∪ li′ is coplanar
2. For i ≠ j, (li ∪ li′ ) ∩ (lj ∪ lj′ ) = ∅.
14
Proof . If Π is a plane of P3 containing l then Π ∩ S = l + conic, (since f ∣Π is divisible by the
equation of l). This conic can either be singular or non-singular.
I have to prove that there are exactly 5 distinct planes Πi ⊂ l for which the singular case occurs.
Part 2 of this theorem then follows from Proposition 6.1, part 1.
Suppose that l ∶ (Z = T = 0). Then expand f out as
f = AX 2 + BXY + CY 2 + DX + EY + F
where A, B, C, D, E, F ∈ k[Z, T ], A, B, C linear forms, D, E quadratic forms and F a cubic form.
Considering this equation as a variable conic in X, Y , it is singular if and only if
∆(Z, T ) = 4ACF + BDE − AE 2 − B 2 F − CD2 = 0
To be more precise, any plane through l is given by Π ∶ (µZ = λT ); if µ ≠ 0 we can assume
that µ = 1 so that Z = λT . Then in terms of the homogeneous coordinates (X, Y, T ) on Π,
f ∣Π = T Q(X, Y, T ) where
Q = A(λ, 1)X 2 + B(λ, 1)XY + C(λ, 1)Y 2 + D(λ, 1)T X + E(λ, 1)T Y + F (λ, 1)T 2
Now ∆(Z, T ) is a homogeneous quintic so by Proposition 3.8, F has exactly 5 roots counted
with multiplicities. To finish the proof, we have to show there are no multiple roots, which is a
consequence of the non-singularity of S.
Suppose (Z = 0) is a root of ∆, and let Π ∶ (Z = 0) be the corresponding plane; I have to prove
that ∆ is not divisible by Z 2 . As the conic is singular, Π ∩ S is a set of 3 distinct lines, and
according to whether they intersect in a single point or not I can choose coordinates so either
1. l ∶ (T = 0), l1 ∶ (X = 0), l1′ ∶ (Y = 0)
2. l ∶ (T = 0), l1 ∶ (X = 0), l1′ ∶ (X = T )
In case 1, then f = XY T + Zg with g quadratic, and therefore B = T + aZ and Z∣A, C, D, E, F .
Therefore
∆ ≡ −T 2 F mod Z 2
in addition, the point P = (0, 0, 0, 1) ∈ S and non-singularity at P means that F must not
contain the term ZT 2 with non-zero coefficient. In particular, Z 2 does not divide F . Therefore
(Z = 0) is a simple root of ∆.
In case 2, then f = X 2 T − XT 2 − Zg2 where g2 is some homogeneous quadratic. Writing f in
the form
AX 2 + BXY + CY 2 + DX + EY + F
tells us that A = T − aZ, D = −T 2 − cZ 2 − dT Z for some constants a, c, d and Z∣B, C, E, F . Then,
if ∆(Z, T ) = 4ACF + BDE − AE 2 − B 2 F − CD2 , we see that ∆(Z, T ) = 0 mod (Z), so Z is
certainly a root of ∆. To show it is a simple root, we work mod Z 2 :
∆(Z, T ) = 0 + 0 + 0 + 0 − CD2 = −C(T 2 + cZ + dT Z)2 = −C(T 4 + 2cT 2 Z + 2dT 3 Z) mod Z 2
but as Z∣C and Z 2 ∣/ C (f is a homogeneous cubic, so CY 2 must have total degree 3. It is
therefore impossible that Z 2 ∣C) we see that ∆(Z, T ) is some multiple of T 4 with no terms in Z
so ∆(Z, T ) is non-zero mod Z 2 and hence the root Z is a simple root.
15
7
Example of finding all 10 lines intersecting a common line in
a cubic surface:
Let S be the surface defined by X 3 +Y 3 = Z 3 +T 3 and L be the line defined by X +Y = Z +T = 0.
We find the 10 lines meeting L explicitly:
We first change coordinates in P3 from X, Y, Z, T to X + Y, Y, Z + T, T . Note that:
X 3 + Y 3 = (X + Y )3 − 3Y (X + Y )2 + 3Y 2 (X + Y )
Z 3 + T 3 = (Z + T )3 − 3T (Z + T )2 + 3T 2 (Z + T )
Let f be the polynomial X 3 + Y 3 − Z 3 − T 3 which defined S. Then, writing X + Y = α and
Z + T = β we have
f = 3αY 2 + 0Y T − 3βT 2 − 3α2 Y + 3β 2 T + α3 − β 3
As in the proof of 7.3 in [UAG], we consider f as a variable conic in Y and T . The corresponding
matrix of this conic is
0
− 23 α2 ⎞
⎛ 3α
3 2
⎟
−3β
A=⎜ 0
2β
3
3
2
2
3
3⎠
⎝− α
α −β
2
2β
and so the conic is singular if and only if det A = 0.
27
63
4
4
3
3
But det A = 9αβ(β 3 − α3 ) − (− 27
4 α β + 4 αβ ) = 4 αβ(β − α ). So det A = 0 when
α = 0, β = 0, β − α = 0, β = αω, β = αω 2
(1)
) is a cube root of unity.
where ω = exp ( 2πi
3
Now, any plane through L is given by µα = λβ. In the plane Π1 defined by µ = 0, we get β = 0,
2
and so f ∣Π1 is defined by 3αY 2 − 3α2 Y + α3 = α(α
− 3αY + 3Y 2 ). For the lines we look at when
√
3± −3
α ≠ 0; by the Quadratic Formula that α = Y 2 . This therefore defines two lines meeting L:
√
√
X = 1±i2 3 Y, Z + T = 0. Observing that 1− 2 −3 = −ω, a cube root of unity, these two lines l1 , l2
are defined by X = −ωY, Z + T = 0 and X = −ω 2 Y, Z + T = 0.
When µ ≠ 0 we may assume µ = 1 so a general plane though L is given by α = λβ. In the plane
Π2 defined when λ = 0, from (1) we expect two more lines corresponding to the case α = 0.
f ∣Π2 is defined√by −3βT 2 + 3β 2 T − β 3 = −β(β 2 − 3βT + 3T 2 ). By the Quadratic formula again,
we get β = 3± 2 −3 T so this defines two more lines l3 , l4 meeting L: X + Y = 0, Z = −ωT and
X + Y = 0, Z = −ω 2 T .
In the plane Π3 defined when λ = 1, from (1) again, we expect two more lines corresponding to
the case α = β. f ∣Π3 is defined by 3αY 2 − 3αT 2 − 3α2 Y + 3α2 T = 3α(Y 2 − T 2 ) + 3α(T − Y ). This is
zero when α = 0 or when α = T + Y ; this gives us one more line l5 intersecting L: X = T, Y = Z.
In the plane Π4 defined when λ = ω 2 , f ∣Π4 is defined by 3αY 2 − 3ωαT 2 − 3α2 Y + 3ω 2 α2 T +
α3 − ω 3 α3 = α(3α(ω 2 T − Y ) + 3(Y 2 − ωT 2 )). We need to solve 3α(ω 2 T − Y ) + 3(Y 2 − ωT 2 ) = 0.
Multiplying this equation throughout by ω:
α=
ωT 2 − Y 2 ω 2 T 2 − ωY 2
=
= Y + ω2T
ω2T − Y
T − ωY
Therefore we get a line l6 described by
Z + T = ω(X + Y ), X = ω 2 T
Repeating for the plane Π5 defined by λ = ω, (that is β = ω 2 α), we get f ∣Π5 is defined by
3αY 2 − 3ω 2 αT 2 − 3α2 Y + 3ωα2 T + α3 − ω 3 α3 = α(3α(ωT − Y ) + 3(Y 2 − ω 2 T 2 )). We need to solve
16
3α(ωT − Y ) + 3(Y 2 − ω 2 T 2 ) = 0.
α=
ω2T 2 − Y 2
= ωT + Y
ωT − Y
Therefore we get a line l7 described by
Z + T = ω 2 (X + Y ), X = ωT
Summarising, from this method we get 7 lines intersecting L which are parameterised by a ∈ P1
as follows:
l1 ∶ (−ω ∶ 1 ∶ a ∶ −a)
l2 ∶ (−ω 2 ∶ 1 ∶ a ∶ −a)
l3 ∶ (a ∶ −a ∶ −ω ∶ 1)
l4 ∶ (a ∶ −a ∶ −ω 2 ∶ 1)
l5 ∶ (a ∶ 1 ∶ 1 ∶ a)
l6 ∶ (ω 2 a ∶ 1 ∶ ω ∶ a)
l7 ∶ (ωa ∶ 1 ∶ ω 2 ∶ a)
Now by permuting the Z and T coordinates of l5 , l6 , l7 , we get 3 more lines of P3 :
l8 ∶ (a ∶ 1 ∶ a ∶ 1),
l9 ∶ (ω 2 a ∶ 1 ∶ a ∶ ω),
l10 ∶ (ωa ∶ 1 ∶ a ∶ ω 2 )
Notice that l8 , l9 , l10 all lie in the surface S defined by X 3 + Y 3 = Z 3 + T 3 and each of them
intersects the line L in the following points respectively:
l8 at (−1 ∶ 1 ∶ −1 ∶ 1),
l9 at (−1 ∶ 1 ∶ −ω ∶ ω),
so we have found the 10 required lines intersecting L.
17
l10 at (−1 ∶ 1 ∶ −ω 2 ∶ ω 2 )