BACKGROUND
Things may be added here as the semester goes on.
The section on Tensor Products has been moved here.
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.1
Exact Sequences
Complexes
Categories
Functors
The Hom Functors
Morphisms of Functors
Tensor Products
Exact Sequences
exactseq
A sequence
dn−1
dn
dn+1
· · · → V n−1 −→ V n −→ V n+1 −→ · · ·
of homomorphisms of abelian groups is exact if the image of dn−1 is equal to the kernel of dn .
For example, a sequence
d
0 → V −→ V ′
is exact if and only if the map d is injective. A sequence
d
V −→ V ′ → 0
is exact if and only if d is surjective, and a sequence
d
0 → V −→ V ′ → 0
is exact if and only if d is bijective.
d
Any homomorphism V −→ V ′ can be embedded into an exact sequence
d
0 → K → V −→ V ′ → C → 0,
where K and C are the kernel and cokernel of d, respectively.
A short exact sequence is an exact sequence
0 → V → V ′ → V ′′ → 0.
The statement that this sequence is exact asserts that the map V → V ′ is injective, and that V ′′ is isomorphic
to the quotient group V ′ /V .
1
kerfunct
0.1.1. Lemma. (functorial property of the kernel and cokernel) Suppose given a (commutative) diagram of
homomorphisms of abelian groups
V −−−−→ V ′




fy
f ′y
W −−−−→ W ′
Let K and C be the kernel and cokernel of f , and let K ′ and C ′ be defined analogously. There are canonical
homomorphisms K → K ′ and C → C ′ .
snake
0.1.2. Proposition. Suppose given a (commutative) diagram
f
(0 −−−−→ )V −−−−→ V ′ −−−−→ V ′′ −−−−→ 0






fy
f ′′ y
f ′y
0 −−−−→ W −−−−→ W ′ −−−−→ W ′′ ( −−−−→ 0)
g
whose rows are short exact sequences. The maps indicated in parentheses are optional. Let K, K ′ , K ′′ and
C, C ′ , C ′′ denote the kernels and cokernels of f, f ′ , f ′′ , respectively.
(i) (left exactness of the kernel) The kernels form an exact sequence
K → K ′ → K ′′ .
If f is injective, the sequence 0 → K → K ′ → K ′′ is exact.
(ii) (right exactness of the cokernel) The cokernels form and exact sequence
C → C ′ → C ′′ .
If g is surjective, the sequence C → C ′ → C ′′ → 0 is exact.
d
(iii) Snake Lemma. There is a canonical homomorphism K ′′ −→ C that combines with the above sequences
to form an exact sequence
d
(0 −→) K → K ′ → K ′′ −→ C → C ′ → C ′′ (−→ 0).
0.2
Complexes
complexes
A complex R• of abelian groups is a sequence of homomorphisms
dn−1
dn
dn+1
· · · → Rn−1 −→ Rn −→ Rn+1 −→ · · ·
with the property that the composition dn dn−1 of adjacent maps is zero, i.e., that the image of dn−1 is contained in the kernel of dn . An exact sequence is a complex.
ϕ
ϕn
•
n
A map R• −→ R′ of complexes is a collection of homomorphisms Rn −→ R′ making a commutative
diagram
dn−1
dn
−−−−→ Rn−1 −−−−→ Rn −−−−→ Rn+1 −−−−→ · · ·






ϕn y
ϕn+1 y
ϕn−1 y
−−−−→ R′
A sequence of maps of complexes
n−1
d′ n−1
−−−−→ R′
ϕ
n
•
d′ n
−−−−→ R′
ψ
•
n+1
· · · → R• −→ R′ −→ R′′ → · · ·
2
−−−−→ . . .
is called exact if the sequence
q
q ψ
ϕq
q
· · · → Rq −→ R′ −→ R′′ → · · ·
is exact for every q.
The cohomology hq of a complex R• is defined by
hq (R• ) = (ker dq )/(im dq−1 ).
chiisaltsum
0.2.1. Proposition. Let 0 → V 0 → V 1 → · · · → V n → 0 be a complex of finite dimensional vector spaces.
Then
X
X
(−1)q dim V q =
(−1)q dim hq (V • ).
0.3
Categories
categ
A category consists of “objects” and “morphisms” between objects, with certain properties that are described
below. Morphisms may also be called maps, and often the objects are sets (with some additional structure) and
the morphisms are maps of sets, but they needn’t be. Two examples of this intuitive concept are:
the category (ab): Its objects are abelian groups and its morphisms are group homomorphisms.
• the category (top spaces): Its objects are topological spaces and its morphisms are continuous maps.
•
As one sees from these examples, the objects and morphisms may form sets large enough to that, if handled
badly, might cause logical problems related to Russell’s Paradox. This will not be an issue for us, so we ignore
logical problems and say that a category C consists of a set (objects)C of objects and a set (morphisms)C of
morphisms.
There are two objects associated to a morphism f : its domain and its range. If X and Y are the domain
f
and range of a morphism f , one writes X −→ Y . (This is symbolic notation, since the morphisms don’t need
to be maps).
The domains and ranges of the morphisms are described by maps of sets
domain
(morphisms)C −→ (objects)C
and
range
(morphisms)C −→ (objects)C
f
g
Furthermore, one can compose morphisms: To a pair f, g of morphisms such that X −→ Y −→ Z, i.e., such
g◦f
that the range of f is the domain of g, there is associated a composed morphism X −→ Z. There are two
axioms for this law of composition:
•
(associative law)
f
g
h
Given morphisms X −→ Y −→ Z −→ W , the composed morphisms h ◦ (g ◦ f ) and (h ◦ g) ◦ f are equal.
•
(identity)
id
Y
Y with domain and range Y , such that, if
For every object Y , there is an identity morphism Y −→
f
g
X −→ Y , then idY ◦ f = f and if Y −→ Z, then g ◦ idY = g.
You needn’t worry too much about the definition. Just remember that a category consists of some objects and
some morphisms between objects. There is nothing peculiar about the rest.
0.4
Functors
functor
The concept of a functor is also intuitive. A functor F from a category C to another category D,
F
C −→ D,
3
is a map that sends objects to objects and morphisms to morphisms. So it consists of a pair of maps
(objects)C → spoint(objects)D
and
(morphisms)C → (morphisms)D
These maps are required to be compatible with the category structures:
•
F (f )
f
If X −→ Y in C, then F (X) −→ F (Y ),
f
g
If X −→ Y −→ Z in C, then F (g ◦ f ) = F (g) ◦ F (f ),
• F (idY ) = idF (Y ) .
•
For example, sending a topological space to its underlying set is a functor (top spaces) → (sets). Sending a
pointed, path-connected topological space X to its fundamental group π1 (X) is a functor from the category
(pt.conn.top) to the category (groups). As is the case for these examples, it often becomes clear what the
functor is supposed to do to morphisms, once the map on objects has been described.
dual
(0.4.1)
contravariant functors
There is a closely related concept, that of a contravariant functor, written as
F
C ◦ −→ D.
f
A contravariant functor is like a functor except that it reverses the direction of a morphism: If X −→ Y in
F (f )
C, then, applying F , one obtains a morphism F (X) ←− F (Y ) in the opposite direction. The fact that F
reverses arrows is indicated by the superscript ◦. As before, F is required to be compatible with composition
of morphisms, and identity morphisms are sent to identity morphisms. The only thing to note is that, because
f
g
F reverses arrows, it also reverses the order of composition of morphisms: If X −→ Y −→ Z are morphisms
F (f )
F (g)
in C, then F (X) ←− F (Y ) ←− F (Z), so the requirement is that F (g ◦ f ) = F (f ) ◦ F (g).
For example, let (vector spaces) denote the category whose objects are complex vector spaces and whose
morphisms are linear transformations. Sending a vector space V to its dual space V ∗ is a contravariant functor
(vector spaces)◦ → (vector spaces) from the category to itself. Sending a topological space X to its qth integer
cohomology group H q (X, Z) is a contravariant functor (top spaces)◦ → (ab).
F
When C ◦ −→ D is a contravariant functor and f is a morphism in C, the map F (f ) is often denoted (ambiguously) by f ∗ , the superscript ∗ indicating that the direction is reversed. (I don’t know why one needs two
notations, ◦ and ∗, for the reversal of arrows.)
F
For emphasis, an ordinary functor is sometimes called a covariant functor. A contravariant functor C ◦ −→
D can be turned into a covariant functor by changing the category C to its dual category. The dual category C ◦
of a category C is a category whose objects correspond bijectively to the objects of C, and if the objects X ◦ , Y ◦
f◦
of C ◦ correspond to the objects X, Y of C, the morphisms X ◦ ←− Y ◦ correspond bijectively to morphisms
f
X −→ Y in C in the other direction. For instance, the dual category of the category of finite-type domains is
the category of affine varieties.
0.5
homfnctr
The Hom Functors
.
If X and Y are objects of a category C, we often denote the set of morphisms X → Y in C by Hom(X, Y )
or by (X, Y ).
α
Let Z −→ Z ′ be a morphism in C and let X be another element of C. Composition on the left with α
f
α
f
defines a map (Z, X) ← (Z ′ , X), that sends a morphism Z ′ −→ X to f α: Z −→ Z ′ −→ X. We denote this
map by ◦α:
◦α
(Z, X) ←− (Z ′ , X).
4
If we fix X, then sending Z
( · , X):
(Z, X) and α
◦α defines a contravariant functor to sets that we denote by
( · ,X)
C ◦ −→ (sets).
The symbol · stands for a variable object of the category C.
Similarly, composition on the right with α defines a map
α◦
(X, Z) −→ (X, Z ′ )
g
g
α
that sends a morphism X −→ Z to α ◦ g, X −→ Z −→ Z ′ , and this gives us a covariant functor
(X, · )
C −→ (sets)
0.6
Morphisms of Functors
morphfunctor
A fact that can be confusing at first is there can be maps between functors. Such a map is called a morphism,
or natural transformation from one functor to another.
Φ
Φ(X)
Let F and G be two functors C → D. A morphism F −→ G consists of morphisms F (X) −→ G(X) in
f
D, one for each object X of C, such that if X −→ Y is a morphism of objects of C, the following square of
morphisms in D commutes
F (f )
F (X) −−−−→ F (Y )




yΦ
yΦ
G(f )
G(X) −−−−→ G(Y )
For example, on the category of sets, let I be the identity functor and let P be the functor that sends a set
X to the product X × X. One can map I(X) to P (X): X → X × X, by the diagonal map ∆(X). The
∆
diagonal is is a morphism of functors I −→ P .
Or, let C denote the category of pointed path connected topological spaces. There are two functors from
this category to the category of groups, namely π1 and H1 . Here π1 (X) denotes the fundamental group of the
space X, and H1 (X) denotes homology with integer coefficients. It is a fact that H1 (X) is the abelianization
of π1 (X), the group obtained by introducing the commutative law into π1 . Abelianization defines a morphism
of functors π1 → H1 .
0.7
tensprod
Tensor Products
We have collected facts about tensor products that will be useful here.
Let U and V be modules over a ring R. The tensor product U ⊗ R V of U and V is an R-module
generated by elements u ⊗v called tensors, with u in U and v in V . Its elements are combinations of tensors
with coefficients in R. Since we can absorb a coefficient
from R into one of the factors of a tensor, every
P
element of U ⊗ R V can be written as a finite sum ui ⊗vi .
The module of relations among the tensors is generated by the following bilinear relations:
bilinrels
(0.7.1)
and
(u1 + u2 ) ⊗ v = u1 ⊗ v + u2 ⊗v ,
u ⊗ (v1 + v2 ) = u ⊗ v1 + u ⊗ v2
ur ⊗ v = u ⊗ rv
for all u in U , v in V , and r in R. The tensor symbol ⊗ is used as a reminder that the elements u ⊗ v are
manipulated using these relations.
colxrow
0.7.2. Example. Let U be the space of m dimensional (complex) column vectors, and let V be the space of
n-dimensional row vectors. Then U ⊗C V identifies naturally with the space of m × n-matrices.
5
There are a few remarks to be made about the definition of tensor product.
(1) The last of the bilinear relations states that scalars move through the tensor symbol. This is why we write
the term on the right of that relation as ur ⊗ v instead of as ru ⊗ v. As written, the relations define the tensor
product of a right module U , a module in which scalars act on the right, and a left module V . Since we are
working with commutative rings, right modules can be made into left modules simply by setting ru = ur.
Let’s agree that unless stated otherwise, scalar multiplication on the two sides supposed to be equal:
ru = ur
One can’t do this when the ring is noncommutative, and there are situations in which a right R-module is also
a left module over a different commutative ring.
(2) The tensor product U ⊗ V is made into a left R-module on using the structure of U as left module:
tensormodule
r(u ⊗ v) = (ru) ⊗ v
(0.7.3)
If there was no left module structure on U , the tensor product wouldn’t be a left module.
β
(3) There is an obvious map U ×V −→ U ⊗ R V from the product set to the tensor product that sends (u, v)
to u ⊗ v. However, this map isn’t a module homomorphism. As modules, the tensor product U ⊗R V and the
product U ×V aren’t closely related. For instance, if U and V are free modules of ranks r and s, then U ⊗R V
is free of rank is rs, while U ×V is free of rank r + s.
Instead, the map U × V → U ⊗ R V is bilinear. It is a universal bilinear map: Any R-bilinear map
fe
f
U×V −→ M to a module M can be obtained from a module homomorphism U ⊗R V −→ M by composition,
β
fe
f = fe ◦ β: U ×V −→ U ⊗ R V −→ M .
canonisom
0.7.4. Proposition. There are canonical isomorphisms
• U ⊗R R ≈ U ,
u ⊗ r ! ur
′
′
• (U ⊕ U ) ⊗R V ≈ (U ⊗R V ) ⊕ (U ⊗R V ),
(u1 + u2 ) ⊗ v ! u1 ⊗ v + u2 ⊗ v
and if R is commutative, then
• U ⊗R V ≈ V ⊗R U ,
u⊗v !v⊗u
• (U ⊗R V ) ⊗R W ≈ U ⊗R (V ⊗R W ),
(u ⊗ v) ⊗ w ! u ⊗ (v ⊗ w)
The proofs are very simple. We verify the “distributive law” (U ⊗R V ) ⊕ (U ′ ⊗R V ) ≈ (U ⊕ U ′ ) ⊗R V as
an example. The left side is generated by tensors u ⊗ v and u′ ⊗ v, and the relations are the bilinear relations in
the two summands. The right side is generated by tensors x ⊗ v, where x = u + u′ , with the bilinear relations
(x1 + x2 ) ⊗ v = x1 ⊗ v + x2 ⊗ v,
x ⊗ (v1 + v2 ) = x ⊗ v1 + x ⊗ v2 ,
xr ⊗ v = x ⊗ rv)
The relations defining the right side hold on the left side, and setting x = u + 0 and x = 0 + u′ shows that the
relations defining the left side hold in the right side.
prodbasis
rexacttensor
0.7.5. Corollary. If U and V are free R-modules with bases {ui } and {vj }, respectively, then U ⊗ R V is a
free R-module with basis {ui ⊗ vj }.
0.7.6. Proposition. The tensor product operation is right exact. If
f
g
U −→ U ′ −→ U ′′ → 0
is an exact sequence of R-modules, then for any R-module V , the sequence
f ⊗id
g⊗id
U ⊗ R V −→ U ′ ⊗ R V −→ U ′′ ⊗ R V → 0
is exact.
6
proof. Let Z be the image of f ⊗ id, and let W = (U ′ ⊗ R V )/Z. The composed map (g ⊗ id)(f ⊗ id)
is zero, so there is an induced map W → U ′′ ⊗ V . We must show that this map is invertible. To define its
inverse, we define a bilinear map U ′′ × V → W . Given a tensor u′′ ⊗ v in U ′′ ⊗ R V , we choose u′ in U ′
such that g(u′ ) = u′′ , and we map u′′ ⊗ v to the residue of u′ ⊗ v in W . This is well-defined because if u′1
and u′2 are elements of U ′ such that g(u′1 ) = g(u′2 ), then u′1 − u′2 is in the image of f , and (u′1 − u′2 ) ⊗ v is
in Z. The bilinear relations hold in W because they hold in U ′ ⊗R V , so this map corresponds to a module
homomorphism U ′′ ⊗ R V → W that inverts g ⊗id.
tensorrels
0.7.7. Corollary. (i) Let U, V be R-modules, and let M be an R-moatrix. Suppose that U is presented as
Rm /M Rn by an exact sequence
M
Rn −→ Rm → U → 0
Then U ⊗R V ≈ V m /M V n .
(ii) With notation as in (i), suppose that V is presented as Rk /N Rℓ by an exact sequence
N
Rℓ −→ Rk → V → 0
Then the map Rm ⊗R Rk → U ⊗R V is surjective. Its kernel is generated by the images of the two maps
I⊗N
M ⊗I
Rn ⊗R Rk −→ Rm ⊗ Rk and Rm ⊗R Rℓ −→ Rm ⊗ Rk .
proof. (ii) We form a diagram with exact rows and columns:
Rn ⊗ Rℓ −−−−→ Rm ⊗ Rℓ −−−−→ U ⊗ Rℓ −−−−→ 0






y
y
y
Rn ⊗ Rk −−−−→ Rm ⊗ Rk −−−−→ U ⊗ Rk −−−−→ 0






y
y
y
Rn ⊗ V −−−−→ Rm ⊗ V −−−−→ U ⊗ V −−−−→ 0






y
y
y
0
0
0
Then the assertion follows from a general fact. For any diagram
A −−−−→


y
A′ −−−−→


y
B −−−−→


y
B ′ −−−−→


y
0
0
C −−−−→


y
C ′ −−−−→


y
A′′ −−−−→ 0


y
B ′′ −−−−→ 0


y
C ′′ −−−−→ 0


y
0
′
whose rows and columns are exact, the composed map B → C ′′ is surjective, and its kernel is the sum of the
images of A′ and B in B ′ . The verification is a diagram chase.
tensornotexact
extendscalars
Example. This example shows that the tensor product operation isn’t exact. Let R = C[x], and let C denote
x
the R-module R/xR. When we tensor the exact sequence 0 → R −→ R → C → 0 with C, the result is the
0
non-exact sequence 0 → C −→ C → C → 0
(0.7.8)
extension of scalars in a module
7
ρ
Let R −→ S be a ring homomorphism. An S-module N can be made into an R-module, in which scalar
multiplication by an element a of R is defined to be multiplication by its image in S:
restrscalar
def
ax = ρ(a)x
(0.7.9)
This operation is called restriction of scalars.
For example, let ρ be the map C[t] → C that evaluates a polynomial p at 0. Restriction of scalars makes a
complex vector space V into a C[t]-module in which scalar multiplication is defined by p(t)v = p(0)v. This
example is trivial, as are all examples of the simple operation of restriction of scalars.
An R, S-bimodule is an abelian group that is a left R-module and a right S-module, and such that left and
right multiplications commute:
leftrightcommute
r(ms) = (rm)s
(0.7.10)
For example, if we are given a homomorphism R → S, the ring S becomes an R, S-bimodule in which
the left operation of R is by restriction of scalars. Then, given a right R-module M , the tensor product
M ′ = M ⊗R S becomes a right S-module, multiplication by s ∈ S being (m ⊗ a)s = m ⊗ (as). This gives a
functor
⊗S
R−modules −→ S −modules
that is called extension of scalars.
locistensor
fibremodule
0.7.11. Corollary. Let U and V be modules over a domain R and let s be a nonzero element of R. Let
Rs , Us , Vs be the (simple) localizations of R, U, V , respectively (see 2.8.11).
(i) There is a canonical isomorphism Us ≈ U ⊗R Rs .
(ii) Localization is compatible with tensor product: Us ⊗Rs Vs ≈ (U ⊗R V )s
(0.7.12)
fibres of a module
Let I be an ideal of a finite-type domain A, and let A = A/I. Also, let U be an A-module, and let U I be
the submodule generated by products uα with u in U and α in I. Tensor product with U givews us a diagram
0 −−−−→
U ⊗A I −−−−→ U ⊗A A −−−−→ U ⊗A A −−−−→ 0




≈
≈
y
y
y
UI
−−−−→
U
−−−−→ U/U I
−−−−→ 0
in which the left vertical arrow is surjective.
If I is the maximal ideal mp at a point p of X = Spec A, then A is the residue field k(p) at p, and
U ⊗A k(p) is the k(p)-module obtained from U by extension of scalars. We call this k(p)-module the fibre of
U at p, and we denote it by U (p). Then there is an exact sequence
Utensorktwo
(0.7.13)
0 → U mp → U → U (p) → 0
that we use to identify the fibre U (p) as U/U mp = U ⊗A k(p).
The support of a finite A-module U is the set of points p such that the fibre U (p) at p isn’t zero. The
support of a finite module is a closed subset of X = Spec A.
tensoralgebras
(0.7.14) tensor product algebras
If A and B are algebras over a ring R, the tensor product module A ⊗ R B is made into an R-algebra with
multiplication law
8
(α1 ⊗ β1 ) · (α2 ⊗ β2 ) = (α1 α2 ) ⊗ (β1 β2 )
and its multiplicative identity is 1⊗1. One must show compatibility of multiplication with the bilinear relations.
Since this is easy, we’ll do one verification as an example. We know that (α1 + α1′ ) ⊗ β1 = α1 ⊗ β1 + α1′ ⊗ β1 ,
so we must show that
(α1 + α1′ ) ⊗ β1 · α2 ⊗ β2 = α1 ⊗β1 · α2 ⊗β2 + α1′ ⊗β1 · α2 ⊗β2
Using the definition of multiplication, what is to be shown is that (α1 + α1′ )α2 ⊗ β1 β2 = α1 α2 ⊗ β1 β2 +
α1′ α2 ⊗β1 β2 , which is true.
tensorproperty
0.7.15. Proposition. (mapping property of tensor product algebras) Let A, B, and S be R-algebras. Algebra
homomorphisms from A ⊗ R B to S correspond bijectively to pairs of algebra homomorphisms from A and B
to S:
HomR (A ⊗ R B, S) ≈ HomR (A, S) × HomR (B, S).
proof. We note first that sending α
α ⊗ 1 defines an R-algebra homomorphism A → A ⊗ R B. This is
pretty clear: (α + α′ ) ⊗ 1 = α ⊗ 1 + α′ ⊗ 1, (αα′ ) ⊗ 1 = (α ⊗ 1)(α′ ⊗ 1), and (rα) ⊗ 1 = r(α ⊗ 1).
Similarly, β
1 ⊗ β defines an R-algebra homomorphism B → A ⊗ R B. This being so, an R-algebra
homomorphism A ⊗ R B → S gives us homomorphisms A → S and B → S by composition. Conversely, let
f
g
ϕ
algebra homomorphisms A −→ S and B −→ S be given. We define A ⊗R B −→ S by ϕ(α ⊗β) = f (α)g(β).
To show that ϕ is well-defined, one must verify the bilinear relations. We verify one as example:
ϕ(a ⊗ b) + ϕ(a′ ⊗ b) = f (a)g(b) + f (a′ )g(b) = f (a + a′ )g(b) = ϕ((a + a′ ) ⊗ b)
Then ϕ is a homomorphism because
ϕ(a1 ⊗ b1 )ϕ(a2 ⊗ b2 ) = f (a1 )g(b1 )f (a2 )g(b2 ) = f (a1 a2 )g(b1 b2 ) = ϕ(a1 a2 ⊗ b1 b2 )
prodaffvar
(0.7.16)
products of affine varieties
Let X = Spec A and Y = Spec B be affine varieties, and say that the coordinate rings are presented as
A = C[x1 , ..., xm ]/(f1 , ..., fk ) and B = C[y1 , ..., yn ]/(g1 , ..., gℓ ). So X and Y are subvarieties of Am and
An , respectively. In the product space Am+n with coordinates x, y, the product X ×Y of the two varieties is
the locus
f1 (x) = · · · = fk (x) = g1 (y) = · · · = gℓ (y) = 0.
It is an affine variety whose coordinate ring is C[x.y]/(f (x), g(y)). The algebra C[x, y]/ (f (x), g(y) is
isomorphic to the tensor product A ⊗C B.
tensorfg
0.7.17. Corollary. Let A = C[x]/(f ) and B = C[y]/(g). Then the product variety Spec A × Spec B is
isomorphic to the affine variety Spec A ⊗C B.
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