Field Coefficients

Notation Let R be a ring with unit element 1. It need not be commutative (yet). Here we use · to denote multiplication in R and for the action of R on an R-module; normally, both are omitted. We use parentheses () for grouping purposes only.

Some R-modules Here we give two examples.

Given an abelian group A, we make A ⊗ R an R-module by defining

s · (a ⊗ r) = a ⊗ (s · r) for a ∈ A, r, s ∈ R.  (1)

[Officially, one should define a pairing ŝ: A × R → A ⊗ R by the formula ŝ(a, r) = a ⊗ (s·r), observe that ŝ is bilinear, and deduce the desired homomorphism s·−: A⊗R → A⊗R. But nobody does this.] One can verify that A⊗R satisfies the module axioms. More generally, the same formula makes A ⊗ M an R-module for any R-module M.

One can also make Hom(A, M) an R-module for any R-module M. Given an element f ∈ Hom(A, M), i. e. homomorphism f: A → M, we define the homomorphism r · f: A → M by

(r · f)a = r · (fa) for a ∈ A, m ∈ M.  (2)

It is easy to see that r · f is indeed a homomorphism, and to verify that this action of R makes Hom(A, M) an R-module.

R-homomorphisms Given two R-modules L and M, a homomorphism f: L → M is R-linear if it satisfies

f(r · l) = r · (fl) for l ∈ L, r ∈ R.  (3)

Such homomorphisms form the subgroup Hom_{R}(L, M) ⊂ Hom(L, M). These are the morphisms in the category R-Mod of left R-modules.

Lemma 4 If R is commutative, Hom_{R}(L, M) is an R-submodule of the R-module Hom(L, M).

Proof Given f ∈ Hom_{R}(L, M) and r ∈ R, we have to show that r · f is R-linear. For any s ∈ R and l ∈ L, we compare

(r · f)(s · l) =^{(2)} r · (f(s · l)) =^{(3)} r · s · (fl)

with

s · ((r · f)l) =^{(2)} s · r · (fl).

Thus for any R-linear homomorphism f: L → M we have

(r · f)l =^{(2)} r · (fl) =^{(3)} f(r · l) for r ∈ R, l ∈ L.

Adjoint functors

Theorem 5 If A is an abelian group and M is an R-module, we have the natural isomorphism of abelian groups

ω: Hom(A, M) ≅ Hom_{R}(A ⊗ R, M).  (6)

If R is commutative, this is an R-linear isomorphism of R-modules.

Remark We have here a pair of adjoint functors: the forgetful functor U: R-Mod → Ab, and the functor F: Ab → R-Mod defined by F A = A ⊗ R, where Ab denotes the category of abelian groups. Then the isomorphism (6) can be stated in more categorical notation as

Ab(A, UM) ≅ R-Mod(F A, M).

Proof Given a homomorphism f: A → M, we define the R-linear homomorphism ωf = g: A ⊗ R → M by

(ωf)(a ⊗ r) = g(a ⊗ r) = r · (fa) for a ∈ A, r ∈ R.  (7)

The homomorphism g is R-linear, because if s ∈ R, we have

g(s · (a ⊗ r)) =^{(1)} g(a ⊗ (s · r)) =^{(7)} s · r · (fa) =^{(7)} s · g(a ⊗ r).

Moreover, ω is a homomorphism because ω(f + f′) = ωf + ωf′.

The inverse to ω is just as easy to define. Given an R-linear homomorphism g: A ⊗ R → M, the homomorphism ω^{−1}f = g: A → M is given by

(ω^{−1}g)a = fa = g(a ⊗ 1) for a ∈ A.  (8)

Then ω and ω^{−1} as defined really are inverse to each other, since

(ω^{−1}ωf)a =^{(8)} (ωf)(a ⊗ 1) =^{(7)} 1 · (fa) = fa

and

\begin{aligned}(ωω^{−1}g)(a ⊗ r) & =^{(7)} r · ((ω^{−1}g)a) =^{(8)} r · (g(a ⊗ 1)) \\ & =^{(3)} g(r · (a ⊗ 1)) =^{(1)} g(a ⊗ (r · 1)) = g(a ⊗ r). \end{aligned}

To check whether ω is R-linear, we have to compare ω(s · f) and s · (ωf), where s ∈ R and f: A → M is a homomorphism. We evaluate on a ⊗ r, and find

(ω(s · f))(a ⊗ r) =^{(7)} r · ((s · f)a) =^{(2)} r · s · (fa)

and

(s · (ωf))(a ⊗ r) =^{(2)} s · ((ωf)(a ⊗ r)) =^{(7)} s · r · (fa).

Vector spaces Now we take the coefficient ring R to be a field F, and the category R-Mod of R-modules becomes the category Vec of vector spaces over F. The contravariant duality functor D = Hom_{F} (−, F): Vec → Vec automatically takes cokernels to kernels. Because all short exact sequences in Vec split, D also takes kernels to cokernels, and so is exact. It therefore takes homology to cohomology: if E is any chain complex of vector spaces over F, D induces the isomorphism

H^{p}(DE) ≅ DH_{p}(E).  (9)

Theorem 10 Let C be any chain complex of abelian groups, and F a field. Then we have the universal coefficient isomorphism

H^{p}(C; F) ≅ Hom_{F} (H_{p}(C; F), F).

Proof First, we have to get into Vec by setting E = C ⊗ F, which is a chain complex in Vec by equation (1). By definition, H_{p}(C; F) = H_{p}(E) and H^{p}(C; F) = H^{p}(Hom(C, F)). But by Theorem 5, Hom(C, F) ≅ Hom_{F} (E, F) = DE, and equation (9) becomes the desired isomorphism.