6

Elements of Homological Algebra with Applications

Homological algebra has become an extensive area of algebra since its introduction in the mid-1940’s. Its first aspect, the cohomology and homology groups of a group, was an outgrowth of a problem in topology that arose from an observation by Witold Hurewicz that the homology groups of a path-connected space whose higher homotopy groups are trivial are determined by the fundamental group π₁. This presented the problem of providing a mechanism for this dependence. Solutions of this problem were given independently and more or less simultaneously by a number of topologists: Hopf, Eilenberg and MacLane, Freudenthal, and Eckmann [see Cartan and Eilenberg (1956), p. 137, and MacLane (1963), p. 185, listed in References]. All of these solutions involved homology or cohomology groups of π₁. The next step was to define the homology and cohomology groups of an arbitrary group and to study them for their own sake. Definitions of the cohomology groups with coefficients in an arbitrary module were given by Eilenberg and MacLane in 1947. At the same time, G. Hochschild introduced cohomology groups for associative algebras. The cohomology theory of Lie algebras, which is a purely algebraic theory corresponding to the cohomology theory of Lie groups, was developed by J. L. Koszul and by Chevalley and Eilenberg. These disparate theories were pulled together by Cartan and Eilenberg (1956) in a cohesive synthesis based on a concept of derived functors from the category of modules over a ring to the category of abelian groups. The derived functors that are needed for the cohomology and homology theories are the functors Ext and Tor, which are the derived functors of the hom and tensor functors respectively.

Whereas the development of homological algebra proper dates from the period of the Second World War. several important precursors of the theory appeared earlier. The earliest was perhaps Hilbert’s syzygy theorem (1890) in invariant theory, which concerned free resolutions of modules for the ring of polynomials in m indeterminates with coefficients in a field. The second cohomology group H²(G. *) with coefficients in the multiplicative group * of non-zero complex numbers appeared in Schur’s work on projective representations of groups (1904). More general second cohomology groups occurred as factor sets in Schreier’s extension theory of groups (1926) and in Emmy Noether’s construction of crossed product algebras (1929). The third cohomology group appeared first in a paper by O. Teichmüller (1940).

In this chapter we shall proceed first as quickly as possible to the basic definition and results on derived functors. These will be specialized to the most important instances: Ext and Tor. In the second half of the chapter we shall consider some classical instances of homology theory: cohomology of groups, cohomology of algebras with applications to a radical splitting theorem for finite dimensional algebras due to Wedderburn, homological dimension of modules and rings, and the Hilbert syzygy theorem. Later (sections 8.4 and 8.5), we shall present another application of homology theory to the Brauer group and crossed products.

6.1   ADDITIVE AND ABELIAN CATEGORIES

A substantial part of the theory of modules can be extended to a class of categories called abelian. In particular, homological algebra can be developed for abelian categories. Although we shall stick to modules in our treatment, we will find it convenient to have at hand the definitions and simplest properties of abelian categories. We shall therefore consider these in this section.

We recall that an object 0 of a category C is a zero object if for any object A of C, hom_c(A, 0) and hom_c(0, A) are singletons. If 0 and 0′ are zero objects, then there exists a unique isomorphism 0 → 0′ (exercise 3, p. 36). If A, B ∈ obC, we define 0_A.B as the morphism 0_0B0_A0 where 0_A0 is the unique element of hom_c(A,0) and 0_0B is the unique element of hom_c(0, B). It is easily seen that this morphism is independent of the choice of the zero object. We call 0_A,B the zero morphism from A to B. We shall usually drop the subscripts in indicating this element.

We can now give the following

DEFINITION 6.1.   A category C   is called additive if it satisfies the following conditions:

              AC1. C   has a zero object.

              AC2. For every pair of objects (A,B) in C, a binary composition + is defined on the set hom_C(A,B) such that (hom_C(A, B), + ,0_{A, B}) is an abelian group.

              AC3. If A,B,C ∈ obC,f,f₁,f₂ ∈ hom_C(A,B), and g, g₁, g₂ ∈ hom_C(B,C), then

              AC4. For any finite set of objects {A₁,…, A_n} there exists an object A and morphisms Pj : A → A_j, i_j : A_j → A, 1 ≤ j ≤ n, such that

We remark that AC2 means that we are given, as part of the definition, an abelian group structure on every hom_C(A , B) whose zero element is the categorically defined 0_{A, B}. AC3 states that the product fg, when defined in the category, is bi-additive. A consequence of this is that for any A, (hom_C(A, A), + , ·, 0, 1 = 1_A) is a ring. We note also that AC4 implies that (A, {p_j}) is a product in C of the A_j, 1 ≤ j ≤ n. For, suppose B ∈ obC and we are given .Then p_kf = f_k by (1) and if p_kf′ = f_k for 1 ≤ k ≤ n, then (1) implies that Hence f is the only morphism from B to A such that p_kf = f_k, 1 ≤ k ≤ n, and (A,{p_j}) is a product of the Aj. In a similar manner we see that (A, {i_j}) is a coproduct of the A_j.

It is not difficult to show that we can replace AC4 by either

              AC4′. C is a category with a product (that is, products exist for arbitrary finite sets of objects of C), or

              AC4″. C is a category with a coproduct.

We have seen that AC4 AC4′ and AC4″ and we shall indicate in the exercises that AC1–3 and AC4′ or AC4″ imply ACR. The advantage of AC4 is that it is self-dual. It follows that the set of conditions defining an additive category is self-dual and hence if C is an additive category, then C^op is an additive category. This is one of the important advantages in dealing with additive categories.

If R is a ring, the categories R-mod and mod-R are additive. As in these special cases, in considering functors between additive categories, it is natural to assume that these are additive in the sense that for every pair of objects A,B, the map F of hom(A,B) into hom(FA,FB) is a group homomorphism. In this case, the proof given for modules (p. 98) shows that F preserves finite products (coproducts).

We define next some concepts that are needed to define abelian categories. Let C be a category with a zero (object), f : A →B in C. Then we call k : K → A a kernel of f if (1) k is monic, (2) fk = 0, and (3) for any g : G → A such that fg = 0 there exists a g′ such that g = kg′. Since k is monic, it is clear that g′ is unique. Condition (2) is that

is commutative and (3) is that if the triangle in

is commutative, then this can be completed by g′ : G → K to obtain a commutative diagram. It is clear that if k and k′ are kernels of f, then there exists a unique isomorphism u such that k′ = ku.

In a dual manner we define a cokernel of f as a morphism c : B → C such that (1) c is epic, (2) cf = 0, and (3) for any h :B → H such that hf = 0 there exists h′such that h = h′c.

If f : A → B in R-mod, let K = ker f in the usual sense and let k be the injection of K in A. Then k is monic, fk = 0, and if g is a homomorphism of G into A such that fg = 0, then gG ⊂ K. Hence if we let g′ be the map obtained from g by restricting the codomain to K, then g = kg′. Hence k is a kernel of f. Next let C = B/fA and let c be the canonical homomorphism of B onto C. Then c is epic, cf = 0, and if h : B → H satisfies hf = 0, then fA ⊂ ker h. Hence we have a unique homomorphism h′ : C = B/fA → H such that

is commutative. Thus C is a cokernel of f in the category R-mod.

We can now give the definition of an abelian category

DEFINITION 6.2.   A category C is abelian if it is an additive category having the following additional properties:

              AC5. Every morphism in C has a kernel and a cokernel.

              AC6. Every monic is a kernel of its cokernel and every epic is a cokernel of its kernel.

              AC7. Every morphism can be factored as f = me where e is epic and m is monic.

We have seen that if R is a ring, then the categories R-mod and mod-R are additive categories satisfying AC5. We leave it to the reader to show that AC6 and AC7 also hold for R-mod and mod-R. Thus these are abelian categories.

EXERCISES

        1. Let C be a category with a zero. Show that for any object A in C, (A, 1_A,0) is a product and coproduct of A and 0.

        2. Let C be a category, (A, p₁, p₂) be a product of A₁ and A₂ in C, (B, q₁, q₂) a product of B₁ and B₂ in C, and let h_i : B_i → A_i. Show that there exists a unique f : B → A such that h_iq_i = p_if. In particular, if C has a zero and we take (B, q₁, q₂) = (A₁, 1_A1, 0) then this gives a unique i₁ : A₁ → A such that p₁i₁ = 1_A1, p₂i₁ = 0. Similarly, show that we have a unique i₂ : A₂ → A such that p₁i₂ = 0, p₂i₂ = 1_A2. Show that (i₁p₁ + i₂p₂)i₁ = i₁ and (i₁p₁ + i₂p₂)i₂ = i₂. Hence concluded that i₁p₁ + i₂p₂ = 1_A. Use this to prove that the conditions AC1 – AC3 and AC4′ AC4. Dualize to prove that AC1 – AC3 and AC4″ AC4.

        3. Show that if A and B are objects of an additive category, then 0_{A, B} = 0_{0, B}0_{A, 0}, where 0_{A, 0} is the zero element of hom(A,0), 0_{0, B} is the zero element of hom(0, B), and 0_{A, B} is the zero element of the abelian group hom(A, B).

        4. Let be a product of the objects and A₁ and A₂ in the category C. If f_j : B → A_j, denote the unique f : B → A₁ Π A₂ such that p_jf = f_j by f₁ Π f₂. Similarly, if (A₁ A₂, i₁, i₂) is a coproduct and g_j : A_j → C, write g₁ g₂ for the unique g : A₁ A₂ → C such that gi_j = g_j. Note that if C is additive with the i_j and p_j as in AC4, then f₁ Π f₂ = i₁f₁ + i₂f₂ and g₁ g₂ = g₁p₁ + g₂p₂. Hence show that if i₁, i₂, p₁, p₂ are as in AC4, so A = A₁ Π A₂ = A₁ A₂, then

(from B →C). Specialize A₁ = A₂ = 0, g₁ = g₂ = 1_c to obtain the formula

for the addition in hom_C(B, C).

        5. Use the result of exercise 4 to show that if F is a functor between additive categories that preserves products and coproducts, then F is additive.

6.2   COMPLEXES AND HOMOLOGY

The basic concepts of homological algebra are those of a complex and homomorphisms of complexes that we shall now define.

DEFINITION 6.3.   If R is a ring, a complex (C, d) for R is an indexed set C = {Ci} of R-modules indexed by together with an indexed set d = {d_i|i ∈ } of R-homomorphisms d_i : C_i → C_{i – 1} such that d_{i – 1}d_i = 0 for all i. If (C, d) and (C′, d′) are R-complexes, a (chain) homomorphism of C into C′ is an indexed set of homomorphisms α_i :C_j → C_i′ such that we have the commutativity of

for every i. More briefly we write αd = d′.α

These definitions lead to the introduction of a category R-comp of complexes for the ring R. Its objects are the R-complexes (C, d), and for every pair of R-complexes (C, d), (C′, d′), the set hom(C, C′) is the set of chain homomorphisms of (C,d) into (C′, d′). It is clear that these constitute a category, and as we proceed to show, the main features of R-mod carry over to R-comp. We note first that hom(C, C′) has a natural structure of abelian group. This is obtained by defining α + β for α, β ∈ hom(C, C′) by (α + β)_i = α_i + β_i. The commutativity α_{i – 1}d_i = d′_iα_i, β_i–1d_i = d′_iβ_i gives (α_{i – 1} + β_{i – 1})d_i = d′_i(α_i + β_i), so α + β ∈ hom(C, C′). Since hom_R(C_i, C′_i) is an abelian group, it follows that hom(C, C′) is an abelian group. It is clear also by referring to the module situation that we have the distributive laws γ(α + β) = γα + γβ, (α + β)δ = αδ + βδ when these products of chain homomorphisms are defined. If (C, d) and (C′, d^r) are complexes, we can define their direct sum (C C′, d d′) by (C + C′)i = C_i C_i′, d_i d_i′ defined component-wise from . It is clear that , so is indeed a complex. This has an immediate extension to direct sums of more than two complexes. Since everything can be reduced to the module situation, it is quite clear that if we endow the hom sets with the abelian group structure we defined, then the category R-comp becomes an abelian category.

The interesting examples of complexes will be encountered in section 4. However, it may be helpful to list some at this point, although most of these will appear to be rather special.

EXAMPLES

1. Any module M becomes a complex in which C_l, = M, i ∈ , and d_i = 0 : C_i → C_{i – 1}.

2. A module with differentiation is an R-module equipped with a module endomorphism δ such that δ² = 0. If (M,δ) is a module with differentiation, we obtain a complex (C, d) in which C_ι = 0 for i ≤ 0, C_l = C₂ = C₂ = C₃ = M, C_j = 0 for j > 3, d₂ = d₃ = δ, and d = 0 if i ≠ 2, 3.

3. Let (M, δ) be a module with a differentiation that is -graded in the following sense: where the M_i are submodules and δ(M_ι) M_{i – 1} for every i. Put C_i = M_ι and d_i = δ|M_ι. Then C = {C_ι}, d = {d_i} constitute an R-complex.

4. Any short exact sequence defines a complex in which if j ≠ 2.3.

We shall now define for each i ∈ a functor, the ith homology functor, from the category of R-complexes to the category of R-modules. Let (C, d) be a complex and let Z_i(C) = ker d_i so Z_i(C) is a submodule of C_i. The elements of z_i are called i-cycles. Since d_id_{i + l} = 0, it is clear that the image d_{i + 1}C_{i + 1} is a submodule of Z_i. We denote this as B_i = B_i(C) and call its elements i-boundaries. The module H_i = H_i(C) = Z_i/B_i is called the ith homology module of the complex (C, d). Evidently, is exact if and only if H_i(C) = 0 and hence the infinite sequence of homomorphisms

is exact if and only if H_i(C) = 0 for all i.

Now let α be a chain homomorphism of (C, d) into the complex (C′,d′). The commutativity condition on (2) implies that and . Hence the map , is a homomorphism of Z_i into H′_i = H_i(C′) = Z′_i/B′_i sending B_i into 0. This gives the homomorphism of H_i(C) into H_i(C′) such that

It is trivial to check that the maps (C, d) H_i(C), hom(C, C′) → hom(H_i(C), H_i(C′)), where the latter is α , define a functor from R-comp to R-mod. We call this the ith homology functor from R-comp to R-mod. It is clear that the map α is a homomorphism of abelian groups. Thus the ith homology functor is additive.

In the situations we shall encounter in the sequel, the complexes that occur will have either C_i = 0 for i < 0 or C_i = 0 for i > 0. In the first case, the complexes are called positive or chain complexes and in the second, negative or cochain complexes. In the latter case, it is usual to denote C_{– i} by Cⁱ and d _{– i} by dⁱ. With this notation, a cochain complex has the appearance

if we drop the C^–i, i > 1. It is usual in this situation to denote ker dⁱ by Zⁱand d^{i – 1}C^{i – 1} by Bⁱ. The elements of these groups are respectively i-cocycles and i-coboundaries and Hⁱ = Zⁱ/Bⁱ is the ith cohomology group. In the case of if H⁰, we have H⁰ = Z⁰. A chain complex has the form . In this case H₀ = C₀/d₁C₁ = coker d₁

EXERCISES

        1. Let α be a homomorphism of the complex (C, d) into the complex (C′, d′). Define and if , define Verify that (C″, d″) is a complex.

        2. Let (C, d) be a positive complex over a field F such that Σdim C_i < ∞ (equivalently every C_i is finite dimensional and C_n = 0 for n sufficiently large). Let r_i = dim C_i, ρ_i = dim H_i(C). Show that Σ(–1)ⁱρ_i = Σ(–1)ⁱr_i.

        3. (Amitsur’s complex.) Let S be a commutative algebra over a commutative ring K. Put , n factors, where means _K. Note that for any n we have n + 1 algebra isomorphisms , of Sⁿ into Sⁿ⁺¹ such that

For any ring R let U(R) denote the multiplicative group of units of R. Then . Define , by

(e.g., d²u = (δ¹u)^–1(δ²u)(δ^-3u)^–1). Note that if i ≥ j, then δⁱ⁺¹δ^j = δ^jδⁱ and use this to show that d^{n + 1}dⁿ = 0, the map u 1. Hence conclude that {U(Sⁿ), dⁿ|n ≥ 0} is a cochain complex.

6.3   LONG EXACT HOMOLOGY SEQUENCE

In this section we shall develop one of the most important tools of homological algebra: the long exact homology sequence arising from a short exact sequence of complexes. By a short exact sequence of complexes we mean a sequence of complexes and chain homomorphisms such that is exact for every i ∈ , that is, α_i is injective, β_i is surjective, and ker β_i = im α_i. We shall indicate this by saying that 0 → C′ → C → C′ → 0 is exact. We have the commutative diagram

in which the rows are exact. The result we wish to prove is

THEOREM 6.1.   Let be an exact sequence of complexes. Then for each i ∈ we can define a module homomorphism so that the infinite sequence of homology modules

is exact.

Proof. First, we must define Δ_i. Let z″_i ∈ Z_i(C″), so d″_iz″_i = 0. Since β_i is surjective, there exists a c_i ∈ C_i such that β_ic_i = z″_i. Then β_{i – l}d_ic_i = d″_iβ_ic_i = d″_iz″_i = 0. Since ker β_{i – 1} = im α_{i – 1} and α_{i – 1} is injective, there exists a unique z′_{i – 1} ∈ C′_{i – 1} such that α_{i – 1}z′_{i – 1} = d_ic_i. Then α_{i – 2}d′_{i – 1}z′_{i – 1} = d_{i – 1} α_{i – 1} z′_{i – 1} = d_{i– 1}d_ic_i = 0. Since α_{i – 2} is injective, d′_{i – 1}z′_{i – 1} = 0 so z′_{i – 1} ∈ Z_{i – 1}(C′). Our determination of z′_{i – 1} can be displayed in the formula

where β^{– 1}_i( ) and α^{– 1}_{i– 1}( )denote inverse images. We had to make a choice of c_i ∈ β_i^{– 1}(z″_i) at the first stage. Suppose we make a different one, say, . Then implies that c. Then . Thus the replacement of c_i by replaces z′_{i – 1} by z′_{i– 1} + d′_ic′_i. Hence the coset z′_{i– 1} + B_{i– 1}(C′) in H_{i– 1}(C′) is independent of the choice of c_i and so we have a map

of Z_i(C″) into H_{i – 1}(C′). It is clear from (5) that this is a module homomorphism. Now suppose z″_i ∈ B_i(C″), say, . Then we can choose c_i+1 ∈ C_i+1 so that β_{i + 1}c_{i + 1} = c″_i+1 and then . Hence and since , we have z′_{i – 1} = 0 in (5). Thus B_i(C″) is in the kernel of the homomorphism (6) and

is a homomorphism of H_i(C″)into H_{i– 1}(C′).

We claim that this definition of Δ_i makes (4) exact, which means that we have .

I. It is clear that , so Suppose z_i ∈ Z_i(C) and . There exists a c_{i + 1} ∈ C_{i + 1} such that β_{i + 1}c_{i + 1} = C″_{i + 1} and so = 0. Then there exists z′_i ∈ C′_i such that α_iz′_i = z_i-d_{i + 1}c_{i + 1}. Then α_{i– 1}d′_iz′_i. = = 0. Hence d′_iz′_i = 0 and z′_i ∈ Z_i(C′). Now . Thus z_i + B_i(C) ∈ im and hence ker ⊂ im and hence ker = im .

II. Let z_i ∈ Z_i(C) and let z″ = β_iz_i. Then Δ_i(z″_i+B_i(C″)) = 0 since z_i ∈ β_i ^–1(z″_i) and d_iz_i = 0, so α_{i– 1}0 = d_iZ_i. Thus Δ_i(z_i + B_i(C)) = 0 and im ⊂ ker Δ_i. Now suppose z″_i ∈Z_i(C″) satisfies Δ_i(z″_i + B_i(C″)) = 0. This means that if we choose c_i ∈ C_i so that β_ic_i = z″_i and z′_{i – 1} ∈C′_{i – 1} so that α_{i– 1}z′_{i – 1} = d_ic_i, then z′_{i – 1} = d′_ic′_i for some c′ ∈ C′_ii. Then and d_i(c_i – α_ic′_i) = 0. Also . Hence, if we put z_i = c_i – α_ic′_i, then we shall have z_i ∈Z_i(C) and . Thus ker Δ_i ⊂ im and hence ker Δ_i = im .

III. If z′_{i – 1} ∈Z_{i – 1} and z′_{i – 1} + B_{i – 1}(C′) ∈ im Δ_i, then we have a z″_i ∈ Z_i(C″) and a c_i ∈ C_i such that β_ic_i = z″_i and α_i–1z′_i–1 = d_ic_i. Then . Hence im Δ_i⊂ker _–1. Conversely, let z′_i–1 + . Then . Put z″ = β_ic_i. Then d″_iz″_i = = 0, so z″_i ∈ Z_i(C″). The definition of Δ_i shows that . Thus ker _–1 ⊂ im Δ_i and hence ker _{– 1} = im Δ_i.

The homomorphism Δ_i that we constructed is called the connecting homomorphism of H_i(C″) into H_{i – 1}(C′) and (4) is the long exact homology sequence determined by the short exact sequence of complexes . An important feature of the connecting homomorphism is its naturality, which we state as

THEOREM 6.2.   Suppose we have a diagram of homomorphisms of complexes

which is commutative and has exact rows. Then

is commutative.

By the commutativity of (8) we mean of course the commutativity of

for every i. The proof of the commutativity of (9) is straightforward and is left to the reader.

EXERCISE

        1. (The snake lemma.) Let

be a commutative diagram of module homomorphisms with exact columns and middle two rows exact. Let x″ ∈ K″ and let y ∈ M satisfy µy=f″x″. Then and there exists a unique z′ ∈ N′ such that v′z′ = gy. Define Δx″ = h′z′. Show that Δx″ is independent of the choice of y and that Δ : K″ → C′ is a module homomorphism. Verify that

is exact. Show that if µ′ is a monomorphism, then so is K′ and if v is an epimorphism, then so is γ.

6.4   HOMOTOPY

We have seen that a chain homomorphism α of a complex (C,d) into a complex (C′,d′) determines a homomorphism of the ith homology module Hi(C) into H_i(C′) for every i ∈ . There is an important relation between chain homomorphisms of (C,d) to (C′,d′) that automatically guarantees that the corresponding homomorphisms of the homology modules are identical. This is defined in

DEFINITION 6.4.   Let α and β be chain homomorphisms of a complex (C, d) into a complex (C′, d′). Then α is said to be homotopic to β if there exists an indexed set s = {s_i} of module homomorphisms s_i : C_i → C′_{i + 1}, i ∈ , such that

We indicate homotopy by α ~ β.

If α ~ β, then = for the corresponding homomorphisms of the homology modules H_i(C) → H_i(C′). For, if z_i ∈ Z_i(C), then and . Hence

.

It is clear that homotopy is a symmetric and reflexive relation for chain homomorphisms. It is also transitive, since if α ~ β is given by s and β ~ γ is given by t, then

Hence

Thus s + t = {s_i + t_i} is a homotopy between α and γ.

Homotopies can also be multiplied. Suppose that α ~ β for the chain homomorphisms of (C, d) → (C′, d′) and γ ~ δ for the chain homomorphisms (C′, d′) → (C″, d″). Then γα ~ δβ. We have, say,

Multiplication of the first of these by γ_i on the left and the second by β_i on the right gives

(by (2)). Hence

Thus γ α ~ δ β via u = {u_i} where u_i = γi + 1s_i + t_iβ_i,

6.5   RESOLUTIONS

In the next section we shall achieve the first main objective of this chapter : the definition of the derived functor of an additive functor from the category of modules of a ring to the category of abelian groups. The definition is based on the concept of resolution of a module that we now consider.

DEFINITION 6.5.   Let M be an R-module. We define a complex over M as a positive complex C = (C, d) together with a homomorphism ε : C₀ → M, called an augmentation, such that εd₁ = 0. Thus we have the sequence of homomorphisms

where the product of any two successive homomorphisms is 0. The complex C, ε over M is called a resolution of M if (10) is exact. This is equivalent to H_i(C) =0 for i > 0 and H₀(C) = C₀/d₁C₁ = C₀/ker ε M. A complex C,ε over M is called projective if every C_i is projective.

We have the following important

THEOREM 6.3.   Let C, ε be a projective complex over the module M and let C, ε′ be a resolution of the module M′, μ a homomorphism of M into M′. Then there exists a chain homomorphism α of the complex C into C′ such that µε = ε′α₀. Moreover, any two such homomorphisms α and β are homotopic.

Proof.   The first assertion amounts to saying that there exist module homomorphisms α_i, i > 0, such that

is commutative. Since C₀ is projective and is exact, the homomorphism µε of C₀ into M′ can be “lifted” to a homomorphism α₀ : C₀ → C′₀ so that µε = ε′α₀. Now suppose we have already determined α₀, …, α_n–1 so that the commutativity of (11) holds from C₀ to C_{n – 1}. We have . Hence α_{n – 1}d_nC_n ⊂ ker d′_{n – 1} = im d′_n = d′_nC′_n. We can replace C′_{n– 1} by d′_nC′_n for which we have the exactness of C′_n → d′_nC′_n → 0. By the projectivity of C_n we have a hommomorphism α_n : C_n → C′_n such that d′_n α_n = α_n–1d_n. This inductive step proves the existence of α. Now let α and β satisfy the conditions. Let γ = α – β. Then we have

We have the diagram

Since ε′ γ₀ = 0, γ₀C₀ ⊂ d′₁C′₁ We have the diagram

with exact row. As before, there exists a homomorphism s₀ : C₀ → C′₁ such that γ₀ = d′₁s₀. Suppose we have determined s₀, …, s_n–1 such that s_i : C_i → C′_{i + 1} and

Consider γ_n–s_n–1d_n. We have . It follows as before that there exists a homomorphism s_n : C_n → C_{n + 1} such that d′_{n + 1}s_n = γ_n – s_{n – 1}d_n. The sequence of homomorphisms s₀, s₁, … defines a homotopy of α to β. This completes the proof.

The existence of a projective resolution of a module is easily established. In fact, as we shall now show, there exists a resolution (10) of M that is free in the sense that the modules C_i are free. First, we represent M as a homomorphic image of a free module C₀, which means that we have an exact sequence ker ε C₀ M → 0 where C₀ is free and i is the injection of ker ε. Next we obtain a free module C₁ and an epimorphism π of C₁ onto ker ε. If we put d₁ = iπ : C₁ → C₀, we have the exact sequence C₁ C₀ M → 0. Iteration of this procedure leads to the existence of an exact sequence

where the C_i are free. Then (C, d) and ε constitute a free resolution for M.

All of this can be dualized. We define a complex under M to be a pair D, η where D is a cochain complex and η is a homomorphism M → D⁰ such that d⁰η = 0. Such a complex under M is called a coresolution of M if

is exact. We have shown in section 3.11 (p. 159) that any module M can be embedded in an injective module, that is, there exists an exact sequence 0 → M D⁰ where D⁰ is injective. This extends to 0 → M → D⁰ coker η where coker η = D⁰/ηM and π is the canonical homomorphism onto the quotient module. Next we have a monomorphism η₁ of coker η into an injective module D¹ and hence we have the exact sequence 0 → M D⁰ D¹ where d⁰ = η₁π. Continuing in this way, we obtain a coresolution (12) that is injective in the sense that every Dⁱ is injective. The main theorem on resolutions can be dualized as follows.

THEOREM 6.4.   Let (D, η) be an injective complex under M, (D′, η′) a coresolution of M′, λ a homomorphism of M′ into M. Then there exists a homomorphism g of the complex D′ into the complex D such that ηλ = g⁰η′. Moreover, any two such homomorphisms are homotopic.

The diagram for the first statement is

The proof of this theorem can be obtained by dualizing the argument used to prove Theorem 6.3. We leave the details to the reader.

6.6   DERIVED FUNCTORS

We are now ready to define the derived functors of an additive functor F from a category R-mod to the category Ab. Let M be an R-module and let

be a projective resolution of M. Applying the functor F we obtain a sequence of homomorphisms of abelian groups

Since F(0) = 0 for a zero homomorphism of a module into a second one and since F is multiplicative, the product of successive homomorphisms in (13) is 0 and so FC = {FC_i}, F(d) = {F(d_i)} with the augmentation Fε is a (positive) complex over FM. If F is exact, that is, preserves exactness, then (13) is exact and the homology groups H_i(FC) = 0 for i ≥ 1. This need not be the case if F is not exact, and these homology groups in a sense measure the departure of F from exactness. At any rate, we now put

This definition gives

since we are taking the terms FC_i = 0 if i < 0.

Let M′ be a second R-module and suppose we have chosen a projective resolution 0 ← M′ C₀ C′₁ … of M′. from which we obtain the abelian groups H_n(FC′), n ≥ 0. Let μ be a module homomorphism of M into M′. Then we have seen that we can determine a homomorphism a of the complex (C, d) into (C′, d′) such that με = ε′α₀. We call α a chain homomorphism over the given homomorphism μ. Since F is an additive functor, we have the homomorphism F(α) of the complex FC into the complex FC′ such that F(μ)F(ε) = F(ε′)F(α₀). Thus we have the commutative diagram

Then we have the homomorphism of H_n(FC) into H_n(FC′). This is independent of the choice of α. For, if β is a second homomorphism of C into C′ over μ, β is homotopic to α, so there exist homomorphisms s_n : C_n → C′_{n + 1}, n ≥ 0, such that α_n – β_n = d′_{n + 1}s_n + s_{n − 1} d_n. Since F is additive, application of F to these relations gives

Thus F(α) ~ F(β) and hence = . We now define L_nF(μ) = . Thus a homomorphism μ : M → M′ determines a homomorphism L_nF(μ) : H_n(FC) → H_n(FC′). We leave it to the reader to carry out the verification that L_nF defined by

is an additive functor from R-mod to Ab. This is called the nth left derived functor of the given functor F.

We now observe that our definitions are essentially independent of the choice of the resolutions. Let be a second projective resolution of M. Then taking μ = 1 in the foregoing argument, we obtain a unique isomorphism η_n of H_n(FC) onto H_n(F). Similarly, another choice ^′ of projective resolution of M′ yields a unique isomorphism η′_n of H_n(FC′) onto H_n(F^′) and L_nF is replaced by η′_n(L_nF)η_n ^{– 1}.

From now on we shall assume that for every R-module M we have chosen a particular projective resolution and that this is used to determine the functors L_nF. However, we reserve the right to switch from one such resolution to another when it is convenient to do so.

We consider next a short exact sequence 0 → M′ M M″ → 0 and we shall show that corresponding to such a sequence we have connecting homomorphisms

such that

is exact. For this purpose we require the existence of projective resolutions of short exact sequences of homomorphisms of modules. By a projective resolution of such a sequence 0 → M′ → M → M″ → 0 we mean projective resolutions C′, ε′, C, ε, C″, ε″ of M′, M, and M″ respectively together with chain is homomorphisms i: C′ → C, p : C → C″ such that for each n, 0 → C′_n → C_n → C″_n → 0 is exact and

is commmutative.

We shall now prove the existence of a projective resolution for any short exact sequence of modules 0 → M′ M M″ → 0. We begin with projective resolutions C′, ε′, and C″, ε″ of M′ and M″ respectively:

We let C_n = C′_n C″_n, n = 1, 2, 3, …, i_nx′_n = (x′n, 0) for x′_n ∈ C′_n, p_nx_n = x″_n for x_n = (x′_n, x″_n). Then 0 → C′_n → C′_n C″_n → C″_n → 0 is exact and C_n is projective. We now define εx₀ = αε′x′₀ + σx″₀, d_nx_n = (d′_nx′_n + θ_nx″_n, d_n″x_n″) where σ : C″₀ → M′, θ_n : C′_n → C′_n–1 are to be determined so that C, ε is a projective resolution of M which together with C′, ε′ and C″, ε″ constitutes a projective resolution for the exact sequence 0 → M′ M M″ → 0.

We have εi₀x′₀ = ε(x′₀, 0) = αε′x′₀. Hence commutativity of the left-hand rectangle in (16) is automatic. Also ε″p₀x₀ = ε″x″₀ and βεx₀ = β(αε′x′₀ + σx″₀) = βσx″₀. Hence commutativity of the right-hand rectangle in (16) holds if and only if

We have εd₁x₁ = ε(d′₁x′₁ + θx″₁, d₁″x₁″) = αε′θ₁x″ + σd″₁x₁″. Hence εd₁ = 0 if and only if

Similarly, the condition d_{n– 1}d_n = 0 is equivalent to

Now consider the diagram

Since C′₀ is projective there exists a σ: C″0 → M′ such that (18) holds. Next we consider

Since ε′C₀′ = M′ the row is exact, and since C₁″ is projective and βσd′₁ = ε″d″₁ = 0, there exists a θ₁: C₁″ → C′₀ such that (19) holds (see exercise 4, p. 100). Next we consider

Here again the row is exact, C₂″ is projective and ε′θ₁d″₂ = 0 since αε′θ₁d₂″ = – σd″₁d″₂ = 0 and ker α = 0. Hence there exists θ₂: C₂″ → C′₁ such that (19) holds for n = 2. Finally, the same argument establishes (20) for n > 2 using induction and the diagram

It remains to show that … C₂ C₁ C₀ M → 0 is a resolution. For this purpose we regard this sequence of modules and homomorphisms as a complex and similarly we regard the modules and homomorphisms in (17) as complexes and ″. Then we have an exact sequence of complexes 0 → ′ → → ″ → 0. Since H_i(′) = 0 = H_i(′) it follows from (4) that H_i() = 0. Then provides a resolution for M which together with the resolutions for M′ and M″ gives a projective resolution for 0 → M′ → M → M″ → 0.

We can now prove

THEOREM 6.5.   Let F be an additive functor from R-mod to Ab.   Then for any short exact sequence of R-modules 0 → M′ M M″ → 0 and any n = 1, 2, 3,… there exists a connecting homomorphism Δ_n:L_nFM″ L_{n– 1}FM′ such that

is exact

Proof. We construct a projective resolution C′, ε′, C, ε, C″, ε″, i, p for the given short exact sequence of modules. For each n ≥ 0, we have the short exact sequence of projective modules 0 → C′_n C_n C″n → 0. Since C″_n is projective, this splits and consequently, 0 → FC′_n FC′_n FC″_n → 0 is split exact. Thus we have a short exact sequence of complexes 0 → F(C′) F(C) F(C″) → 0. The theorem follows by applying the long exact homology sequence to this short exact sequence of complexes.

Everything we have done can be carried over to coresolutions, and this gives the definition and analogous results for right derived functors. We shall now sketch this, leaving the details to the reader.

Again let F be a functor from R-mod to Ab. For a given R-module M, we choose an injective coresolution 0 → M D⁰ D¹ D² → …, Applying F, we obtain 0 → FM FD⁰ FD¹ FD² → … and we obtain the complex FD = {FDⁱ}, F(d) = {F(dⁱ)}. Then we put (RⁿF)M = Hⁿ(FD), n ≥ 0. If M′ is a second R-module, (D′, η′) a coresolution of M′, then for any homomorphism λ:M′ → M we obtain a homomorphism RⁿF(λ): (RⁿF)M′ → (RⁿF)M. This defines the right derived functor of the given functor F. The results we obtained for left derived functors carry over. In particular, we have an analogue of the long exact sequence given in Theorem 6.5. We omit the details.

EXERCISES

        1. Show that if M is projective, then L₀FM = FM and L_nFM = 0 for n > 0.

        2. Show that if F is right exact, then F and L₀F are naturally equivalent.

6.7   EXT

In this section and the next we shall consider the most important instances of derived functors. We begin with the contravariant hom functor hom( –, N) defined by a fixed module N, but first we need to indicate the modifications in the foregoing procedure that are required in passing to additive contravariant functors from R-mod to Ab. Such a functor is a (covariant) functor from the opposite category R-mod^op to Ab, and since arrows are reversed in passing to the opposite category, the roles of injective and projective modules must be interchanged. Accordingly, to define the right derived functor of a contravariant functor G from R-mod to Ab, we begin with a projective resolution 0 ← M C₀ C₁ ← … of the given module M. This gives rise to the sequence 0 → GM GC₀ GC₁ →… and the cochain complex (GC,G(d)) where GC = {GC_i} and G(d) = (G(d_i)}. We define (RⁿG)M = Hⁿ(GC). In particular, we have (R⁰G)M = ker (GC₀ → GC₁). For any μ ∈ hom_R(M′, M) we obtain a homomorphism (R_nG) (μ): (RⁿG)M → (RⁿG)M′ and so we obtain the nth rigrto derived functor RⁿG of G, which is additive and contravariant. Corresponding to a short exact sequence 0 → M′ → M → M″ → 0 we have the long exact cohomology sequence

where RⁿGM′ → R^{n + 1}GM″ is given by a connecting homomorphism. The proof is almost identical with that of Theorem 6.5 and is therefore omitted.

We now let G = hom(–, N) the contravariant hom functor determined by a fixed R-module N. We recall the definition: If M ∈ ob R-mod, then hom(–, N)M = hom_R(M, N) and if α ∈ hom_R(M, M′), hom(–, N)(α) is the map α* of hom_R(M^f, N) into hom_R(M, N) sending any β in the former into βα ∈ hom_R(M, N). hom(– ,N)M is an abelian group and α* is a group homomorphism. Hence hom(– ,N) is additive. Since (α₁α₂)* = α*₂α₁*, the functor hom(– ,N) is contravariant. We recall also that this functor is left exact, that is, if M′ M M″ → 0 is exact, then 0 → hom(M″, N) hom(M, N) hom(M′, N) is exact (p. 105).

The nth right derived functor of hom(–, N) is denoted as Extⁿ(– ,N); its value for the module M is Extⁿ(M, N) (or Ext_Rⁿ(M, N) if it is desirable to indicate R). If C, ε is a projective resolution for M, then the exactness of C₁ → C₀ M → 0 implies that of 0 → hom(M, N) hom(C₀, N) → hom(C₁, N). Since Ext⁰(M, N) is the kernel of the homomorphism of hom(C₀, N) into hom(C₁, N) it is clear that

under the map ε*

Now let 0 → M′ → M → M″ → 0 be a short exact sequence. Then we obtain the long exact sequence

If we use the isomorphism (23), we obtain an imbedding of the exact sequence 0 → hom(M″, N) → hom(M, N) → hom(M′, N) in a long exact sequence

We can now prove

THEOREM 6.6.   The following conditions on a module M are equivalent:

            (1) M is projective.

            (2) Extⁿ(M, N) = 0 for all n ≥ 1 and all modules N.

            (3) Ext¹ (M, N) = 0 for all N.

Proof.   (1) (2). If M is projective, then 0 ← M C₀ = M ← 0 ← … is a projective resolution. The corresponding complex to calculate Extⁿ(M, N) is 0 → hom(M, N) → hom(M, N) → 0 → …. Hence Extⁿ(M, N) = 0 for all n ≥ 1. (2) (3) is clear. (3) (1). Let M be any module and let 0 → K P M → 0 be a short exact sequence with P projective. Then (25) and the fact that Ext¹ (P, N) = 0 yield the exactness of

Now assume Ext¹ (M, N) = 0. Then we have the exactness of 0 → hom(M, N) → hom(P, N) → hom(K, N) → 0, which implies that the map η* of hom(P, N) into hom(K, N) is surjective. Now take N = K. Then the surjectivity of η* on hom(K,K) implies that there exists a ζ ∈ hom(P,K) such that 1_K = ζη. This implies that the short exact sequence 0 → K → P M → 0 splits. Then M is a direct summand of a projective module and so M is projective.

The exact sequence (24) in which 0 → K P M → 0 is exact, M is arbitrary and P is projective gives the following formula for Ext¹ (M, N):

We shall use this formula to relate Ext¹ (M, N) with extensions of the module M by the module N. It is this connection that accounts for the name Ext for the functor.

If M and N are modules, we define an extension of M by N to be a short exact sequence

For brevity we refer to this as “the extension Two extensions E₁ and E₂ are said to be equivalent if there exists an isomorphism γ: E₁ → E₂ such that

is commutative. It is easily seen that if γ is a homomorphism from the extension E₁ to E₂ making (29) commutative, then γ is necessarily an isomorphism. Equivalence of extensions is indeed an equivalence relation. It is clear also that extensions of M by N exist; for, we can construct the split extension

with the usual i and p.

We shall now define a bijective map of the class E(M, N) of equivalence classes of extensions of M by N with the set Ext¹(M, N) More precisely, we shall define a bijective map of E(M, N) onto coker η* where 0 → K P M → 0 is a projective presentation of M and η* is the corresponding map hom(P, N) → hom(K, N) (η*(λ) = λη). In view of the isomorphism given in (27), this will give the bijection of E(M, N) with Ext¹ (M, N).

Let 0 → N E M → 0 be an extension of M by N. Then we have the diagram

without the dotted lines. Since P is projective, there is a homomorphism λ: P → E making the triangle commutative. With this choice of λ there is a unique homomorphism μ :K → N making the rectangle commutative. For, if x ∈ K, then βληx = εηx = 0. Hence ληx ∈ ker β and so there exists a unique y ∈ N such that αy = ληx. We define μ by x y. Then it is clear that μ is a homomorphism of K into N making the rectangle commutative and μ is unique. Next, let λ′ be a second homomorphism of P into E such that βλ′ = ε. Then β(λ′ – λ) = 0, which implies that there exists a homomorphism τ:P → N such that λ′ – λ = ατ. Then λ′η = (λ + ατ)η = α(λ + τη). Hence μ′ = μ + τη makes

commutative. Since τ ∈ hom (P,N), τη ∈ imη*. Thus μ and μ′ determine the same element of coker η* and we have the map sending the extension E into the element η + imη* of coker η*. It is readily seen, by drawing a diagram, that the replacement of E by an equivalent extension E′ yields the same element μ+im η*. Hence we have a map of E(M, N) into coker η*.

Conversely, let μ ∈ hom(K, N). We form the pushout of η and η (exercise 8, p. 37). Explicitly, we form N P and let I be the submodule of elements (– μ(x), η(x)), x ∈ K. Put E = (N P)/I and let α be the homomorphism of N into E such that α(y) = (y, 0) + I. Also we have the homomorphism of N P into M such that (y, z) ∈(z),y ∈ N, z ∈ P. This maps I into 0 and so defines a homomorphism β of E = (N P)/I into M such that (y, z) + I ε(z). We claim that 0 → N E M → 0 is exact. First, if α(y) = (y, 0) + I = 0, then (y, 0) = (– μ(x), η(x)), x ∈ K, so η(x) = 0 and x = 0 and y = 0. Thus α is injective. Next, βαy = β((y, 0) + I) = ε(0) = 0, so (βα = 0. Moreover, if β((y, z) + I) = 0, then ∈(z) = 0 so z = η(x), x ∈ K. Then (y, z) + I = (y + μ(x), 0) + I = α(y + μ(x). Hence ker β = im α. Finally, β is surjective since if u ∈ M, then u = ε(z), z ∈ P, and β((0, z) + I) = ε(z) = u. If we put λ(z) = (0, z) + I, then we have the commutativity of the diagram (31). Hence the element of coker η* associated with the equivalence class of the extension E is μ + imη*. This shows that our map is surjective. It is also injective. For, let E be any extension such that the class of E is mapped into the coset μ + imη*. Then we may assume that E μ under the original map we defined, and if we form the pushout E′ of μ and η, then E′ is an extension such that E′ μ. Now since E′ is a pushout, we have a commutative diagram (29) with E₁ = E′ and E₂ = E. Then E and E′ are isomorphic. Evidently, this implies injectivity.

We state this result as

THEOREM 6.7.   We have a bijective map of Ext¹ (M, N) with the set E(M, N) of equivalence classes of extensions of M by N.

We shall study next the dependence of Extⁿ(M, N) on the argument N. This will lead to the definition of a functor Extⁿ(M, –) and a bifunctor Extⁿ. Let M, N, N′ be R-modules and β a homomorphism of N into N′. As before, let C = {C_i}, ε be a projective resolution of M. Then we have the diagram

where the horizontal maps are as before and the vertical ones are the left multiplications β_L by β. It is clear that (32) is commutative and hence we have homomorphisms of the complex hom(C, N) into the complex hom(C, N′); consequently, for each n ≥ 0 we have a homomorphism of the corresponding cohomology groups. Thus we have the homomorphism : Extⁿ(M, N) → Extⁿ(M, N′). It is clear that this defines a functor Extⁿ(M, –) from R-mod to Ab that is additive and covariant.

If α ∈ hom(M′, M), we have the commutative diagram

which gives the commutative diagram

This implies as in the case of the hom functor that we can define a bifunctor Extⁿ from R-mod to Ab (p. 38).

Now suppose that we have a short exact sequence 0 → N′ → N → N″ → 0. As in (32), we have the sequence of homomorphisms of these complexes: hom(C, N′) → hom(C, N) - hom(C, N″). Since C_i, is projective and 0 → N′ → N → N″ → 0 is exact, 0 → hom(C_i, N′) → hom(C_i, N) → hom(C_i, N″) → 0 is exact for every i. Thus 0 → hom(C, N′) → hom(C, N) → hom(C, N″) → 0 is exact. Hence we can apply Theorem 6.1 and the isomorphism of hom(M, N) with Ext⁰(M, N) to obtain a second long exact sequence of Ext functors:

We shall call the two sequences (24) and (33) the long exact sequences in the first and second variables respectively for Ext.

We can now prove the following analogue of the characterization of projective modules given in Theorem 6.6.

THEOREM 6.8. The following conditions on a module N are equivalent:

            (1) N is injective.

            (2) Extⁿ(M, N) = 0 for all n ≥ 1 and all modules M.

            (3) Ext¹ (M, N) = 0 for all M.

Proof.   (1) (2). If N is injective, the exactness of 0 ← M ← C₀ ← C₁ ← … implies that of 0 → hom(M, N) → hom(C₀, N) → hom(C₁, N) → …. This implies that Extⁿ(M, N) = 0 for all n ≥ 1. The implications (2) (3) are trivial, and (3) (1) can be obtained as in the proof of Theorem 6.6 by using a short exact sequence 0 → N → Q → L → 0 where Q is injective. We leave it to the reader to complete this argument.

The functors Extⁿ(M, –) that we have defined by starting with the functors Extⁿ(– ,N) can also be defined directly as the right derived functors of hom(M, –). For the moment, we denote the value of this functor for the module M as ⁿ(M, N). To obtain a determination of this group, we choose an injective coresolution 0 → N D⁰ D¹ → … and we obtain the cochain complex hom(M, D):0 → hom(M, D⁰) → hom(M, D¹) → …. Then ⁿ(M, N) is the nth cohomology group of hom(M, D). The results we had for Ext can easily be established for . In particular, we can show that ⁰(M, N) ≅ hom(M, N) and we have the two long exact sequences for analogous to (24) and (33). We omit the derivations of these results. Now it can be shown that the bifunctors Extⁿ and ⁿ are naturally equivalent. We shall not prove this, but shall be content to prove the following weaker result:

THEOREM 6.9. Extⁿ(M, N) ≅ ⁿ (M, N)for all n, M, and N.

Proof.   If n = 0, we have Ext⁰(M, N) ≅ hom(M, N) = ⁰(M, N). Now let 0 → K → P → M → 0 be a short exact sequence with P projective. Using the long exact sequence on the first variable for and ¹ (P, N) = 0 we obtain the exact sequence 0 → hom(M, N) → hom(P, N) → hom(K, N) → ¹(M, N) → 0. This implies that ¹ (M, N) ≅ hom(K, N)/im hom(P, N). In (27) we showed that Ext¹(M, N) ≅ hom(K, N)/im hom(P, N). Hence Ext¹(M, N) ≅ ¹(M, N). Now let n > 1 and assume the result for n – 1. We refer again to the long exact sequence on the first variable for Ext and obtain

from which we infer that Extⁿ(M, N) ≅ Ext^{n– 1}(K, N). Hence Extⁿ(M, N) ≅ Ext^{n– 1}(K, N) ≅ ^{n– 1}(K, N). The same argument gives ⁿ(M, N) ≅ ^{n– 1} (K, N). Hence Extⁿ(M, N) ≅ ⁿ(M, N).

EXERCISES

        1. Let R = D, a commutative principal ideal domain. Let M = D/(a), so we have the projective presentation 0 → D → D → M → 0 where the first map is the multiplication by a and the second is the canonical homomorphism onto the factor module. Use (27) and the isomorphism of hom_D(D, N) with N mapping η into η1 to show that Ext¹(M, N) ≅ N/aN. Show that if N = D/(b), then Ext¹ (M, N) ≅ D/(a, b) where (a, b) is a g.c.d. of a and b. Use these results and the fundamental structure theorem on finitely generated modules over a p.i.d. (BAI, p. 187) to obtain a formula for Ext¹(M, N) for any two finitely generated modules over D.

        2. Give a proof of the equivalence of Ext_n(M, –) and ⁿ(M, –).

        3. Show that in the correspondence between equivalence classes of extensions of M by N and the set Ext¹ (M,N) given in Theorem 6.7, the equivalence class of the split extension corresponds to the 0 element of Ext¹ (M, N).

        4. Let N E_i M, i = 1, 2, be two extensions of M by N. Form E₁ E₂ and let F be the submodule of E₁ E₂ consisting of the pairs (z₁, z₂), z_i ∈ E_i such that β₁z₁ = β₂z₂. Let K be the subset of E₁ E₂ of elements of the form (α₁y, – α₂y), y ∈ N. Note that K is a submodule of F. Put E = F/K and define maps α: N → E, β: E → M by αy = (α₁y, 0) + K = (0, –α₂y) + K, β((z₁, z₂) + K) = α₁z₁ = β₂z₂. Show that α and β are module homomorphisms and N E M, so we have an extension of M by N. This is called the Baer sum of the extensions N E_i M. Show that the element of Ext¹ (M, N) corresponding to the Baer sum is the sum of the elements corresponding to the given extensions. Use this and exercise 3 to conclude that the set of equivalence classes of extensions of M by N form a group under the composition given by Baer sums with the zero element as the class of the split extension.

6.8   TOR

If M ∈ mod-R, the category of right modules for the ring R, then M_R is the functor from R-mod to Ab that maps a left R-module N into the group M_rN and an element η of hom_R(N, N′) into 1η. M_R is additive and right exact. The second condition means that if N′ → N → N″ → 0 is exact, then MN′ → MN → MN″ → 0 is exact. The nth left derived functor of M(= M_R) is denoted as Tor_n(M, –) (or Tor^R_n(M, –)). To obtain Tor_n(M, N) we choose a projective resolution of N: 0 ← N C₀ ← … and form the chain complex MC = {MC_i}. Then Tor_n(M, N) is the nth homology group H_n(MC) of MC. By definition, Tor₀(M, N) = (MC₀)/im(MC₁). Since M is right exact, MC₁ → MC₀ → MN → 0 is exact and hence M N ≅ (M C₀)/im(M C₁) = Tor₀(M, N).

The isomorphism Tor₀(M, N) ≅ MN and the long exact sequence of homology imply that if 0 → N′ → N → N″ 0 is exact, then

is exact.

We recall that a right module M is flat if and only if the tensor functor M from the category of left modules to the category of abelian groups is exact (p. 154). We can now give a characterization of flatness in terms of the functor Tor. The result is the following analogue of Theorem 6.6 on the functor Ext.

THEOREM 6.10.   The following conditions on a right module M are equivalent:

            (1) M is flat.

            (2) Tor_n(M, N) = 0 for all n ≥ 1 and all (left) modules N.

            (3) Tor ₁(M, N) = 0 for all N.

Proof.   (1) (2). If M is flat and 0 ← N ← C₀ ← C₁ ← … is a projective resolution of N, then 0 ← MN ← MC₀ ← MC₁ ← … is exact. Hence Tor_n(M, N) = 0 for any n ≥ 1. (2) (3) is clear. (3) (1). Let 0 → N′ → N ′ N″ ′ 0 be exact. Then the hypothesis that Tor₁(M, N′) = 0 implies that 0 → MN → → MN → MN″ → 0is exact. Hence M is flat.

We consider next the dependence of Tor_n(M, N) on M. The argument is identical with that used in considering Extⁿ. Let α be a homomorphism of the right module M into the right module M′ and as before let 0 ← N ← C₀ ← C₁ ← … be a projective resolution for the left module N. Then we have the commutative diagram

where the vertical maps are α1_N, α1_C0, α1_Ci, etc. Hence we have a homomorphism of the complex {MC_i into the complex and a corresponding homomorphism of the homology groups Tor_n(M, N) into Tor_n(M′, N).In this way we obtain a functor Tor_n(–, N)> from mod-R, the category of right modules for the ring R, to the category Ab that is additive and covariant.

We now suppose we have a short exact sequence of right modules 0 → M′ → M → M″ → 0 and as before, let C, ε be a projective resolution for the left module N. Since the C_i are projective, 0 → M′ C_i → MC_i M″C_i → 0 is exact for every i. Consequently, by Theorem 6.1 and the isomorphism of Tor₀(M, N) with M N we obtain the long exact sequence for Tor in the first variable:

Finally, we note that as in the case of Ext, we can define functors _n(M, N) using a projective resolution of the first argument M. Moreover, we can prove that Tor_n(M, N) ≅ _n(M, N). The argument is similar to that we gave for Ext and Ext and is left to the reader.

EXERCISES

        1. Determine Tor₁(M, N) if M and N are cyclic groups.

        2. Show that Tor₁(M, N) is a torsion group for any abelian groups M and N.

6.9   COHOMOLOGY OF GROUPS

In the remainder of this chapter we shall consider some of the most important special cases of homological algebra together with their applications to classical problems, some of which provided the impetus to the development of the abstract theory.

We begin with the cohomology of groups and we shall first give the original definition of the cohomology groups of a group, which, unlike the definition of the derived functors, is quite concrete. For our purpose we require the concept of a G-module, which is closely related to a basic notion of representation theory of groups. If G is a group, we define a G-module A to be an abelian group (written additively) on which G acts as endomorphisms. This means that we have a map

of G × A into A such that

for g, g₁, g₂ ∈ G, x,y ∈ A. As in representation theory, we can transform this to a more familiar concept by introducing the group ring [G], which is the free -module with G as base and in which multiplication is defined by

where α_g, β_h ∈ . Then if A is a G-module, A becomes a [G]-module if we define

The verification is immediate and is left to the reader. Conversely, if A is a [G]-module, then A becomes a G-module if we define gx as (1g)x.

A special case of a G-module is obtained by taking A to be any abelian group and defining gx = x for all g ∈ G, x ∈ A. This action of G is called the trivial action. Another example of a G-module is the regular G-module A = G[] in which the action is h(∑α_qg) = ∑ α_ghg

Now let A be a G-module. For any n = 0, 1,2, 3,…, let Cⁿ(G, A) denote the set of functions of n variables in G into the module A. Thus if n > 0, then Cⁿ(G, A) is the set of maps of into A and if n = 0, a map is just an element of A. Cⁿ(G, A) is an abelian group with the usual definitions of addition and 0: If f, f′ ∈ Cⁿ(G, A), then

In the case of C⁰(G, A) = A, the group structure is that given in A.

We now define a map δ( = δ_n) of C_n(G, A) into C^{n + 1}(G, A). If f ∈ Cⁿ(G,A), then we define δf by

For n = 0, f is an element of A and

For n = 1 we have

and for n = 2 we have

It is clear that δ is a homomorphism of Cⁿ(G, A) into C^{n + 1}(G, A). Let Zⁿ(G, A) denote its kernel and B^{n + 1}(G, A) its image in C^{n + 1}(G, A). It can be verified that δ²f = 0 for every f ∈ Cⁿ(G, A). We shall not carry out this calculation since the result can be derived more simply as a by-product of a result that we shall consider presently. From δ(δf) = 0 we can conclude that Zⁿ(G, A) ⊃ Bⁿ(G, A). Hence we can form the factor group Hⁿ(G, A) = Zⁿ(G, A)/Bⁿ(G, A). This is called the ftth cohomology group of G with coefficients in A.

The foregoing definition is concrete but a bit artificial. The special cases of H¹(G, A) and H²(G, A) arose in studying certain natural questions in group theory that we shall consider in the next section. The general definition was suggested by these special cases and by the definition of cohomology groups of a simplicial complex. We shall now give another definition of Hⁿ(G, A) that is functorial in character. For this we consider as trivial G-module and we consider Extⁿ(, A) for a given G-module A. We obtain a particular determination of this group by choosing a projective resolution

of as trivial [G]-module. Then we obtain the cochain complex

whose nth homology group is a determination of Extⁿ(, A).

We shall now construct the particular projective resolution (45) that will permit us to identify Extⁿ(, A) with the nth cohomology group Hⁿ(G, A) as we have defined it. We put

Since [G] is a free -module with G as base, C_n is a free -module with base g ₀ g₁ … g_n, g_i ∈ G. We have an action of G on C_n defined by

which makes C_n a [G]-module. This is [G]-free with base

We now define a [G]-homomorphism d_n of C_n into C_{n − 1} by its action on the base {(g₁, …, g_n)}:

where it is understood that for n = 1 we have d₁(g₁) = g₁ − 1 ∈ C₀ = [G]. Also we define a [G]-homomorphism ε of C₀ into by ε(1) = 1. Then ε(g) = ε(g1) = g1 = 1 and ε(∑α_gg) = ∑α_g. We proceed to show that 0 ← C₀ C₁ ← ^… is a projective resolution of . Since the C_i are free [G]-modules, projectivity is clear. It remains to prove the exactness of the indicated sequence of [G]-homomorphisms. This will be done by defining a sequence of contracting homomorphisms:

By this we mean that the s_i are group homomorphisms such that

We observe that {1} is a -base for , G is a -base for C₀ = [G], and {g₀(g₁, …, g_n) = g₀ g₁ ^… g_n|g_i ∈ G} is a -base for C_n, n ≥ 1. Hence we have unique group homomorphisms s _{– 1} : → C₀, s_n : C_{n– 1} → C_n such that

If n > 0, we have

Hence d_{n + 1}s_ng₀(g₁, …, g_n) + s_{n– 1}d_ng₀(g₁, …, g_n) = g₀(g₁, …, g_n). This shows that the third equation in (51) holds. Similarly, one verifies the other two equations in (51).

We can now show that the [G]-homomorphisms ε, d_n satisfy εd₁ = 0, d_nd_{n + 1} = 0, n ≥ 1. By (49), C₁ is free with base {(g) = 1 g|g ∈ G}. Since εd₁(g) = ε(g – 1) = 1 – 1 = 0, the first equality holds. Thus if we put ε = d₀, then we have d_nd_{n + 1} = 0 for n = 0. We note also that s_nC_n for n > 0 contains the set of generators {(g₀, g₁, …, g_n)} for C_{n + 1} as [G]-module. Hence it suffices to show that d_nd_{n + 1}s_n = 0 if n > 0, and we may assume d_{n – 1}d_n = 0.

Then

We can regard (45) as a -complex and the sequence of maps s_{− 1}, s₀, s₁, … as a homotopy between the chain homomorphism of this complex into itself that is the identity on every C_i with the chain homomorphism that is 0 on every C_i. Then these chain homomorphisms define the same homomorphisms of the homology groups. It follows that the homology groups of the -complex (45) are all 0. This means that (45) is exact and hence we have proved

THEOREM 6.11.   Let C_n = , n ≥ 0, and let ε : C₀ → be the [G]-homomorphism such that ε1 = 1, d_n : C_n → C_{n – 1}, the [G]- homomorphism such that (50) holds. Then C = {C_n}, and ε constitutes a free resolution for regarded as trivial G-module.

To calculate Extⁿ(, A) for any [G]-module we can use the resolution C, ε. Then Extⁿ(, A) is the nth homology group of the complex

Since {(g₁, …, g_n)|g_i ∈ G} is a [G]-base for C_n, we have a bijection of hom(C_n, A) with the set Cⁿ(G, A) of functions of n variables in G into A, which associates with any f ∈ hom(C_n, A) its restriction to the base {(g₁, …, g_n)}. The map hom(C_n, A) → hom(C_{n + 1}, A) is right multiplication by d_{n + 1}; that is, if f is a [G]-homomorphism of C_n into A, then its image under hom(C_n, A) → hom(C_{n + 1}, A) is the homomorphism x f(d_{n + 1}x). If we take x to be the element (g₁, …, g_{n + 1}) of C_{n + 1}, then this map is

Thus we have the following commutative diagram

where the vertical arrows are group isomorphisms. Since the product of hom(C_{n– 1}, A) → hom(C_n, A) and hom(C_n, A) → hom(C_{n + 1}, A) is 0, we have δ² = 0. Hence 0 → C⁰(G, A) → C¹(G, A) → C²(G, A) → ^… is a cochain complex and this is isomorphic to the cochain complex 0 → hom(C₀, A) → hom(C₁, A) → hom(C₂, A) → ^…. It follows that these two complexes have isomorphic homology groups. We therefore have

THEOREM 6.12.   Bⁿ(G, A) is a subgroup of Zⁿ(G, A) and Hⁿ(G, A) = Zⁿ(G, A)/Bⁿ(G, A) ≅ Extⁿ(, A).

We shall now switch from the original definition of the cohomology groups of G with coefficients in A to the groups Extⁿ(, A). From now on we use the definition Extⁿ(, A) for the nth cohomology group of G with coefficients in A. This definition has the advantage that it makes available the functorial results on Ext for the study of cohomology groups of a group. Also it offers considerably more flexibility since it permits us to replace the resolution of that we have used by others. Some instances of this will be given in the exercises.

We shall now look at the cohomology group Hⁿ(G, A) for n = 0, 1, 2. We prove first

THEOREM 6.13.   H⁰(G, A) ≅ A^G, the subgroup of A of elements x satisfying gx = x, g ∈ G.

Proof.     We recall that Ext⁰(M, N) ≅ hom(M, N). Hence H⁰(G, A) ≅ hom_[G](, A), the group of [G]-module homomorphisms of into A. If η is such a homomorphism, η is determined by η(1) and if η(1) = x ∈ A, then x = η(1) = η(g1) = gη(1) = gx, g ∈ G. Conversely, if x ∈ A satisfies gx = x, g ∈ G, then the map η such that η(n) = nx is a [G]-homo- morphism of into A. It follows that hom(, A) is isomorphic (under η η(1)) to A^G.

We remark that this proposition can also be proved easily by using the definitions of Z⁰(G, A), B⁰(G, A), and Z⁰(G, A)/B⁰(G, A). We leave it to the reader to carry out such a proof.

If A is a G-module, a map f of G into A is called a crossed homomorphism of G into A if

If x ∈ A, then the map f defined by

is a crossed homomorphism of G into A since

A crossed homomorphism defined by (56) is called principal. It is clear that the crossed homomorphisms form an abelian group under addition of maps and that the principal ones form a subgroup. Comparison with (42) and (43) shows that the first of these groups is Z¹(G, A) and the second is B¹(G, A). Hence the factor group is (isomorphic to) the first cohomology group of G with coefficients in A.

We have encountered crossed homomorphisms in Galois theory in considering Speiser’s equations and their additive analogue (BAI, pp. 297–299). We recall these results, the first of which constituted a generalization of Hilbert’s Satz 90. Let E be a finite dimensional Galois extension field of the field F, G the Galois group of E/F, so G is finite and |G| = [E : F]. We have the natural actions of G on the additive group E of E and on the multiplicative group E* of non-zero elements of E. If we consider the additive group of E as G-module, then a crossed homomorphism is a map f of G into E such that f(gh) = f(g) + gf(h). Theorem 4.32 of BAI (p. 297) states that any such crossed homomorphism is principal. Thus we have the result H¹(G, E) = 0: The first cohomology group of the Galois group G of E/F with coefficients in E is 0. Now consider the action of G on E*. Since the composition in E* is multiplication, a crossed homomorphism of G into E* is a map f of G into E* such that

These are Speiser’s equations as given in (75), p. 297 of BAI. Speiser’s theorem is that such an f is principal, that is, it has the form f(g) = (gu)u^{– 1} for some u ∈ E*. Thus Speiser’s theorem is the homological result that H¹(G, E*) = 1 (using multiplicative notation).

If G is a group and A is an abelian group written multiplicatively on which G acts by automorphisms, then the group C²(G, A) is the group of functions of two variables in G to A with multiplication as composition. Z²(G, A) is the subgroup of f ∈ C²(G, A) such that

This is clear from (44). Such a map is called a factor set. The subgroup B²(G, A) is the subgroup of maps of the form f where f(g, h) = u(g)gu(h)u(gh)^{– 1} where u is a map of G into A. The group Z²(G, A)/B²(G, A) is the second cohomology group of G with coefficients in A. We shall give an interpretation of this group in the next section.

We shall conclude this section by proving the following result on cohomology groups of finite groups.

THEOREM 6.14.   If G is a finite group, A a G-module, then every element of Hⁿ(G, A), n > 0, has finite order a divisor of |G|.

Proof.   Let f ∈ Cⁿ(G, A) and consider the formula (41) for δf. We let g_{n + l} range over G and sum the corresponding formulas. If we denote by u(g₁, …, g_{n – 1}), then since , the result we obtain is

Hence if δf = 0, then |G| f (g_l, …, g_n) = ± δu(g_l, …, g_n) ∈ Bⁿ(G, A). Then |G|Zⁿ(G, A) ⊂ Bⁿ(G, A), so |G|Hⁿ(G, A) = 0, which proves the theorem.

An immediate consequence of this result is the

COROLLARY.   Let G be a finite group, A a finite G-module such that (|G|, |A|) = 1. Then Hⁿ(G, A) = 0 for every n > 0.

This is clear since |A| f = 0 for every f ∈ Cⁿ(G, A).

EXERCISES

        1. Let B be a right module for [G]. Define the nth homology group of G with coefficients in B, n ≥ 0, as H_n(G, B) = Tor_n(B, ) where is regarded as a trivial G-module. Show that H₀(G, B) ≅ B/B_G where B_G is the subgroup of B generated by the elements xg – x, x ∈ B.

        2. Let ε be the homomorphism of [G] into defined in the text and let I = ker ε. Show that I is a free -module with base {g – l |g ∈ G, g ≠ 1}.

        3. Let A and B be G-modules. Show that A B is a G-module with the action such that g(x y) = gx gy.

        4. Let A be a G-module and let A_t denote the abelian group A with the trivial G-action. Show that [G] A and [G] A_t are isomorphic as G-modules. (Hint: Show that there is a module isomorphism of [G] A onto [G] A_t such that gx → gg^{– 1} x, g ∈ G, x ∈ A.)

        5. Use exercise 4 to prove that if A is a G-module that is -free, then [G] A is [G]-free.

        6. Let I = ker ε, ε : [G] → , as in the text and put , n factors. Show that the short exact sequence , where the first map is inclusion, yields a short exact sequence . be the homomorphism of that is the composite of . Note that

g_i ∈ G. Show that

is a free resolution for .

        7. Let G = , the cyclic group of finite order m generated by the element g. Note that , t an indeterminate. Let D = g – 1, N = l + g + ^… + g^{m – 1}, and let D′, N′ denote multiplication by D and N respectively in [G]. Show that

is a free resolution for .

        8. Use exercise 7 to show that if A is a G-module for the cyclic group of order m < ∞, then

where Ann_AN = {x ∈ A|Nx = 0}.

6.10   EXTENSIONS OF GROUPS

By an extension of a group G by a group A we shall mean a short exact sequence

Thus i is injective, p is surjective, and ker p = iA. Hence . If is a second extension of G by A, then we say that this is equivalent to (59) if there exists a homomorphism h : E → E′ such that

is commutative. It follows easily, as for modules (p. 348), that in this case h is an isomorphism.

We restrict our attention to extensions of a group G by an abelian group. With such an extension, we can associate an action of G on A by automorphisms and an element of the cohomology group H²(G, A) where A is regarded as G-module by the action we shall define. It will be helpful in these considerations to denote the elements of G by small Greek letters and those of A and E by small Latin letters.

We first define the action of G on A. Let σ ∈ G, x ∈ A. Choose an element s ∈ E such that ps = σ and consider the element s(ix)s^{– 1}. Since , s(ix)s^{– 1} ∈ iA and since i is injective, we have a unique element y ∈ A such that s(ix)s^{– 1} = iy. To obtain y we made a choice of an element s ∈ E such that ps = σ. Let s′ be a second element such that ps′ = σ. Then p(s′s^{– 1}) = 1 and hence s′s^{– 1} = ia, a ∈ A, and s′ = (ia)s. Since iA is abelian, we have . Thus the element y is independent of the choice of s and hence we can put σx = y. Our definition is

It is straightforward to verify that the definition of σx gives an action of G on A by automorphisms: . Except for the fact that A is written multiplicatively, we have defined a G-module structure on A. We shall now call this a G-module even though we retain the multiplicative notation in A.

Our next step is to choose for each σ ∈ G an element s_σ ∈ E such that ps_σ = σ. Thus we have a map s : G → E, σ s_σ. such that ps_σ = σ for all σ (or ps = 1_G).

Let σ, τ ∈ G and consider the element s_σ s_τ s_στ^{– 1} of E. Applying p to this element gives 1. Hence there is a unique element k_{σ, τ} ∈ A such that s_σ s_τ s_στ^{– 1} = ik_{σ, τ} or

If ρ ∈ G also, then

Similarly, . Hence, the associative law in E implies that

These relations show that the map k : G × G → A such that (σ, τ) k_{σ, τ} is an element of Z²(G, A) as defined in the classical definition of H²(G, A) given at the beginning of section 6.9.

We now consider the alteration in k that results from changing the map s to a second one s′ satisfying ps′ = 1_G. Then for any . Hence there exists a unique u_σ ∈ A such that . Thus we have a map u : σ u_σ of G into A such that

Conversely, if u is any map of G into A, then s′ defined by (64) satisfies ps′ = 1_G. By (64), we have . Hence k is replaced by k′ where

This shows that k′ and k determine the same element of H²(G, A). Hence the extension determines a unique element of H²(G, A).

It follows directly from the definitions that if the extensions and are equivalent, then they determine the same module action of G on A and the same element of H²(G, A). To prove the converse we shall show that the multiplication in E is determined by the action of G on A and the map k. Let e ∈ E and put . Then . Hence f = ix for a uniquely determined element x ∈ A. Then we have the factorization

It is clear that the elements x ∈ A, and σ ∈ G are uniquely determined by the given element e. Now let (iy)S_τ, where y ∈ A and τ ∈ G, be a second element of E. Then

Now suppose the extensions and determine the same module structure and the same element of H²(G, A). Then we can choose maps s and s′ of G into E and E′ respectively such that ps = 1_G, p′s′ = 1_G and for any σ, τ ∈ G we have . Then we have (67) and . It follows that the map h : (ix)(s_σ) (i′x) (s′_σ) is a homomorphism of E into E′ so that (60) is commutative. Hence the extensions are equivalent.

We shall now state the following basic

THEOREM 6.15.   Two extensions of G by an abelian group A are equivalent if and only if they determine the same action of G on A and the same element of H²(G, A). Let G be a group, A a G-module, and let M denote the set of extensions of G by A having a given G-module A as associated module. Then we have a 1–1 correspondence between the set of equivalence classes of extensions of G by A contained in M with the elements of H²(G, A).

Proof.     The first statement has been proved. To prove the second, it suffices to show that given a G-module A and a map k of G × G into A satisfying (63) there exists an extension whose associated action of G on A is the given one and whose element of H²(G, A) is the one determined by k. We put E = A × G, the set of pairs (x, σ), x ∈ A, σ ∈ G, and we define a multiplication in E by

Then it is immediate from (63) that this multiplication is associative. If we put p = σ = 1 in (61), we obtain k_{1, 1} k_{1, τ} = k_{1, τ} k_{1, τ}. Hence k_{1, τ} = k_{1, 1}. Then

so is a left unit in E. To prove that E is a group with 1 as unit, it suffices to show that any element (x, σ) of E has a left inverse relative to 1 (BAI, p. 36, exercise 10). This follows from

Hence E is a group. Let . Then i is a homomorphism of A into E and p is a homomorphism of E into G. It is clear that i is injective and p is surjective. Moreover, iA = ker p. Hence is an extension of G by A. To determine the module action of G on A determined by this extension, we note that p(1, σ) = σ so we must calculate , We have seen that k_{1, ρ} = k_{1, 1} and if we put σ = τ = 1 in (63), we obtain . Hence k_{ρ, 1} = ρk_{1, 1}. Now put σ = ρ^{– 1}, τ = ρ in (63). This gives . Thus we have

Now and

(by (69)). Hence the module action is the given one. Now let s be the map σ (l, σ), so ps = 1_G. We have , so s_σs_τ = (ik_{σ, τ})s_στ.Hence the element of H²(G, A) associated with this extension is that determined by k. This completes the proof.

The foregoing result in a slightly different form is classical. It was proved by Schreier, who first considered the problem of extensions as the problem of describing all of the groups that have a given normal subgroup A (not necessarily commutative) with given factor group G.

An extension (59) is said to be split if there exists a group homomorphism s : G → E such that ps = 1_G. Then (sσ)(sτ) = sστ for any σ, τ ∈ G and hence k_{σ, τ} determined by (62) is 1. Thus the element of H²(G, A) associated with a split extension is 1. Conversely, if this is the case, then we have a map s : G → E satisfying ps = 1_G for which the k_{σ, τ} are all 1, which means that s is a homomorphism. Thus the split extensions are those for which the associated element of H²(G, A) is 1. Evidently this implies that H²(G, A) = 1 if and only if all extensions of G by A with the given G-module structure on A split. By the corollary to Theorem 6.14 (p. 361), this is the case if G and A are finite and (|G|, |A|) = 1.

If and E/A = G for arbitrary (not necessarily abelian) A, then we have the extension where i is the injection of A in E and p is the natural homomorphism of E onto G. It is readily seen from the definition that this extension splits if and only if there exists a subgroup S of E such that E = SA, S ∩ A = 1. In this case E is said to be the semi-direct product of S and A. The result just indicated is that if A is abelian and A and G are finite with (|A|, |G|) = 1, then E = SA and S ∩ A = 1 for a subgroup S of E. This result, which was obtained by homological methods, can be supplemented by a little group theory to prove the following theorem due to H. Zassenhaus.

THEOREM 6.16.   Let E be a finite group, A a normal subgroup of E such that (|A|, |E/A|) = 1. Then E is a semi-direct product of A and a subgroup S.

Proof.     Let |A| = m, |E/A| = n. It suffices to show that there exists a subgroup S such that |S| = n. For, if S is such a subgroup, then S ∩ A is a subgroup whose order divides |S| = n and |A| = m. Then S ∩ A = 1. Also since , SA is a subgroup whose order is a multiple of |S| and |A| and so is a multiple of mn = |E|. Let p be a prime divisor of m and let H be a Sylow p-subgroup of A. Then H is also a Sylow p-subgroup of E. The set Syl_p of Sylow p-subgroups of E is {gHg^{– 1}|g ∈ E} (BAI, p. 80). Since this is also the set of Sylow p-subgroups of A. Hence if N is the normalizer of H in E then |Syl_p| = [E : N] = [A : N ∩ A]. Thus

and hence

On the other hand, we can use induction on order to conclude that N and hence E contains a subgroup of order n. Now suppose |N| = |E|. Then N = E and . The center Z of H is non-trivial (BAI, p. 76). Since Z is a characteristic subgroup of H, and . Since (E/Z)/(A/Z) ≅ E/A we can apply induction to conclude that E/Z contains a subgroup L/Z of order n. Since Z is abelian and |Z| is a power of p, the result we established by homology implies that L contains a subgroup S of order n. This completes the proof.

EXERCISES

        1. Let be an extension of G by the abelian group A and let H be the set of equivalences h of this extension : the automorphisms of E that make

commutative. Show that H is a subgroup of Aut E that contains the inner automorphisms by the elements of iA. Show that the latter set is a normal subgroup J of H and H/J ≅ H¹(G, A).

        2. (Schreier, Eilenberg and MacLane.) Let be an extension of G by the group A (not necessarily abelian) and let Z be the center of A. Let s be a map σ s_σ of G into E such that ps_σ = σ and let φ(s_σ) be the automorphism x y of A such that s_σ(ix)s_σ^{– 1} = iy. Show that if s′ is a second map of G into E satisfying ps′_σ = σ, σ ∈ G, then φ(s′_σ) ∈ (Inaut A)φ(s_σ) and φ(s′_σ)|Z = φ(s_σ)|Z. Show that φ : σ (Inaut A)φ(s_σ) is a homomorphism of G into Aut A/Inaut A and σc ≅ φ(s_σ)c, c ∈ Z, defines an action of G on Z by automorphisms.

   Show that if σ, τ ∈ G, then s_σs_τ = i(k_{σ, τ})s_στ where k_{σ, τ} is a uniquely determined element of A. Show that if we put φ(σ) = φ(s_σ), then

where I_{kσ, τ} is the inner automorphism , in A. Show that

and

   Conversely, suppose G and A are groups and k is a map of G × G into A and φ a map of G into Aut A such that (70)–(72) hold. Let E = A × G and define a product in E by

Show that E is a group with unit and that if i is the map and p is the map (x, σ) σ, then is an extension of G by A.

   Let and be extensions of G by A such that the associated homomorphisms of G into Aut A/Inaut A and hence the associated actions of G on Z are the same. Show that the maps s and s′ (for G to E') can be chosen so that φ(s_σ) = φ′(s′_σ)(φ′ defined analogously to φ). Let k′_{σ, τ} be defined by and put . Show that f(σ, τ) ∈ Z and is a 2-cocycle for G with values in Z (where the action of G is as defined before). Use this to establish a 1–1 correspondence between the set of equivalence classes of extensions of G by A, all having a fixed associated homomorphism of G into Aut A/Inaut A with H²(G, Z).

        3. Let G be finite and let H²(G, *) be the second cohomology group of G with coefficients in the multiplicative group * of non-zero elements of where the action of G on * is trivial. The group H²(G, *) is called the Schur multiplier of G. Use Theorem 6.14 to show that if [γ] is any element of H²(G, *), then the representative cocycle γ can be chosen to have values that are nth roots of unity, n = |G|. Hence conclude that H²(G, *) is a finite group.

        4. Let ρ be a projective representation of a group G. By definition, ρ is a homomorphism of G into the projective linear group PGL(V) of a finite dimensional vector space V/F (exercise 4, p. 256). As in the exercise cited, for each g ∈ G, let μ(g) denote a representative of the coset ρ(g) ∈ PGL(V), so where γ_{g1, g2} ∈ F*, the multiplicative group of F. Show that is a 2-cocycle of G with coefficients in F* where the action of G on F* is trivial. Show that if we make a second choice of representatives μ′(g) ∈ GL(V) for the ρ(g), g ∈ G, then the resulting 2-cocycle γ′ determines the same element of H²(G, F*) as γ. Hence show that we can associate with ρ a unique element [γ] of H²(G, F*). Note that if [γ] = 1, then we may take γ_g1,g2 = l, g_i ∈ G. Then we have μ(g_lg₂) = μ(g₁)μ(g₂), so ρ is essentially an ordinary representation. In this case we shall say that ρ splits.

        5. Let the notations be as in the last exercise and let A be a subgroup of F* containing the γ_g1,g2 for a particular choice of the cocycle γ. Construct the extension E of G by A corresponding to γ as in the text. Write the elements of E as (g, a) g ∈ G, a ∈ A. Then (g₁, a₁)(g₂, a₂) = (g₁g₂, γ_gl,g2a₁a₂). Note that iA = {(l, a)} is contained in the center of E. Show that defines a representation of E acting on V such that .

        6. Let E be an extension of G by A such that iA ⊂ Z(G), the center of E. Let be a representation of E acting on a finite dimensional vector space over an algebraically closed field F. For g ∈ G define ρ(g) to be the element of PGL(V) having representative . Show that ρ is a projective representation of G.

Note: The preceding exercises 3–6 give a slight indication of a rich connection between the Schur multiplier and projective representations of a finite group. This was developed in three papers by Schur. The reader may consult Huppert’s Endliche Gruppen, Springer-Verlag, Berlin-Heidelberg-New York, 1967, pp. 628–641 for an account of Schur’s theory, with references.

6.11   COHOMOLOGY OF ALGEBRAS

The definitions of homology and cohomology modules for an (associative) algebra, which are due to Hochschild, are based on the concept of a bimodule for an algebra. Let A be an algebra over a commutative ring K. We have defined a (left) module M for A (as K-algebra) to be an abelian group written additively that is both a K-module and an A-module in the usual ring sense such that k(ax) = (ka)x = a(kx), k ∈ K, a ∈ A, x ∈ M (p. 211). One has a similar definition for a right (algebra) A-module M : M is a K-module, a right A-module in the usual sense such that k(xa) = (kx)a = x(ka), k ∈ K, a ∈ A, x ∈ M. Now let A and B be algebras over K. Then we define an (algebra) A-B-bimodule M to be a left A-module and a right B-module such that the K-module structures given by A and by B are the same and (ax)b = a(xb), a ∈ A, b ∈ B, x ∈ M.

There is a simple device for reducing the study of A-B-bimodules to that of modules for another algebra. Let B^op be the opposite algebra of B and form the algebra AB^op where stands for _K. Let M be an A-B-bimodule, x an element of M. Then we have the map of A × B into M sending (a, b) axb ∈ M. This is K-bilinear, so we have a K-linear map of A B into M such that . The main point of this is that for a given and a given x ∈ M we have a well-defined product . Direct verification shows that this renders M an algebra A B^op-module. Conversely, if M is given as an A B^op-module, then ax = (a l)x, xb = (l b)x, kx = kx for a ∈ A, b ∈ B, k ∈ K, make M an (algebra)A-B-bimodule. It is clear from this that we can pass freely from the point of view of A-B-bimodules to that of A B^op-modules and conversely.

If M and N are modules for the algebra A, a homomorphism η of M into N is a homomorphism of abelian groups satisfying η(kx) = k(ηx), η(ax) = a(ηx), k ∈ K, a ∈ A. Since kx = (kl)x, the first of these conditions is superfluous, so the notion coincides with that of homomorphisms of M into N in the sense of modules for the ring A. On the other hand, it is natural to endow hom_A(M, N) with a K-module structure rather than just the abelian group structure we have considered hitherto. This is done by defining kη, k ∈ K, η ∈ hom(M, N) by (kη)x = k(ηx). It is clear that in this way hom(M, N) becomes a K-module. Similarly, if M is a right A-module and N is a left A-module for the algebra A, then M _AN is a K-module. In place of the usual functors hom_A(M, –), hom_A( –, N), _AN, etc., to the category of abelian groups, we now have functors to K-modules. Moreover, these are not only additive but also K-linear in the sense that the maps between the K-modules involved in the definitions are K-homomorphisms. Similar remarks apply to the derived functors. All of this is quite obvious and would be tedious to spell out in detail. We shall therefore say nothing more about it and shall replace abelian groups by modules in what follows when it is appropriate to do so.

We now consider a single algebra A and the category of A-A-bimodules (homomorphisms = A-A-homomorphisms). Equivalently we have the category of A_KA^op-modules. We shall now write A^e for A A^op. Evidently A itself is an A-A-bimodule relative to the left and right multiplications in A. Thus A is an A^e-module in which we have

for a_i, x ∈ A, b_i ∈ A^op. Evidently A is cyclic as A^e-module with 1 as generator. Hence we have an A^e-homomorphism

of A^e onto A.

We are now ready to define the homology and cohomology modules of an algebra A. Let M be an A-A-bimodule (= A^e-module). Then we define the nth cohomology module Hⁿ(A, M) of A with coefficients in M as Extⁿ_A^e(A, M) and the nth homology module H_n(A, M) of A with coefficients in M as Tor_n^Ae(A, M). In both cases A is regarded as A^e-module in the manner defined above.

We now assume that A is K-free (projective would suffice). We shall define a free resolution of A as A^e-module such that the determination of Hⁿ(A, M) by this resolution can be identified with the original definition of Hⁿ(A, M) as given by Hochschild for algebras over fields. Let X₀ = A_KA, X₁ = A _K A _K A and, in general, X_n = A _K … _K A with n + 2 factors. If M and N are A-A-bimodules, then M_KN is an A-A-bimodule in which a(xy) = axy, (xy)a = xya, a ∈ A, x ∈ M, y ∈ N (Proposition 3.6, p. 135). It follows that X_n is an A-A-bimodule in which

It is clear from the definitions that and for as A^e-modules. The isomorphism maps . Since A is K-free, the X_n are K-free. It follows that and are A^e-free.

It is clear from the usual argument with tensor products that we have a unique K-homomorphism d_n of X_n into X_{n – 1} such that

It is clear from the definition of the left and right A-actions that this is an A-A-bimodule, hence, A^e-homomorphism of X_n into X_{n – 1}. Together with the augmentation ε: X₀ → A we have the sequence of A^e-homomorphisms

We shall show that this is exact, which will prove that (78) is a free resolution of A as A^e-module. We obtain a proof of exactness in a manner similar to that given for the complex employed in the group case (p. 357). We define a contracting homomorphism

that is, a sequence of K-homomorphisms s_i such that

We define s_n, n ≥ – 1, to be the K-homomorphism such that

Then it follows directly from the definition that (80) holds. As in the group case (p. 358), this implies that (78) is exact and hence this is a free resolution of A with e as augmentation.

For a given A-A-bimodule M we now have the cochain complex

whose cohomology groups are the cohomology groups of A with coefficients in M. Now we have the sequence of isomorphisms (see p. 136) . We can also identify hom_K(X_{n – 2}, M) with the K- module of n-linear maps of A × … × A, n times, into M. Such a map has A × … × A as domain and M as codomain and is a K-homomorphism of A into M if all but one of the arguments is fixed. Hence the isomorphism above becomes an isomorphism onto the K-module Cⁿ(A, M) of n-linear maps f of A × … × A into M. We now define for by

x_i ∈ A. Then one can check the commutativity of

where the vertical maps are the indicated isomorphisms. It follows that

is a cochain complex isomorphic to (82). Hence it has the same cohomology groups and consequently we have the following

THEOREM 6.17.   Let A be an algebra over K that is K-free, M an A-A-bimodule, Cⁿ(A, M), n ≥ 0, the K-module of n-linear maps of A × … × A, n times, into M. For f ∈ Cⁿ(A, M) define δf ∈ C^{n + 1}(A, M) by (83). Let Zⁿ(A, M) = ker δ on Cⁿ(A, M), Bⁿ(A, M) = δC^{n – 1}(A, M). Then Bⁿ(A, M) is a submodule of Zⁿ(A, M) and Zⁿ(A, M)/Bⁿ(A, M) ≅ Hⁿ(A, M), the nth cohomology module of A with coefficients in M.

Although some of the results we shall now indicate are valid without this restriction, we continue to assume that A is K-free. Following the pattern of our discussion of the group case, we now consider Hⁿ(A, M) for n = 0, 1, 2, using the determination of these modules given in Theorem 6.17.

As usual, it is understood that C⁰(A, M) is identified with the module M. Taking u ∈ M, the definition of δu gives (δu)(x) = xu – ux, x ∈ A. Hence Z⁰(A, M) is the submodule of M of u such that ux = xu, x ∈ A. Since C^{– 1}(A, M) = 0, we see that H⁰(A, M) is isomorphic to the submodule of M of u such that ux = xu, x ∈ A.

Next let f ∈ C¹ (A, M). Then and δf = 0 if and only if f is a K-homomorphism of A into M such that

It is natural to call such an f a derivation of A into the A-A-bimodule M. If u ∈ M, u determines the inner derivation δu such that

These form a submodule Inder(A, M) of the module Der(A, M) of derivations of A into M. The special case n = 1 of Theorem 6.17 gives the isomorphism

Now let f ∈ C²(A, M). Then

and δf = 0 if and only if

x, y, z ∈ A. The set of f ∈ C²(A, M) satisfying this condition constitutes Z²(A, M). This contains the submodule of maps δg, g ∈ C^l(A, M) and (δg)(x, y) = xg(y) – g(xy) + g(x)y. The quotient of Z²(A, M) by this submodule is isomorphic to H²(A, M).

The second cohomology group of an algebra in the form Z²(A, M)/B²(A, M) made its first appearance in the literature in proofs by J. H. C. Whitehead and by Hochschild of a classical structure theorem on finite dimensional algebras over a field: the so-called Wedderburn principal theorem. We shall give a sketch of a cohomological proof of this theorem, leaving the details to be filled in by the reader in a sequence of exercises at the end of the chapter.

Let A be a finite dimensional algebra over a field F, N = rad A. Then N is a nilpotent ideal in A and = A/N is semi-primitive. The Wedderburn principal theorem asserts that if is separable in the sense that ^E is semi-primitive for every extension field E/F, then A contains a subalgebra S such that A = S + N and S ∩ N = 0, that is, A = S N as vector space over F (not as algebra direct sum!).

To prove the theorem one first reduces the proof to the case in which N² = 0. This is done by introducing B = A/N² (N² is an ideal) whose radical is N/N² and (N/N²)² = 0. If N² ≠ 0, then the dimensionality [B : F] < [A : F] so we may assume that the theorem holds for B. It follows easily that it holds also for A.

Now assume N² = 0. We can choose a subspace V of A such that A = V N. This is equivalent to choosing a linear map s of into A such that ps = 1 for p, the canonical map x x + N of A onto . Then V = s and s is injective. Any element of A can be written in one and only one way as . Since N is an ideal and N² = 0, we have the multiplication

If we define , then N becomes an A-A-bimodule. Since , we have

where . The map f ∈ Z²(, N). Replacing s by the linear map t: → A such that pt = 1 replaces f by a cohomologous cocycle. Moreover, if f = 0 in (91), then S = s() is a subalgebra such that A = S N. Then Wedderburn’s principal theorem will follow if we can prove that H²(, N) = 0 for any separable algebra and any --bimodule N. A proof of this is indicated in the following exercises.

EXERCISES

        1. First fill in the details of the foregoing argument: the reduction to the case N² = 0, the reduction in this case to the proof of H²(, N) = 0.

        2. Let A be a finite dimensional separable algebra over a field F. Show that there exists an extension field E/F such that . (The easiest way to do this is to use the algebraic closure of F as defined in section 8.1. However, it can be done also without the algebraic closure.)

        3. Use exercise 2 to show that A^e = A_FA^op is finite dimensional semi-simple.

        4. Show that any module for a semi-simple artinian ring is projective.

        5. Use exercises 3 and 4 to show that if A is finite dimensional separable, then A is a projective A^e-module. Hence conclude from Theorem 6.6, p. 347, that Hⁿ(A, M) = 0 for any n ≥ l and any M. (This completes the proof of the theorem.)

        6. (A. I. Malcev.) Let A = S N where N = rad A and S is a separable subalgebra of A. Let T be a separable subalgebra of A. Show that there exists a z ∈ N such that (1 – z)T(l – z)^{– 1} ⊂ S.

        7. Let A be an arbitrary algebra and let J = ker ε where ε is the augmentation A^e → A defined above. Show that J is the left ideal in A^e generated by the elements a 1 – 1 a.

        8. Show that A is A^e-projective if and only if there exists an idempotent e ∈ A^e such that (a l)e = (1 a)e, a ∈ A.

6.12   HOMOLOGICAL DIMENSION

Let M be a (left) module for a ring R. There is a natural way of defining homological dimension for M in terms of projective resolutions of M. We say lhat M has finite homological dimension if M has a projective resolution C, ε for which C_n = 0 for all sufficiently large n. In this case the smallest integer n such that M has a projective resolution

is called the homological dimension, h.dim M, of M. It is clear from this definition that M is projective if and only if h.dim M = 0. We recall that such a module can be characterized by the property that Extⁿ(M,N) = 0 for all modules N and n ≥ 1. The following result contains a generalization of this criterion.

THEOREM 6.18.   The following conditions on a module M are equivalent:

            (1) h.dim M ≤ n.

            (2) Ext^{n + 1}(M, N) = 0 for all modules N.

            (3) Given an exact sequence 0 η C_n → C_{n − 1} → C₀ → M → 0 in which every C_k, k < n, is projective, then C_n is projective.

Proof.   (1) (2). The hypothesis is that we have a projective resolution . Then we have the complex 0 → hom(C₀, N) → hom(C₁, N) → … → hom(C_n, N) → 0 → …. The cohomology groups of this cochain complex are the terms of the sequence Ext⁰(M, N), Ext ¹(M,N), Evidently we have Ext^{n + 1}(M, N) = 0 and this holds for all N.

(2) (3). If we are given an exact sequence with the properties stated in (3) , we obtain from it a sequence of homomorphisms

where D_k = im(C_k → C_{k - 1}) for k > 0, D₀ = im(C₀ → M) = M, C_k → D_k, is obtained from C_k → C_{k − 1} by restricting the codomain and D_k → C_{k − 1} is an injection. Then 0 → D_k → C_{k −} → D_{k − 1} → 0 is exact. Hence the long exact sequence for Ext in the first variable gives the exactness of

for i = 1, 2,…, 1 ≤ k ≤ n. Since C_{k − 1} is projective, the first and last terms are 0. Thus Extⁱ(D_k, N) ≅ Ext^{i + 1}(D_{k − 1}, N) and hence

Assuming (2), we have Ext^{n + 1} (D₀, N) = 0. Hence Ext¹(D_n, N) = 0. Since 0 → C_n → C_{n − 1} → … is exact, D_n ≅ C_n. Thus Ext¹ (C_n, N) = 0 for all N, which implies that C_n is projective.

(3) (1). The construction of a projective resolution for M gives at the (n − l)-st stage an exact sequence 0 ← M ← C₀ ← … ← C_{n − 1} where all of the C_i are projective. Let C_n = ker(C_{n − 1} → C_{n − 2}). Then we have the exact sequence 0 ← M → C₀ ← … ← C_n → 0. Assuming (3), we can conclude that C_n is projective. Then 0 → M ← C₀ ← ≠ ← C_n ← 0 ← 0 ≠ is a projective resolution, which shows that h.dim M ≤ n.    

Remarks. The proof of the implication (1) (2) shows also that if h.dim M ≤ n, then Ext ^k(M, N) = 0 for every k < n and every module N. In a similar manner the condition implies that Tor_k(M, N) = 0 for all k > n and all N. It is clear also that if h.dim M = n, then for any k ≤ n there exists a module N such that Ext^k(M, N) ≠ 0.

It is clear from the fact that Extⁿ( −, N) is an additive functor from R-mod to Ab that Ext^k . An immediate consequence of this and Theorem 6.18 is that M = M′ M″ has finite homological dimension if and only if this is the case for M′ and M″. Then h.dim M is the larger of h.dim M′ and h.dim M″. We now consider, more generally, relations between homological dimensions of terms of a short exact sequence 0 → M′ → M → M″ → 0. For any module N we have the long exact sequence

Suppose h.dim M ≤ n. Then and hence . Similarly, if h.dim M′ ≤ n, then and if h.dim M″ ≤ n, then . These relations imply first that if any two of the three modules M, M′, M” have finite homological dimension, then so hasd the third. Suppose this is the case and let h.dim M = n, h.dim M’ = n′, h.dim M″ = n″. It is readily seen that the facts we have noted on the Ext’s imply that we have one of the following possibilities:

            I. n ≤ n′ n″. Then either n = n′ = n″ or n ≤ n′ and n″ = n′ +1.

         II. n′ ≤ n″, n’ < n. Then n = n″

III. n″ ≤ n′, n″ < n. Then n = n′.

From this it follows that if n > n′, then n″ = n; if n < n′, then n″ = n + 1; and if n = n′, then n″ < n′ +1. We state these results as

THEOREM 6.19.   Let 0 → M′ → M → M″ → 0 be exact. Then if any two of h.dimM′, h.dimM, h.dimM″ are finite, so is the third. Moreover, we have h.dim M″ = h.dim M if h.dim M′ < h.dim M, h.dim M″ = h.dim M′ +1 if h.dim M < h.dim M′, and h.dim M″ ≤ h.dim M + 1 if h.dim M = h.dim M′.

The concept of homological dimension of a module leads to the definition of homological dimensions for a ring. We define the left (right) global dimension of a ring K as sup h.dim M for the left (right) modules for R. Thus the left (right) global dimension of R is 0 if and only if every left (right) R-module is projective. This is the case if and only if every short exact sequence of left (right) modules 0 → M′ → M → M″ → 0 splits (p. 150) and this happens if and only if every left (right) module for R is completely reducible. It follows that R has left (right) global dimension 0 if and only if R is semi-simple artinian (p. 208). Thus R has left global dimension 0 if and only if it has right global dimension 0. Otherwise, there is no connection between the left and right global dimensions of rings in general. We shall be interested primarily in the case of commutative rings where, of course, there is no distinction between left and right modules and hence there is only one concept of global dimension.

A (commutative) p.i.d. R that is not a field has global dimension one. For, any submodule of a free K-module is free (exercise 4, p. 155). Hence if M is any R-module, then we have an exact sequence 0 → K → F → M → 0 in which F and K are free. Hence h.dim M ≤ 1 for any R-module M and the global dimension of R is ≤ 1. Moreover, it is not 0, since this would imply that R is semi-simple artinian and hence that R is a direct sum of a finite number of fields. Since R has no zero divisors ≠ 0, this would imply that R is a field, contrary to assumption.

EXERCISE

        1. Let M be a module over a commutative ring K, L a commutative algebra over K that is K-free. Show that h.dim_kM = h.dim_LM_L (h.dim_kM = homological dimension as K-module).

6.13   KOSZUL′S COMPLEX AND HILBERT′S SYZYGY THEOREM

We shall now consider homological properties of the ring r = F[x₁,…, x_m] of polynomials in indeterminates x_i with coefficients in a field F. Our main objective is a theorem of Hilbert that concerns graded modules for the ring R, graded in the usual way into homogeneous parts. We consider first the decomposition R = F J where J is the ideal in R of polynomials with 0 constant term, or, equivalently, vanishing at 0. This decomposition permits us to define an R-module structure on F by (a + f)b = ab for a,b ∈ F,f ∈ J. Note that this module is isomorphic to R/J. An important tool for the study of the homological properties of R is a certain resolution of F as R-module, which was first introduced by Koszul in a more general situation that is applicable to the study of homology of Lie algebras—see the author’s Lie Algebras, pp. 174–185).

Koszul’s complex, which provides a resolution for F, is based on the exterior algebra E(M) for a free module M of rank m over R = F[x₁,… ,x_m]. We need to recall the definitions and elementary facts on exterior algebras that we obtained in Chapter 3 (p. 141). Let K be an arbitrary commutative ring, M a K-module, and E(M) the exterior algebra defined by M. We recall that M is embedded in E(M) and E(M) is graded so that E(M) = K M M² … We recall also the basic universal property of E(M), namely, x² = 0 for every x ∈ M, and if f is a K-homomorphism of M into an algebra A over K such that f(x)² = 0, then f can be extended in one and only one way to a K-algebra homomorphism of E(M) into A. In particular, the map x − x in M has a unique extension to a homomorphism l of E(M) into itself and since x − x is of period two, i² = 1 E(M).

We shall call a K-endomorphism D of E(M) an anti-derivation if

We require the following

LEMMA 1. Let D be a K-homomorphism of M into E(M) such that

Then D can be extended in one and only one way to an anti-derivation of E(M).

Proof.   Consider the map

of M into the algebra A = M₂(E(M)). The condition on D implies that f is a K-homomorphism such that f(x)² = 0. Hence f has a unique extension to an algebra homomorphism of E(M) into A. It is clear that the extension has the form

where = la and D is a K-endomorphism of E(M) extending the given D. The condition f(ab) = f(a) f(b) implies that D is an anti-derivation. The uniqueness of the extension follows from the following readily verified facts:

        1. The difference of two anti-derivations is an anti-derivation.

        2. The subset on which an anti-derivation is 0 is a subalgebra.

        3. M generates E(M).

Now if D₁ and D₂ are anti-derivations such that D₁ | M = D₂ | M = D, then D₁ – D₂ is an anti-derivation such that (D₁ – D₂) | M = 0. Since M generates E(M), we have D₁ – D₂ = 0 and D₁ = D₂.

A particular case in which the lemma applies is that in which D = d ∈ M*, the K-module of K-homomorphisms of M into K (the dual module of M). Since K ⊂ E(M), d can be regarded as a K-homomorphism of M into E(M). Since K is contained in the center of E(M), it is clear that the condition (dx)x = x(dx) of the lemma is fulfilled. Hence we have the extension d that is an anti-derivation of E(M). Since dM ⊂ K and d is an anti-derivation, we can prove by induction that dMⁱ ⊂ M^{I – 1}, i > 1. We prove next that if d ∈ M*, the anti-derivation extension d satisfies

LEMMA 2.   d² = 0.

Proof.   It is clear from (90) that D1 = 0 for any anti-derivation. Hence dk = 0 for k ∈ K. Then d²M = 0 since dM ⊂ K. We note next that = dx = – d for x ∈ M and if a ∈ E(M) satisfies = – d, then = –d() follows from the fact that d is an anti-derivation. It follows that = – d for all a. This relation implies that d²Mⁱ = 0, by induction on i. Hence d² = 0.

We now have the chain complex

determined by the element d ∈ M*.

We shall require also a result on change of base rings for exterior algebras.

LEMMA 3.   Let M be a module over a commutative ring K, L a commutative algebra over K, and let E(M) be the exterior algebra over M. Then E(M)_L = E(M_L).

Proof.   Since M is a direct summand of E(M), we can identify M_L with the subset of E(M)_L of elements . We have x_i² = 0 and hence for x_i, x_j ∈ M. This implies that (∑l_i x_i)² = 0 for every ∑l_i x_i ∈ M_L. Let be an L-homomorphism of M_L into an algebra / L such that ()² = 0, ∈ M_L. Now becomes an algebra over K if we define . If x ∈ M, then is a K-homomorphism of M into / K such that f(x)² = 0. Hence this has an extension to a K-homomorphism f of E(M) into / K. Then we have the homomorphism 1 f of E(M)_L into ( / K)_L and we have the L-algebra homomorphism of ( / K)_L into / L such that l . Taking the composite we obtain an L-algebra homomorphism of E(M)_L into / L. This maps the element 1 x, x ∈ M, into f(x) = (1 x). Hence it coincides with the given L-homomorphism on M_L. Thus we have obtained an extension of to an algebra homomorphism of E(M)_L into / L. Since M generates E(M), M_L generates E(M)_L. Hence the extension is unique. We have therefore shown that E(M)_L has the universal property characterizing E(M)_L and so we may identify these two algebras.

We now specialize K to R = F[x₁,…, x_m] where F is a field and the x_i are indeterminates. Let V be an m-dimensional vector space over F, (y₁,…, y_m) a base for V/ F, E(V) the exterior algebra determined by V. Then , and V^r has the base of elements y_i1 … y_ir where i₁ < i₂ < … < i_r (BAI, p. 415). Let M = V_R. Then by Lemma 3, where M^r has the base {y_i1 … y_ir} over R. Thus this module is free of rank over R. (This can also be seen directly by using the same method employed in the case V/ F.) Let d be the element of M* = hom_R(M, R) such that d_yi = x_i, 1 ≤ i ≤ m, and let d also denote the anti-derivation in E(M) extending d. Then we have the chain complex (94) and we wish to show that if ε is the canonical homomorphism of R into F obtained by evaluating a polynomial at (0,…, 0), then

is a resolution of F as R-module with ε as augmentation. Since F ⊂ R ⊂ E(M) and E(M) is a vector space over F, we can extend ε to a linear transformation ε in E(M)/ F so that ε(Mⁱ) = 0, i ≥ 1. Then εE(M) = εR = F and so dε = 0. Also since dE(M) ⊂ dM + ∑_{i ≥ 1}Mⁱ and dM is the ideal in R generated by the elements x_i = dy_i, we have εdM = 0 and so εdE(M) = 0. Thus we have

Also since ∑_{i ≥ 1}Mⁱ is an ideal in E(M) and ε | R is an F-algebra homomorphism of R into F, we have

for a, b ∈ E(M). Thus ε is an F-algebra homomorphism of E(M).

The proof that (95) is exact is similar to two other proofs of exactness that we have given (p. 359 and p. 373). It is based on the following

LEMMA 4.   There exists a linear transformation s in E(M)/ F such that

Proof We use induction on m. If m = 1, then M = Ry and E(M) = R1 Ry. Since (1, x, x²,…) is a base for R = F[x] over F, (1, x, x²,…, y, xy, x²y,…) is a base for E(M)/ F. We have dxⁱ = 0, dxⁱy = x^{i + 1}, i ≥ 0, and ε1 = 1, εx^{i + 1} = 0, εxⁱy = 0. Let s be the linear transformation in E(M)/ F such that s1 = 0, sxⁱ = x^{i – 1}y, sxⁱy = 0. Then it is readily checked that (98) holds. We note also that

if l is the automorphism defined before. Now assume the lemma for m – 1 > 0. Let E₁ be the F-subspace of E(M) spanned by the elements , and E₂ the F-subspace spanned by the elements where k_j ≥ 0 and . Then it is clear by looking at bases that E₁ is a subalgebra isomorphic to E(M₁), M₁ = Ry₁, E₂ is a subalgebra isomorphic to E(M₂), M₂ = ∑^m₂Ry_j, and we have a vector space isomorphism (not algebra isomorphism) of E_{1 F}E₂ onto E(M) such that u v uv. We note also that E₁ and E₂ are stabilized by l and by d and the induced maps are as defined in E(M). Using induction we have a linear transformation s₂ in E₂ such that s₂d + ds₂ = l – ε. Let s₁ be the linear transformation in E_l as defined at the beginning of the proof. Since E(M) = , there exists a unique linear transformation s in E(M) such that

for u ∈ E₁, v ∈ E₂. Then

Hence

since ε is an algebra homomorphism. This completes the proof.

We can now prove that the sequence (95) is exact. First, we know that ε is surjective, so R → F → 0 is exact. We have seen also that dM is the ideal in R generated by the x_i. Since this is the ideal of polynomials vanishing at (0, 0, …, 0), we have dM = J = ker ε. Hence is exact. It remains to show that if z_i ∈ Mⁱ, i ≥ 1, and dz_i = 0, then there exists a w_{i + 1} ∈ M^{i + 1} such that dw_{i + 1} = Z_i. We have z_i = 1z_i = (1 – ε)z_i = (sd + ds)z_i = d(sz_i) = dw, w = sz_i. Now we can write , then dw_j ∈ M^{j – 1} so dw = z_i gives dw_{i + 1} = z_i. Thus (95) is exact.

We summarize our results in

THEOREM 6.20.   Let R = F[x₁,…, x_m], x_i indeterminates, F a field, and let M be the free R-module with base (y₁,…, y_m), E(M) the exterior algebra defined by M. Let d be the anti-derivation in E(M) such that dy_i = x_i, 1 ≤ i ≤ m, ε the ring homomorphism of R into F such that εf = f(0,…, 0). Then dMⁱ ⊂ M^{i – 1} and is a free resolution of F as R-module.

We call this resolution the Koszul resolution of F as jR-module and the complex the Koszul complex for R.

Since M^k = 0 for k > m, we evidently have h.dim F ≤ m. We claim that, in fact, h.dim F = m. This will follow from one of the remarks following Theorem 6.18, by showing that Tor_m^R(F, F) ≠ 0. More generally, we determine Tor_r^R(F, F) by using the Koszul resolution of the second F. Then we have the complex

whose homology groups give Tor ₀^R(F, F), Tor₁^R(F, F),…. Now where V is the m-dimensional vector space over F (as above) and . If we use the base for , as before, we can follow through the chain of isomorphisms and see that the isomorphism of F _RM^r onto V^r / F sends for α ∈ F, f ∈ R into . The definition of d gives

Hence under our isomorphism. Thus the boundary operators in (101) are all 0 and the isomorphism F _R M^r → V^r gives the F-isomorphism

In particular, Tor_m^R(F, F) V^m F.

The result just proved that h.dimF = m implies that R = F[x₁,…, x_m] has global dimension ≥ m. It is not difficult to supplement this result and prove that the global dimension of R is exactly m. We shall indicate this in the exercises. In the remainder of the section we shall consider a somewhat similar result on free resolutions of graded modules for R that is due to Hilbert. Of course, Hilbert had to state his theorem in a cruder form than we are able to, since the concepts that we shall use were not available to him.

We recall first the standard grading of R = F[x₁,…, x_m] as

where R⁽ⁱ⁾ is the subspace over F of i-forms, that is, F-linear combinations of monomials of total degree i in the x’s. We have R⁽ⁱ⁾R^(j) ⊂ R^{(i + j)}.

If R is a graded ring, graded by the subgroups R⁽ⁱ⁾ of the additive group then an R-module M is said to be graded by the subgroups M⁽ⁱ⁾, i = 0, 1,…, of its additive group if and R⁽ⁱ⁾M^(j) ⊂ M^{(i + j)} for every i, j. The elements of M⁽ⁱ⁾ are called homogeneous of degree i. A submodule N of M is called homogeneous if N = ∑(N ∩ M⁽ⁱ⁾), or equivalently, if v = ∑v⁽ⁱ⁾ ∈ N where v⁽ⁱ⁾ ∈ M⁽ⁱ⁾, then every v⁽ⁱ⁾ ∈ N. Then N is graded by the submodules N⁽ⁱ⁾ = N ∩ M⁽ⁱ⁾. Moreover, M / N is graded by the subgroups (M⁽ⁱ⁾ + N)/ N and . If M and N are graded R-modules, a homomorphism η of graded modules (of degree 0) is a homomorphism in the usual sense of R-modules such that ηM⁽ⁱ⁾ ⊂ N⁽ⁱ⁾ for every i. Then the image η(M) is a homogeneous submodule of N and ker η is a homogeneous submodule of M.

If M is a graded module for the graded ring R, we can choose a set of generators {u_x} for M such that every u_x is homogeneous (e.g., {u_x} = M⁽ⁱ⁾). Let {e_x} be a set of elements indexed by the same set I = {α} as {u_x} and let L be a free R-module with base {e_x}. Then we have the homomorphism ε of L onto M such that e_x u_x. Let L⁽ⁱ⁾ be the set of sums of the elements of the form a^(j)e_x where a^(j) ∈ R^(j) and j + deg u_x = i. Then it is readily seen that L is graded by the L⁽ⁱ⁾ and it is clear that the epimorphism e is a homomorphism of graded modules. It is easily seen also that if N is a graded module and η : is a homomorphism (of graded modules), then there exists a homomorphism ζ such that

is commutative.

We now suppose that M is a graded module for R = F[x₁,…, x_m]. Let L = L₀ be a free graded module for which we have an epimorphism ε of L₀ onto M and let K₀ = ker ε. Then K₀ is graded and hence we can determine a free graded module L₁ with an epimorphism ε₁ of L₁ onto K₀. Combining with the injection of K₀ into L₀ we obtain d₁:L₁ -→ L₀. Again, let K₁ = ker d₁ and let L₂, ε₂ be a free graded module and epimorphism of L₂ onto K₁, d₂ the composite of this with the injection of K₁ into L₁. We continue this process. Then Hilbert’s syzygy theorem states that in at most m steps we obtain a kernel K_i that is itself free, so that one does not have to continue. In a slightly different form we can state

HILBERT’S SYZYGY THEOREM.   Let M be a graded module for R = F[x₁,…, x_m], the ring of polynomials in m indeterminates with coefficients in afield F. Let be an exact sequence of homomorphisms of graded modules such that every L_i is free. Then K_m is free.

The proof will be based on the result that h.dim F ≤ m and two further results that we shall now derive. As before, F is regarded as an R-module in which JF = 0, where J is the ideal of polynomials vanishing at (0,…, 0). Since J = ∑_{i > 0} R⁽ⁱ⁾, F becomes a graded R-module if we put F⁽⁰⁾ = F, F⁽ⁱ⁾ = 0 for i ≥ 1. We now prove

LEMMA 5. 7/M is a graded R-module and M _RF = 0, then M = 0.

Proof. We have the exact sequence 0 → J → F → 0. This gives the exact sequence . Since and M _RF = 0, by hypothesis, we have the exactness of M _R J → M → 0. The map here sends an element ∑u_i f_i into ∑f_iu_i, so the exactness means that every element of M can be written in the form ∑f_iu_i where the u_i ∈ M and the f_i ∈ J. It follows that M = 0. For, otherwise, let u be a non-zero of M of minimum degree ( = highest degree of the homogeneous parts). Then u = ∑f_iu_i, f_i, u_i ∈ M, f_i and u_i homogeneous. Since the degrees of the f_i are > 0 we have a u_i ≠ O with deg u_i < deg u, contrary to the choice of u. Hence M = 0.

The key result for the proof of Hilbert’s theorem is

THEOREM 6.21.   If M is a graded module such that Tor₁^R(M, F) = 0, then M is free.

Proof We can regard M _RF as vector space over F by restricting the action of R to F. Then if αβ ∈ F and u ∈ M, . It follows that the elements of the form u 1, u ∈ M, u homogeneous, span M _RF as vector space over F and so we can choose a base for M _RF of the form {u_i 1 | u_i ∈ M}. Let L be a free R-module with base {b_i}. We shall prove that M is free with base {u_i} by showing that the homomorphism η of L into M such that ηb_i = u_i is an isomorphism. Let C be the cokernel of η. Then we have the exact sequence and hence we have the exact sequence . The map η 1 sends b_i 1 into u_i 1 and since {u_i 1} is a base for M _RF, η 1 is an isomorphism. Hence C _RF = 0 and so, by Lemma 5, C = 0. This means that η is surjective and is exact. Let K now denote ker η. Then we have the exact sequence 0 → K → L → M → 0, which gives the exact sequence Tor₁^R(M, F) → K _RF → L _RF M _RF. By hypothesis, Tor₁^R(M, F) = 0, and we have seen that η 1 is an isomorphism. Hence K _R F = 0 and so, again by Lemma 5, K = 0. Then, ker η = 0 and η is an isomorphism.

Since any projective module satisfies the hypothesis Tor₁^R(M, F) = 0 (Theorem 6.10, p. 354), we have the following consequence of Theorem 6.21.

COROLLARY.   Any projective R-module for R = F[x₁,…, x_m] that is graded free.

We can now give the

Proof of Hilberfs syzygy theorem. The argument used in the proof of the implication (2) =>(3) in Theorem 6.18, p. 376, shows that Tor₁^R(K_m, F) Tor^R_m(M, F). Since h.dim F ≤ m, by one of the remarks following Theorem 6.18. Hence Tor₁^R{K_m, F) = 0 and so K_m is free by Theorem 6.21.

EXERCISES

        1. Let K be a commutative ring of finite global dimension m and let K[x] be the polynomial ring in an indeterminate x over K. Let M be a K[x]-module and = K[X] _KM. Note that any element of can be written in one and only one way in the form . Note that there is a K[x]-homomorphism η of onto M such that f(x) u f(x)u for f(x) ∈ K[x], u ∈ M. Show that N = ker η is the set of elements (x u₀ – 1 xu₀) and that the map is a K[x]-isomorphism, so we have an exact sequence of K[x]-homomorphisms By exercise 1, p. 278, h.dim_K[x] = h.dim_K M ≤ m. Hence, by Theorem 6.19, h.dim_K[x]M ≤ m + 1 and the global dimension of K[x] ≤ m +1.

        2. Prove that if F is a field and x_i are indeterminates, then the global dimension of F[x₁,…, x_m] is m.

REFERENCES

D. Hilbert, Uber die Theorie der Algebraischen Formen, Mathematische Annalen, vol. 36 (1890), pp. 473–534.

H. Cartan and S. Eilenberg, Homological Algebra, Princeton University Press, Princeton, N.J., 1956.

S. MacLane, Homology, Springer, New York, 1963.

P. J. Hilton and U. Stammbach, A Course in Homological Algebra, Springer, New York, 1970.