Algebra IV

Alexei Skorobogatov

April 7, 2020

Abstract

This course is an introduction to homological algebra and group cohomology.

Contents

1 Modules over a ring 2

1.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Injective and projective modules . . . . . . . . . . . . . . . . . . . . . 5

1.3 Hom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Tensor product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.5 Semisimple modules . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Modules over integral domains 13

2.1 Torsion and divisible submodules . . . . . . . . . . . . . . . . . . . . 13

2.2 Modules over principal ideal domains . . . . . . . . . . . . . . . . . . 14

2.3 Enough injectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Resolutions and derived functors 18

3.1 Projective and injective resolutions . . . . . . . . . . . . . . . . . . . 18

3.2 Derived functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.3 Tor and Ext . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.4 Ext as the group of extensions . . . . . . . . . . . . . . . . . . . . . . 26

4 Group cohomology and the Brauer group 29

4.1 Group ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.2 Cohomology of cyclic groups . . . . . . . . . . . . . . . . . . . . . . . 31

4.3 Standard resolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.4 Change of the group . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.5 Central simple algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 37

1

1 Modules over a ring

1.1 Definitions and examples

[2, Ch.1, §1]

Let R be an associative ring with 1. Recall that this means that R is an abeliangroup with respect to addition (the group operation is denoted by +) that has anassociative operation of multiplication (denoted by ×) such that

x(y + z) = xy + xz, (x+ y)z = xz + yz,

for any x, y, z ∈ R. We require that R has a unit 1 for multiplication, i.e. 1 × x =x× 1 = x for any x ∈ R.

We do not require R to be commutative with regard to multiplication.

R∗ denotes the set of elements x ∈ R for which there exists y ∈ R such thatxy = 1 and yx = 1. Such elements are called invertible. The set R∗ is a group. IfR∗ = R \ {0}, then R is called a skew-field. If, moreover, R is commutative, then Ris called a field.

Key examples of rings: Z; fields Q, R, C, Fq; polynomial rings k[x1, . . . , xn], wherek is a field; the ring of square matrices Matn(k); the skew-field of quaternions H.

The central concept of this course is that of a module.

Definition 1.1 A left R-module M is an abelian group (written additively) with afunction ? : R×M →M (called the left action of R) satisfying

r ? (m1 +m2) = r ? m1 + r ? m2, (r1 + r2) ? m = r1 ? m+ r2 ? m,

1 ? m = m, (r1r2) ? m = r1 ? (r2 ? m),

for any r, r1, r2 ∈ R and m,m1,m2 ∈M .

Unless explicitly mentioned, all our modules are left modules.

Examples

• M = Rn with coordinate-wise left action of R, i.e. r ∈ R sends (r1, . . . , rn) to(rr1, . . . , rrn). These left R-modules are called free of finite rank.

• Any left ideal I ⊂ R (i.e. a subgroup of R closed under left multiplication bythe elements of R) is a left R-module.

• Given a linear transformation L : V → V of a vector space V over a field k

one turns V into a k[x]-module by making x act by L.

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• Take R = Matn(k) and M = kn identified with column vectors. The leftaction of R on M is defined by applying a square matrix to a column vector.

If there are left modules, there should be also right modules! Their definition isobtained by replacing the property (r1r2) ? m = r1 ? (r2 ? m) by

(r1r2) ? m = r2 ? (r1 ? m).

An example of a right Matn(k)-module is the vector space of row vectors kn, onwhich square matrices act by right multiplication.

From now on we shall denote the action of R on a left module by rm and the actionof R on a right module by mr. Thus for the left modules we have (r1r2)m = r1(r2m),whereas for the right modules we have m(r1r2) = (mr1)r2.

The opposite ring Rop of R is obtained by replacing the multiplication xy byyx. It is trivial that a right R-module is the same as a left Rop-module. If R iscommutative, then Rop is the same as R, and there is no difference between left andright modules.

Some known classes of objects can be interpreted as modules over a specific ring:

• Abelian groups = modules over the (commutative) ring Z.

• Vector spaces over a field k = modules over the (commutative) ring k.

• Representations of a group G over a field k = modules over the group algebrak[G]. This is the vector space over k with basis eg, g ∈ G, with multiplicationgiven by eg · eh = egh.

A homomorphism (or simply a map) of left R-modules f : L → M is a homo-morphism of groups that preserves the action of R: f(rx) = rf(x) for any r ∈ Rand x ∈ L. An isomorphism (resp. an injective, a surjective map) of R-modules isa bijective (resp. an injective, a surjective) map of R-modules.

A subgroup L ⊂M stable under the action of R is caled a submodule of M . SinceM is an abelian group under +, we can form the quotient group M/L. But L isR-stable, so we have a well defined action of R on M/L, which is called the quotientmodule.

For example, a left R-submodule of R is the same as a left ideal of R.

Let S be a set. Suppose we are given a left R-module Ms for each s ∈ S. The directproduct

∏s∈SMs is defined as the set of “vectors” (ms)s∈S, where ms ∈ Ms. The

addition is defined coordinate-wise. The action of r ∈ R multiplies each coordinateby r. For a fixed element s0 ∈ S the canonical surjective map

∏s∈SMs → Ms0 is a

map of R-modules. If all Ms are isomorphic to a given module M , we denote thedirect product by MS, because this is the same as the set of all functions S →M .

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The direct sum ⊕s∈SMs is the submodule of∏

s∈SMs given by the condition thatall but finitely many coordinates are zero. For a fixed element s0 ∈ S the canonicalinjective map Ms0 → ⊕s∈SMs is a map of R-modules. If S is finite, then there is nodifference between the direct sum and the direct product.

A free module is a direct sum of copies of the left R-module R. In the particularcase when S is finite, a free module is isomorphic to Rn for some positive integer n.

A sequence (finite or infinite) of modules and maps

. . .fn−1−→ An

fn−→ An+1fn+1−→ An+2

fn+1−→ . . .

is called a complex if fn+1fn = 0 whenever this is defined. A complex is called exactif Ker(fn+1) = Im(fn) for all n. An exact complex

0 −→ Aα−→ B

β−→ C −→ 0 (1)

is called a short exact sequence. To give a short exact sequence is the same as to givean injective map α : A → B (then C = B/α(A)), or a surjective map β : B → C(then A = Ker(β)). The map α gives an isomorphism of A with α(A), so we usuallyidentify A with the submodule α(A) ⊂ B, and C with the quotient module B/α(A).

An example of an exact sequence is

0 −→ Aα−→ A⊕ C β−→ C −→ 0,

where α(x) = (x, 0) and β((x, y)) = y. Such exact sequences are called split. Notall sequences are split, for example this sequence of abelian groups (=Z-modules) isnot:

0 −→ Z/2 −→ Z/4 −→ Z/2 −→ 0. (2)

Proposition 1.2 An exact sequence of left R-modules is split if and only if there isa map of R-modules σ : C → B such that βσ = idC. (Such a map is called a sectionof β.) Equivalently, if there is a map ρ : B → A such that ρα = idA. (Such a mapis called a retraction of α.)

Proof. If the sequence is split, then σ and ρ are the obvious maps.

Now suppose that σ exists. We need to identify B with A ⊕ C so that α and β

are the natural injective and surjective maps, respectively. Let f : B → A ⊕ C bethe map

f(b) = (α−1(b− σβ(b)), β(b)).

It is well defined because b−σβ(b) ∈ Ker(β) = Im(α) and α induces an isomorphismA−̃→Im(α). The kernel of f is contained in Ker(β) ⊂ B. However, the restrictionof f to Ker(β) = Im(α) is an isomorphism (α−1, 0), hence f is injective.

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Let us prove that f is surjective. Let a ∈ A and c ∈ C. Since β is surjective, wecan find x ∈ B such that β(x) = c. For any a′ ∈ A we have β(x+ a′) = c, so we canmodify x by any element of A without changing c. For example, we can replace xby x− a′, where a′ = x− σβ(x)− a ∈ A. Then f(x− a′) = (a, c). �

Exercises 1. Prove the second claim of Proposition 1.2.

2. Show that (1) is split when C = R.

3. Show that (1) is split when C is a free R-module.

4. Show that (2) is not split.

1.2 Injective and projective modules

[2, Ch. 1, §2,3]

An R-module M is called projective if for any exact sequence (1) and for any mapof R-modules f : M → C there is a map of R-modules g : M → B such that f = βg.One says that f ‘lifts’ to g, or that g is a ‘lifting’ of f .

If C is projective, then (1) has a section and so is split.

Lemma 1.3 A direct sum is projective if and only if each summand is projective.

Proof. Let M = ⊕s∈SMs and f : M → C. If Ms is projective for each s ∈ S,then the restriction of f to each Ms (embedded into M in the obvious way), call itfs : Ms → C, lifts to gs : Ms → B. By definition each element of the direct sumhas only finitely many non-zero coordinates, hence g :=

∑s∈S gs is a well defined

function M → B. It is clear that βg = f .

Conversely, assume that M is projective. A map fs : Ms → C for one elements ∈ S extends trivially to a map f : M → C whose restriction to any other summandis zero. It lifts to g : M → B. Restricting to Ms we get a lifting of fs. �

Proposition 1.4 (Criterion of projectivity) A module is projective if and onlyif it is a direct summand of a free module.

Proof. Any module M is a quotient of the free module F freely generated as anR-module by the elements of M . In other words, F consists of (rm)m∈M , whereeach rm ∈ R and only finitely many rm can be non-zero. There is a natural mapof R-modules F → M sending (rm)m∈M ∈ F to

∑m∈M rmm ∈ M . When M is

projective, there is a section. By Proposition 1.2 M is a direct summand of F .

The R-module R is projective, because one element (namely, 1) can be lifted. ByLemma 1.3 all free modules are projective. Again by Lemma 1.3 a direct summandof a free module is projective. �

The fact that every R-module is a quotient of a projective R-module is usuallystated as ‘the category of left R-modules has enough projectives’.

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Example Projective modules are not always free. Let R = Z/6 and let M bethe principal ideal of R generated by 3, so M ∼= Z/2. The abelian group Z/6 isisomorphic to Z/2×Z/3. One checks that this gives an isomorphism of Z/6-modulesZ/6 ∼= Z/2 ⊕ Z/3. Hence Z/2 is a projective Z/6-modules. Since 2 is not a powerof 6, it is certainly not free.

An R-module M is called injective if for any exact sequence (1) and for any mapof R-modules f : A→M there is a map of R-modules g : B →M such that f = gα.

The theory of injetive modules is in a certain sense ‘dual’ to the theory of projec-tive models. If A is injective, then (1) has a retraction and so is split.

Lemma 1.5 A direct product is injective if and only if each factor is injective.

Proof. Let M =∏

s∈SMs and f : A→M . If Ms is injective for each s ∈ S, then thes-coordinate of f , which is a map fs : A→Ms, is such that there exists gs : B →Ms

with the property gsα = fs. The product (which can be infinite) g :=∏

s∈S gs is awell defined function B →M . It is clear that gα = f .

Conversely, assume that M is injective. A map fs : A → Ms for one elements ∈ S extends trivially to a map f : A → M . Then there is g : B → M such thatgα = f . The s-coordinate gs of g has the desired property gsα = fs. �

Theorem 1.6 (Baer’s criterion of injectivity) A module M is injective if andonly if for any map of left R-modules f : I →M , where I is a left ideal of R, thereis an element m ∈M such that f has the form f(x) = xm.

Equivalently, any map of left R-modules I → M extends to a map of left R-modules R→M .

Proof. To see the equivalence of two properties we note that to give a map of leftR-modules R → M is the same as to specify the image of 1 ∈ R. So if f(x) = xm

for any x ∈ I, then f can be extended to R by sending 1 to m. Conversely, any mapI →M which is a restriction of some map F : R→M has the form F (x) = xF (1).

Now let’s prove the theorem. If M is injective, then any f : I →M extends to amap of R-modules R→M , so one direction is clear.

The proof of the converse uses Zorn’s lemma. Let A ⊂ B be left R-modules andlet g : A → M be a map of left R-modules. To show that M is injective we needto show that g extends to a map of R-modules B → M . All pairs (A′, g′), whereA′ is a submodule of B containing A and g′ : A′ → M is a map of R-modules,form a partially ordered set: a pair (A′′, g′′) dominates (A′, g′) if A′ ⊂ A′′ and therestriction of g′′ to A′ is g′. To apply Zorn’s lemma we need to check that eachtotally order subset of this set of pairs has a maximal element, but this is clear asone takes the union of these sets. By Zorn’s lemma this set has a maximal element.Call it (A0, g0).

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Let us prove that A0 = B. Otherwise there is an b ∈ B, b /∈ A0. To deduce acontradiction we want to show that g0 extends to A0 + Rb 6= A0. We need to useour assumption, so we need an ideal. Consider the set of elements x ∈ R such thatxb ∈ A0. This is a left ideal of R. Let us call it I. For x ∈ I let f(x) = g0(xb); this isa map of R-modules. By the assumption there is an m ∈M such that g0(xb) = xm.Now consider the map A0+Rb→M which sends a+yb to g0(a)+ym, where a ∈ A0

and y ∈ R. We need to check that this map is well defined, i.e. if a+ yb = a′ + y′b

for a′ ∈ A0 and y′ ∈ R, then g0(a) + ym = g0(a′) + y′m. But a − a′ = (y′ − y)b,

hence y − y′ ∈ I and in this case

g0(a)− g0(a′) = g0(a− a′) = g0((y′ − y)b) = f(y′ − y) = (y′ − y)m,

so our map is indeed well defined. It is clearly a map of left R-modules, so we getthe desired contradiction. �

1.3 Hom

[2, Ch. 2, §2,4]

Let A and B be left R-modules. We denote by HomR(A,B) the set of maps of R-modules A→ B. It is an abelian group under addition of maps. We have canonicalisomorphisms

HomR(A1 ⊕ A2, B) ∼= HomR(A1, B)⊕ HomR(A2, B)

andHomR(A,B1 ⊕B2) ∼= HomR(A,B1)⊕ HomR(A,B2).

This can be rephrased by saying that HomR(A,B) can be viewed as an additivebi-functor from the category of R-modules to the category of abelian groups that iscovariant in the second argument and contravariant in the first argument.

We note a few examples. The group HomR(R,M) is canonically isomorphic to Mby the map that associates to f : R→M the value f(1) ∈M .

If Z is given the trivial structure of an R-module, that is, rn = n for every r ∈ Rand n ∈ Z, then HomR(Z,M) is the subgroup of M consisting of the elements onwhich R acts trivially. E.g., if R is the group ring Z[G], then HomZ[G](Z,M) = MG

is the subgroup of G-invariant elements of M .

Let0→ A1 → A2 → A3 → 0 (3)

be an exact sequence of left R-modules.

Lemma 1.7 (i) The sequence of abelian groups

0→ HomR(A3, B)→ HomR(A2, B)→ HomR(A1, B)

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is exact.

(ii) The R-module B is injective if and only if

0→ HomR(A3, B)→ HomR(A2, B)→ HomR(A1, B)→ 0

is exact for any short exact sequence (1.8).

Proof. (i) is easy. (ii) is direct form the definition of an injective module. �

Lemma 1.8 (i) The sequence of abelian groups

0→ HomR(B,A1)→ HomR(B,A2)→ HomR(B,A3)

is exact.

(ii) The R-module B is projective if and only if

0→ HomR(B,A1)→ HomR(B,A2)→ HomR(B,A3)→ 0

is exact for any short exact sequence (1.8).

Proof. (i) is easy. (ii) is direct form the definition of a projective module. �

Now let R = Z and let M be an abelian group. The group HomZ(Z/n,M) isthe subgroup of M consisting of the elements x such that nx = 0. We denote thissupgroup by M [n]. We get the exact sequence

0→ A1[n]→ A2[n]→ A3[n].

The last map in this sequence is not always surjective. A typical case is whenB = Z/2 and the exact sequence is

0→ Z/2→ Z/4→ Z/2→ 0. (4)

Exercises 1. Show that0→ Z→ Q→ Q/Z→ 0

is exact. Is it split?

2. Let M denote the quotient of M by its subgroup nM = {nx|x ∈M}. Use thesnake lemma to prove that the following sequence is exact:

0→ A1[n]→ A2[n]→ A3[n]→ A1/n→ A2/n→ A3/n→ 0.

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1.4 Tensor product

Now we would like to define the tensor product. This is an abelian group associatedto two R-modules. More precisely, the general definition of tensor product is for aright R-module A and a left R-module B. (This subtlety can be ignored when R iscommutative and there is no difference between left and right modules.)

Let F be the free abelian group, i.e. the direct sum of copies of Z, generated byall pairs (a, b), where a ∈ A and b ∈ B. (So in general F will have infinitely manygenerators.) Let S ⊂ F be the subgroup generated by all elements of the form

(a+ a′, b)− (a, b)− (a′, b), (a, b+ b′)− (a, b)− (a, b′),

and also by the elements(ar, b)− (a, rb),

for all r ∈ R. In particular, we have relations n(a, b) = (na, b) = (a, nb), wherea ∈ A, b ∈ B, n ∈ Z.

Define A⊗RB = F/S. The image of the generator (a, b) ∈ F in A⊗RB is denotedby a⊗ b. From the definition we obtain the relations

(a+ a′)⊗ b = a⊗ b+ a′ ⊗ b, a⊗ (b+ b′) = a⊗ b+ a⊗ b′, ar ⊗ b = a⊗ rb,

for all a, a′ ∈ A, b, b′ ∈ B, and r ∈ R.

Lemma 1.9 Let M be a left R-module. The map m 7→ 1 ⊗ m gives a canonicalisomorphism of abelian groups M−̃→R⊗RM . (Here R is a right module over itself.)

Proof. It is clear that our map is a homomorphism. The abelian group R ⊗R M isgenerated by r ⊗m = 1⊗ rm, so M → R ⊗R M is surjective. Let us prove that itis injective too.

Recall that R ⊗R M is the quotient of a free abelian group F freely generatedby all pairs (r,m). The function of sets R ×M → M sending (r,m) to rm canbe extended by linearity to a homomorphism F → M . All the defining relationsof the tensor product are contained in its kernel, so we obtain a homomorphismR⊗RM →M . The composition M → R⊗RM →M is the identity map. It followsthat M → R⊗RM is injective. �

In particular, by taking R = Z we obtain the tensor product of abelian groups G1

and G2, denoted by G1⊗ZG2. Note that this is completely different from the usualproduct of two abelian groups G1 × G2. For example, as a particular case of thecanonical isomorphismR⊗RM ∼= M , we obtain a canonical isomorphism Z⊗ZZ ∼= Z.Similarly, Z/2 ⊗Z Z/2 ∼= Z/2 ⊗Z/2 Z/2 ∼= Z/2. Sometimes the result of tensormultiplication can be counter-intuitive. For example, Z/2⊗Z Z/3 = 0. This follows

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from the relations 2(a⊗ b) = 2a⊗ b = 0⊗ b = 0 and 3(a⊗ b) = a⊗ 3b = a⊗ 0 = 0,hence a⊗ b = 0 for all a ∈ A and b ∈ B.

Exercise. Show that Q/Z⊗Z Q/Z = 0.

We note that tensor product is covariant in each argument. Both these functorsare additive, that is, there are obvious canonical isomorphisms

(A1 ⊕ A2)⊗R B ∼= (A1 ⊗R B)⊕ (A2 ⊗R B)

andA⊗R (B1 ⊕B2) ∼= (A⊗R B1)⊕ (A⊗R B2).

Let φ : A×B → A⊗RB be the set-theoretic function mapping (a, b) to a⊗b. Theuniversal property of the tensor product states that if there is an abelian group C anda set-theoretic function f : A×B → C which is linear in each argument and satisfiesf(ar, b) = f(a, rb), then f is uniquely written as f = gφ, where g : A ⊗R B → Cis a homomorphism of abelian groups. (For the proof note that φ is the restrictionof the canonical surjective homomorphism F → A ⊗R B to the basis A × B of F .Similarly, f uniquely extends to a homomorphism F → C. By the assumptions onf , the kernel of F → A ⊗R B is contained in the kernel of F → C. A standardisomorphism theorem for abelian groups then produces a homomorphism g.)

Exercise. Let A be an abelian group. How that the map A → A ⊗Z Z/n givenby a 7→ a ⊗ 1 defines an isomorphism A/n−̃→A ⊗Z Z/n. (Hint: Prove that themap is surjective, then construct a homomorphism A⊗Z Z/n→ A/n such that thecomposition A/n→ A⊗Z Z/n→ A/n is the identity on A/n.)

Lemma 1.10 Let0→ A1 → A2 → A3 → 0

be an exact sequence of right R-modules. For any left R-module B the sequence ofabelian groups

A1 ⊗R B → A2 ⊗R B → A3 ⊗R B → 0

is exact.

Proof. Let Q be the quotient of A2 ⊗R B by the image of A1 ⊗R B. We have anatural map

u : Q→ A3 ⊗R B.

It is enough to show that it is an isomorphism. We choose a set-theoretic sections of A2 → A3, i.e. a function s : A3 → A2 such that βs = idA3 . Now define afunction A3 × B → Q by sending a pair (a, b), where a ∈ A3 and b ∈ B, to theelement of Q which is the coset of s(a) ⊗ b. This map does not depend on thechoise of s, because a different set-theoretic section will change s(a) ⊗ b by (x, b),

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where x ∈ A1. Similar calculations show that the function A3 × B → Q is bilinearand sends (ar, b) and (a, rb) to the same element of Q. By the universal propertyof tensor product, the function A3 × B → Q factors through a homomorphism ofabelian groups v : A3 ⊗R B → Q. One checks that uv and vu are identities, so u isinjective and surjective, hence an isomorphism. �

Let A be an abelian group. With B = Z/n we obtain an exact sequence

A1/n→ A2/n→ A3/n→ 0.

The example of (4) shows that this sequence cannot in general be extended by 0 onthe left.

Definition 1.11 A left R-module is caled flat if ⊗RB preserves exact sequences ofright R-modules.

We see that Z/n is not flat as an abelian group.

Proposition 1.12 All projective modules are flat.

Proof. There is a canonical isomorphism A ⊗R (B ⊕ C) ∼= (A ⊗R B) ⊕ (A ⊗R C),and similarly for any number of summands. It follows that a direct sum of modulesis flat if and only if each summand is flat. Thus all free modules are flat, and so allprojective modules are flat by Proposition 1.4. �

We shall see that the converse is not true: Q is flat as an abelian group (seeTherem 2.6 below) though visibly not a direct summand of a power of Z (since it isdivisible), hence not projective.

Lemma 1.13 If M is a flat left R-module, then for a non-zero divisor a ∈ R andnon-zero m ∈M we have am 6= 0.

Proof. Consider the injective map R→ R of right R-modules sending 1 to a. If Mis flat, then the induced map R⊗RM → R⊗RM is also injective. The identificationm 7→ 1⊗m of M with R ⊗RM allows us to rewrite this map as the map M → Msending m to am. �

1.5 Semisimple modules

[2, Ch. 1, §4] (non-examinable)

A non-zero left R-module M is simple if M and 0 are the only submodules. Asemisimple module is a direct sum of simple modules.

We will use the easy fact that if A,B ⊂M are submodules such that A+B = Mand A ∩ B = 0, then the natural map A⊕ B →M is an isomorphism. In this casewe identify M with A⊕B.

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Lemma 1.14 A module is semisimple if and only if every submodule is a directsummand.

Proof. Assume that M = ⊕Mi, where each Mi is a simple module. Let A ⊂ M

be a submodule. Let I be the maximal (by inclusion) subset of indices such thatA ∩⊕i∈IMi = 0. For each j /∈ I the module Mj ⊕⊕i∈IMi has non-zero intersectionwith A. We deduce that A⊕⊕i∈IMi has non-zero intersection with Mj. But Mj issimple, hence Mj ⊂ A⊕⊕i∈IMi. Therefore, A⊕⊕i∈IMi = M , and we are done.

Conversely, assume that the property holds for M . Then it holds for any non-zero submodule A ⊂ M . Each such A contains simple submodules. Indeed, leta ∈ A, a 6= 0. Using Zorn’s lemma we find a maximal submodule B ⊂ A thatdoes not contain a. By assumption we can write A = B ⊕ C. We claim that Cis simple. Indeed, otherwise C = D ⊕ E, where D and E are non-zero, so thatA = B ⊕D⊕E. If a ∈ B ⊕D and a ∈ B ⊕E, we get a contradiction with the factthat the sum of B,D,E is a direct sum (if a = b+ d = b′ + e, then since a /∈ B thesum (b− b′) + d− e = 0 is a zero sum of non-zero summands and this is not possiblein a direct sum). To fix ideas, assume that a /∈ B ⊕D. This is a contradiction withthe choice of B. We conclude that A contains simple submodules.

There exist sums of simple submodules of M which are direct (e.g. one simplesubmodule). Then any maximal such sum is M . Indeed, otherwise the complementto it in M contains a simple submodule of M , so our direct sum is not actuallymaximal. �

Theorem 1.15 The following statements are equivalent:

(a) The ring R is semisimple as a left R-module;

(b) each left ideal of R is a direct summand;

(c) each left ideal of R is an injective module;

(d) all left R-modules are simisimple;

(e) all short exact sequences of R-modules are split;

(f) all left R-modules are projective;

(g) all left R-modules are injective.

Proof. (a)⇔(b) follows from Lemma 1.14.

(d)⇔(e) follows from Lemma 1.14.

If all modules are projective, then all short exact sequences are split, becausethere is a section. If all exact sequences are split, then every module is a direct sumof a free module that maps surjectively onto it (see the proof of Proposition 1.4),hence it is projective. This gives (e)⇔(f).

In Lemma 2.7 below we shall prove that any module M can be embedded into aninjective module, say M ⊂ I. If all short exact sequences are split, then I = M⊕M ′

for some R-module M ′. But all direct factors of injective modules are injective by

12

Lemma 1.5, hence M is injective. Conversely, if all modules are injective, then allshort exact sequences are split, because then there is a retraction of the injectivemap in the sequence. This shows that (e)⇔(g).

It is clear that (g)⇒(c). If I is a left ideal of R which is injective, then theinclusion of I in R has a retraction, so I is a direct summand. Thus (c)⇒(b).

If (b) holds, then any map of R-modules f : I → M extends to R → M . ByBaer’s criterion of injectivity (Theorem 1.6), M is injective, so we get (g).

We have proved the equivalence of all the statements. �

A ring R satisfying the equivalent conditions of Theorem 1.15 is called semisimple.Later we shall prove that the ring of n×n-matrices Matn(D) with entries in a skew-field (=division algebra) D is a semisimple ring.

2 Modules over integral domains

2.1 Torsion and divisible submodules

[2, Ch. VII, §1]

Let R be an integral domain, i.e. a commutative ring with 1 such that ab = 0implies a = 0 or b = 0.

An element m ∈ M is a torsion element if rm = 0 for some r ∈ R, r 6= 0. DefineMtors as the subset of torsion elements in M ; this is a submodule. If Mtors = 0, thenM is called torsion-free. If M = Mtors, then M is called a torsion module.

Example Abelian groups Z and Q are torsion-free, whereas Z/n, Q/Z and Z/n aretorsion groups.

Exercise Let M be the quotient of Z2 by the subgroup generated by (m,n), wherem,n ∈ Z. Determine Mtors. What is (R/Z)tors?

Lemma 2.1 Any projective module is torsion-free.

Proof. Free modules are torsion-free since R has no zero divisors. Hence theirsubmodules are torsion-free too. Now use Proposition 1.4. �

An element x ∈ M is infinitely divisible if for any r ∈ R, r 6= 0, we can writex = ry for some y ∈ M . Define Mdiv as the subset of infinitely divisible elementsin M ; this is a submodule. If M = Mdiv, then M is called an infinitely divisiblemodule. Often the word ‘infinitely’ is omitted.

Example Abelian groups Q/Z and Q are divisible, whereas Z and Z/n are not.What is (R/Z)div?

Lemma 2.2 Any injective module is divisible.

13

Proof. Let m ∈ M and let r ∈ R, r 6= 0. We need to show that there is an s ∈ Msuch that m = rs.

We note that the principal ideal I = rR and the ring R are isomorphic as R-modules: x 7→ rx is an isomorphism R−̃→I since R has no zero divisors.

The map of R-modules R→M sending x to xm composed with the inverse of theisomorphism R−̃→I gives a map of R-modules f : I → M such that f(rx) = xmfor any x ∈ R.

Consider the natural inclusion of the principal ideal I into R. Since M is injective,f : I → M extends to a map of R-modules F : R → M . We claim that s = F (1)does the job. Indeed, rs = rF (1) = F (r) = f(r) = m. �

2.2 Modules over principal ideal domains

A principal ideal domain (PID) is an integral domain where every ideal is principal.The crucial example for this section is the category of Z-modules, that is, abeliangroups.

Theorem 2.3 If R is a PID, then every submodule of a free module is free. Everysubmodule of Rn is isomorphic to Rm for some m ≤ n.

Proof. According to Zermelo’s theorem any set S can be well-ordered. This meansthat there is a total order a ≤ b on S such that every subset of S has a least element.We fix such an order. In a well-ordered set we can use transfinite induction, whichsays that a statement is true for any s ∈ S if it is true for the least element of Sand if it holds for all x < s, then it also holds for s.

Let S be the indexing set of the free R-module ⊕s∈SR(s), where each R(s) is a copyof R. Write 1(s) for 1 considered as an element of R(s). Let M be a submodule of⊕s∈SR(s). Choose a total order on S. For each t ∈ S we define Mt = M ∩⊕s≤tR(s).The projection to the t-th summand ⊕s∈SR(s) → R restricted to Mt is a map ofR-modules ft : Mt → R. Since R-submodules of R are precisely the ideals of R, andall ideals are principal by assumption, we see that ft(Mt) = Rat for some at ∈ R.Whenever at 6= 0 choose mt ∈Mt such that ft(mt) = at.

Define M ′t as the R-submodule of Mt generated by the ms for s ≤ t. We claim

that Mt = M ′t for all t ∈ S. This is clear for the least element of S. By transfinite

induction it is enough to prove that if Ms = M ′s for all s < t, then Mt = M ′

t . Butany m ∈ Mt can be written as m = rmt + (m − rmt), where m − rmt is a finitelinear combination of 1(s) ∈ R(s) for s < t. Let q be the largest index in this finiteset. Then m − rmt ∈ Mq and, by inductive assumption, m − rmt ∈ M ′

q. But thenm ∈M ′

t . This proves that Mt = M ′t for all t ∈ S, and hence M is generated by the

ms for s ∈ S.

We claim that (ms)s∈S is a free basis of M ; in other words, every element of Mis uniquely written as a finite linear combination of the ms with coefficients in R.

14

Otherwise 0 can be written as∑

s∈T rsms, where T ⊂ S is finite and all rsms 6= 0(which implies as 6= 0). If t is the largest element of T , then the projection to thet-th coordinate kills all terms except possibly rtmt which it sends to rtat ∈ R. Thisis zero, hence rt = 0 since at 6= 0. This contradiction finishes the proof of the firststatement. The second statement is clear, because we proved that M has a basiswhose elements are in a bijection with a subset of S. �

Corollary 2.4 If R is a PID, then projective modules and free modules are the samething.

Proof. Use Proposition 1.4. �

Theorem 2.5 If R is a PID, then injective modules and divisible modules are thesame thing.

Proof. By Lemma 2.2 all injective modules are infinitely divisible, so we need toshow that any divisible module M is injective. Let us use Baer’s criterion (Theorem1.6). So let f : I →M be a map of R-modules, where I ⊂ R is an ideal. Since R isa PID, we have I = Rr for some r ∈ R. As M is divisible we can write f(r) = rsfor some s ∈ M . Then f : I → M extends to a map of R-modules R → M whichsends 1 to s. �

In particular, the abelian groups Q/Z, R/Z and Q are divisible, hence injective.

Theorem 2.6 If R is a PID, then flat modules and torsion-free modules are thesame thing.

Proof. We know from Lemma 1.13 that flat modules are torsion-free, so let us provethe converse. Let A ⊂ B be a submodule and let M be a torsion-free R-module.We need to prove that A⊗RM → B ⊗RM is an injective map.

Step 1: It is enough to prove our claim in the case when B is free. Indeed, weknow that any module is a quotient of a free module F , so there is an exact sequence

0→ K → F → B → 0.

If we denote the inverse image of A in F by A′, we obtain a commutative diagramof R-modules

0 → K → A′ → A → 0↓ ↓ ↓

0 → K → F → B → 0

This gives rise to a commutative diagram of abelian groups

K ⊗RM → A′ ⊗RM → A⊗RM → 0|| ↓ ↓

K ⊗RM → F ⊗RM → B ⊗RM → 0

15

By assumption the middle vertical map is injective. Now a digram chase gives usthat the right vertical map is injective too.

Step 2: It is enough to prove our claim in the case when B = R. Indeed, supposethat some element x ∈ A ⊗R M goes to a non-zero element of F ⊗R M . Anyelement of A⊗RM is a finite sum of elements of the form a⊗m, so x ∈ A0 ⊗RM ,where A0 ⊂ A is a submodule generated by the relevant a. The finitely generatedsubmodule A0 ⊂ F is contained in some Rn ⊂ F , which is free of finite rank. Thusit is enough to show that A0 ⊗R M → Rn ⊗R M is injective for any submoduleA0 ⊂ Rn.

Write Rn = R⊕Rn−1 and A1 = A0∩R. Let A2 be the image of A0 in Rn−1 underthe projection map. We obtain a commutative diagram

0 → A1 → A0 → A2 → 0↓ ↓ ↓

0 → R → Rn → Rn−1 → 0

and hence a commutative diagram

A1 ⊗RM → A0 ⊗RM → A2 ⊗RM → 0↓ ↓ ↓

0 → R⊗RM → Rn ⊗RM → Rn−1 ⊗RM → 0

because the bottom exact sequence of the previous diagram is split. By assumptionthe left vertical map is injective. Arguing by induction we assume that the rightvertical map is injective. Then a diagram chase (or the snake lemma) gives theinjectivity of the middle vertical map.

Step 3: end of proof. Now A is an ideal in R. Since R is a PID, we have A = aR.We need to prove the injectivity of the natural map α : aR⊗RM → R⊗RM . Themap of R-modules R → aR, r 7→ ar, gives rise to β : R ⊗R M → aR ⊗R M . Thismap is clearly surjective. The composition αβ is the map R⊗RM → R⊗RM , whichis identified with the map M →M given by m 7→ am. Since M is torsion-free, thismap is injective. Now αβ is injective and β is surjective, and this implies that α isinjective. �

Now we can justify the example of a module that is flat but not projective givenabove: the abelian group Q is torsion-free, hence flat. But Q is divisible, so it is nota direct summand of a free abelian group, hence Q is not projective.

2.3 Enough injectives

We are now ready to show that the category of R-modules has enough injectives.

Theorem 2.7 Let R be a (not necessarily commutative) ring. Every left R-moduleis isomorphic to a submodule of an injective module.

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Proof. Let us first prove this in the case of abelian groups, i.e. for R = Z.

For every non-zero abelian group G we have HomZ(G,Q/Z) 6= 0. Moreover, forany non-zero x ∈ G there is an f ∈ HomZ(G,Q/Z) such that f(x) 6= 0. Indeed, letC ⊂ G be the cyclic subgroup generated by x. It can be finite or infinite. If C isfinite, then there is an injective homomorphism C → Q/Z (e.g., sending x to 1/|C|).If C is infinite, then consider the homomorphism sending x to 1/2 ∈ Q/Z. In eachcase the image of x is non-zero, so it is an non-zero homomorphism C → Q/Z.By the injectivity of Q/Z (Theorem 2.5) it extends to a non-zero homomorphismG→ Q/Z.

Let I(G) = (Q/Z)HomZ(G,Q/Z) be the product of copies of Q/Z indexed by theelements of HomZ(G,Q/Z). Since any product of injective modules is injective,I(G) is injective. There is a canonical map G → I(G) that sends x ∈ G to theelement of I(G) whose coordinate corresponding to f ∈ HomZ(G,Q/Z) is f(x). Bythe previous paragraph this homomorphism is injective.

Here is the proof for an arbitrary ring R.

Step 1. Consider R as a right R-module. The abelian group S = HomZ(R,Q/Z)can be equipped with the structure of a left R-module by making r ∈ R send ahomomorphism f(x) to f(xr). Indeed, the action of b ∈ R sends f(x) to f(xb), andthen the action of a ∈ R sends f(xb) to f(xab), which is the same as the action ofab.

Step 2. Claim: the R-module S is injective.

Proof. For any R-module M we have a canonical isomorphism of abelian groups

HomR(M,S) = HomR(M,HomZ(R,Q/Z)) ∼= HomZ(M,Q/Z).

The map from left to right is obtained by evaluating at 1 ∈ R. The map from rightto left sends φ(x) : M → Q/Z to the map which sends x ∈M to φ(rx).

Since Q/Z is injective as an abelian group, for any injective map of R-modulesM → N any map M → Q/Z extends to a map N → Q/Z. In other words,the natural map HomZ(N,Q/Z) → HomZ(M,Q/Z) (the restriction to M ⊂ N) issurjective. The above isomorphism for N in place of M now gives that the naturalmap HomR(N,S)→ HomR(M,S) is surjective. This means that S is injective as aleft R-module.

Step 3. For any M we have HomR(M,S) 6= 0; moreover, for any non-zero m ∈Mthere exists an f ∈ HomR(M,S) such that f(m) 6= 0. This follows directly from theprevious steps.

Now let I(M) be the product of copies of S indexed by the elements of HomR(M,S).Since any product of injective modules is injective, I(M) is injective. There is acanonical map M → I(M) that sends m ∈ M to the element of I(M) whose coor-dinate corresponding to f ∈ HomR(M,S) is f(m). Then rm is sent to the elementof I(M) whose f -coordinate is f(rm) = rf(m), so this is a map of left R-modules.

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Step 4. Claim: this map is injective. Indeed, let m ∈ M be non-zero. By Step3 there exists an f ∈ HomR(M,S) such that f(m) 6= 0. Hence this m goes to anon-zero element of I(M). �

3 Resolutions and derived functors

3.1 Projective and injective resolutions

Let R be a ring (not necessarily commutative). A chain complex of R-modules is

a (finite, infinite or semi-infinite) complex . . .d−→ An+1

d−→ And−→ An−1

d−→ . . .The fact that it is a complex is equivalent to d2 = 0. A map of chain complexesf : A• → B• is a family of maps of R-modules fn : An → Bn commuting with thedifferentials d, i.e. such that dfn = fn−1d for all n.

A cochain complex of R-modules is a complex . . .d−→ Cn−1 d−→ Cn d−→ Cn+1 d−→

. . . Maps of cochain complexes are defined as families of map commuting with d.

Let M be a left R-module. We can associate to M chain and cochain complexeswith good properties, as follows.

Definition 3.1 A left resolution of M is a complex of R-modules . . .d−→ Pn

d−→. . .

d−→ P1d−→ P0 together with a map P0 →M such that the complex

. . .d−→ Pn

d−→ Pn−1d−→ . . .

d−→ P1d−→ P0 −→M −→ 0

is exact. A left resolution is denoted by P• → M . If each Pn is projective, theresolution P• →M is called projective.

Definition 3.2 A right resolution of M is a complex of R-modules I0d−→ I1

d−→. . .

d−→ Ind−→ . . . together with a map M → I0 such that the complex

0 −→M −→ I0d−→ I1

d−→ . . .d−→ In

d−→ . . .

is exact. A right resolution is denoted by M → I•. If each In is injective, theresolution M → I• is called injective.

Lemma 3.3 Every module has a projective and injective resolutions.

Proof. We know that for any M there is a surjective map ε : P0 → M , where P0

is free, hence projective. Define M0 = Ker(ε). Take a surjective map P1 → M0

with P1 projective. Now let d : P1 → P0 be the composition P1 → M0 → P0. Itis easy to see that Im[P1 → P0] = M0 = Ker[P0 → M ]. This gives the first bit ofour resolution and proves the exactness at P0. Now define M1 = Ker[P1 → P0] andrepeat the procedure to construct P2, and so on.

An injective resolution is built in exactly the same way using Theorem 2.7. �

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Proposition 3.4 Suppose that we are given a projective resolution P• →M and amap M → N . Then for any left resolution Q• → N there exist maps fn : Pn → Qn,n ≥ 0, making the following diagram commutative

. . . → P2 → P1 → P0 → M → 0↓ ↓ ↓ ↓

. . . → Q2 → Q1 → Q0 → N → 0

If g = (gn) : P• → Q• is another map of complexes making the diagram commutative,then there are maps sn : Pn → Qn+1 such that fn − gn = sn−1dn + dn+1sn for n ≥ 1and f0 − g0 = d1s0.

Proof. Consider the surjective map Q0 → N . By the definition of a projectivemodule the composition P0 → M → N can be lifted to f0 : P0 → Q0. This buildsthe first square of the diagram and shows that it is commutative. Now considerP1 → P0 → Q0. The image of this map goes to 0 in N , so it can be viewed as a mapP1 → Ker[Q0 → N ]. Consider the surjective map Q1 → Ker[Q0 → N ]. Since P1 isprojective, we get a map f1 : P1 → Q1 which builds the second square and showsthat it is commutative. Iterating this procedure constructs all vertical maps one byone and proves the commutativity of respective squares.

To prove the second statement, we consider the zero map M → N and prove thatfor any map of complexes f : P• → Q• making the diagram commutative there aremaps sn : Pn → Qn+1 such that fn = sn−1dn + dn+1sn for n ≥ 1 and f0 = d1s0.Indeed, in this case f0 sends P0 to Ker[Q0 → N ]. Since P0 is projective, f0 lifts to amap s0 : P0 → Q1. This gives f0 = d1s0. Now f1 − s0d1 sends P1 to Ker[Q1 → Q0],hence, by the projectivity of P1, it lifts to s1 : P1 → Q2. We obtain f1−s0d1 = d2s1,which is f1 = s0d1 − d2s1. Now continue by iterating this contruction. �

The second statement of Proposition 3.4 is usually phrased as follows: a map ofcomplexes f = (fn) : P• → Q• making the above diagram commutative is uniqueup to homotopy.

Exercise State and prove the analogous statement for injective resolutions.

3.2 Derived functors

Let A• be a chain complex of R-modules. The n-th homology group Hn(A•) is definedas the quotient of Ker[An → An−1] by Im[An+1 → An]. One defines the cohomologygroups Hn(C•) of a cochain complex C• similarly. A complex is exact if and only ifall its (co-)homology groups are zero.

We note that a map of chain complexes f = (fn) : A• → B• induces maps onhomology groups f∗ = (f∗n) : Hn(A•)→ Hn(B•).

Lemma 3.5 Maps of chain complexes f, g : A• → B• that are equal up to homotopyinduce equal maps of homology groups, i.e. f∗ = g∗.

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Proof. Easy exercise. �

Suppose that we are given maps of chain complexes A• → B• → C•, that is, mapsof R-modules An → Bn → Cn for all n ≥ 0 that commute with the differentials d.We call it a short exact sequence of compexes and write

0→ A• → B• → C• → 0

if for each n ≥ 0 we have a short exact sequence of R-modules

0→ An → Bn → Cn → 0.

Lemma 3.6 A short exact sequence of complexes gives rise to a long exact sequenceof homology groups

. . .→ Hn(A•)→ Hn(B•)→ Hn(C•)→ Hn−1(A•)→ . . .

. . .→ H1(A•)→ H1(B•)→ H1(C•)→ H0(A•)→ H0(B•)→ H0(C•)→ 0.

Proof. We shall use the following form of the snake lemma: a commutative diagramwith exact rows

L → M → N → 0f ↓ g ↓ h ↓

0 → L′ → M ′ → N ′

gives rise to an exact sequence

Ker(f)→ Ker(g)→ Ker(h)→ Coker(f)→ Coker(g)→ Coker(h).

The proof is left to the reader as a (very good and important) exercise.

The idea is to apply the snake lemma to truncated pieces of the short exactsequence of complexes, as follows. We first apply it to the diagram

A1/d(A2) → B1/d(B2) → C1/d(C2) → 0d ↓ d ↓ d ↓

0 → A0 → B0 → C0 → 0

and obtain

H1(A•)→ H1(B•)→ H1(C•)→ H0(A•)→ H0(B•)→ H0(C•)→ 0.

Here we note that H0(B•)→ H0(C•) is surjective because B0 → C0 is surjective.

Next, we apply the snake lemma to

A2/d(A3) → B2/d(B3) → C2/d(C3) → 0d ↓ d ↓ d ↓

0 → Ker[A1 → A0] → Ker[B1 → B0] → Ker[C1 → C0]

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and obtain

H2(A•)→ H2(B•)→ H2(C•)→ H1(A•)→ H1(B•)→ H1(C•).

We can glue this with the previous 6-term exact sequence because the maps betweenthe first homology groups are the same as before. Iterating this procedure we builda long exact sequence of homology groups. �

Let M be a left R-module. For a right R-module A we shall write F (A) = A⊗RM .Then F is a right exact additive functor from the category of right R-modules tothe category of abelian groups. To say that F is a functor means that F (A) is anabelian group for every right R-module A; moreover, to a map of right R-modulesφ : A→ B it associates a homomorphism F (φ) : F (A)→ F (B) in such a way thatF (idA) = idF (A) and F (φ1φ2) = F (φ1)F (φ2). To say that F is additive means thatif φ, φ′ : A → B, then F (φ + φ′) = F (φ) + F (φ′). Finally, to say that F is rightexact means that we have the conclusion of Lemma 1.10.

We are going to define left derived functors LiF from the category of right R-modules to the category of abelian groups, i ≥ 0. They will have these properties:L0F = F , and for any short exact sequence of right R-modules

0→ A→ B → C → 0

we have a (semi-infinite) long exact sequence of derived functors

. . .→ LnF (A)→ LnF (B)→ LnF (C)→ Ln−1F (A)→ Ln−1F (B)→ Ln−1F (C)→ . . .

. . .→ L1F (A)→ L1F (B)→ L1F (C)→ F (A)→ F (B)→ F (C)→ 0.

Definition 3.7 DefineLnF (A) = Hi(F (P•)),

where P• → A is a projective resolution of A.

In other words, LnF (A) is the n-th homology group of the complex

. . .→ F (Pn)→ F (Pn−1)→ . . .→ F (P1)→ F (P0)→ 0.

Theorem 3.8 (i) L0F = F ;

(ii) the abelian group LiF (A) does not depend on the choice of a projective reso-lution of A, for any i ≥ 0;

(iii) LiF is an additive functor, for any i ≥ 0;

(iv) for any short exact sequence of R-modules there is an associated long exactsequence of derived functors.

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Proof. (i) We have a short exact sequence

0→ Im[P1 → P0]→ P0 → A→ 0,

which, since F is right exact (Lemma 1.10), gives an exact sequence

F (Im[P1 → P0])→ F (P0)→ F (A)→ 0.

Since F (P1) surjects onto F (Im[P1 → P0]), we deduce an exact sequence

F (P1)→ F (P0)→ F (A)→ 0,

whence F (A) = H0(F (P•)).

(ii) If Q• → A is another projective resolution, then Proposition 3.4 together withLemma 3.5 gives well defined maps Hn(F (P•)) → Hn(F (Q•)). Reversing the rolesof P• and Q• produces well defined maps in the opposite direction. Moreover, bothcompositions are the maps induced by maps of complexes which up to homotopyare the identity maps. Hence both compositions are the identity maps. Therefore,the canonical maps Hn(F (P•))−̃→Hn(F (Q•)) are isomorphisms.

(iii) Proposition 3.4 together with Lemma 3.5 shows that for any map of R-modules f : A→ B we have associated well defined maps Lif : LiF (A)→ LiF (B).It is clear that Li(idA) is the identity map id : LiF (A) → LiF (A). It is also clearthat Li(fg) = (Lif)(Lig). We conclude that each LiF is a functor from the categoryof R-modules to the category of abelian groups; moreover, it is an additive functor.

(iv) Finally, the existence of a long exact sequence of derived functors follows fromLemma 3.6 and the following Horseshoe lemma. Its idea is that since the LiF donot depend on the choice of a projective resolution, we can construct a resolution ofB from given resolutions of A and C, which will have certain additional properties.

Lemma 3.9 (Horseshoe lemma) Let P• → A and R• → C be projective resolu-tions. Define Qn = Pn ⊕ Rn for all n ≥ 0. One can define maps Qn → Qn−1 forn ≥ 1 and a map Q0 → B to make Q• a projective resolution of B in such a waythat P• → A, Q• → B and R• → C form a short exact sequence of exact complexes.

Remark The maps Qn → Qn−1 that are going to be defined are not the direct sumsof differentials d : Pn → Pn−1 and d : Rn → Rn−1. Nevertheless, each short exactsequence

0→ Pn → Qn → Rn → 0

is split because Pn → Qn = Pn⊕Rn is the natural injective map and Qn = Pn⊕Rn →Rn is the natural surjective map. We can rephrase this by saying that

0→ P• → Q• → R• → 0

is a split short exact sequence of complexes.

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Proof. We need to construct a map Q0 = P0 ⊕ R0 → B. The first componentP0 → A is the augmentation map of P• → A. Since R0 is a projective R-module,the map R0 → C can be lifted to a map R0 → B. It is easy to see that this definesa surjective map Q0 → B.

The snake lemma now gives an exact sequence

0→ Ker[P0 → A]→ Ker[Q0 → B]→ Ker[R0 → C]→ 0.

We repeat the previous argument for Q1 = P1 ⊕ R1 to construct a map Q1 →Ker[Q0 → B] compatible with d : P1 → P0 and d : R1 → R0. Iterating the sameprocedure we prove the lemma. �

Lemma 3.10 (i) An R-module M is a non-tivial direct sum if and only if we canwrite idM = p1 + p2, where p1 and p2 are non-zero endomorphisms of the R-moduleM such that p21 = p1 and p22 = p2.

(ii) If F is an additive functor, then F (A ⊕ B) is canonically isomorphic to thedirect sum F (A)⊕ F (B).

Proof. (i) If M = A⊕B, then let p1 be the projector to A and let p2 be the projectorto B.

Let us prove the converse. Note that idM = p1 + p2, p21 = p1, p

22 = p2 formally

imply p1p2 = p2p1 = 0. Define A = p1(M) and B = p2(M). There is a natural mapA⊕B →M sending a⊕ b to a+ b. It is surjective since idM = p1 +p2. It is injectivesince A ∩B = 0, because p2(A) = 0, p1(B) = 0, hence idM(A ∩B) = 0.

(ii) Write M = A ⊕ B; this defines projectors p1 and p2. An additive functorpreserves identities, sums and compositions, so idF (M) = F (idM) = F (p1) + F (p2).Moreover, F (p1)

2 = F (p1) and F (p2)2 = F (p2). By (i), F (M) is the direct sum of

F (p1)F (M) = F (A) and F (p2)F (M) = F (B). �

Now we can complete the proof of part (iv) of Theorem 3.8. We choose anyprojective resolutions P• → A and R• → C and construct a projective resolutionQ• → B as in the Horseshoe lemma. Apply our functor F to the split short exactsequence of resolutions. By Lemma 3.10 the fact that this sequence is split impliesthat the resulting sequence of complexes of abelian groups

0→ F (P•)→ F (Q•)→ F (R•)→ 0

is exact. (Here F (P•) denotes the complex . . .→ F (Pn)→ F (Pn−1)→ . . .) ApplyingLemma 3.6 to this exact sequence of complexes gives rise to a long exact sequenceof derived functors. �

Exercise Let G be a left exact additive functor from the category of left R-modulesto the category of abelian groups. Using the symmetry between injective and projec-tive modules (which reverses the direction of maps) define the right derived functorsRiG, i ≥ 0. Prove their properties analogous to the properties of left derived func-tors established above.

23

Definition 3.11 The left derived functors of F (A) = A ⊗R B are denoted byLiF (A) = TorRi (A,B). The right derived functors of G(A) = HomR(B,A) aredenoted by RiG(A) = ExtiR(B,A).

Remark Given a right R-module A one can also consider the functor A⊗R− fromthe category of left R-modules to the category of abelian groups, call it F ′(B) =A ⊗R B. This functor is right exact, and so has left derived functors which arecomputed using projective resolutions. A balancing theorem (which we will notprove in this course) states that the left derived functors LiF

′ give rise to the sameabelian groups, i.e. LiF

′(B) = TorRi (A,B) = LiF (A).

Similarly, given a left R-module B consider the contravariant functor HomR(−, B)from the category of left R-modules to the category of abelian groups, call it G′(A) =HomR(A,B). This functor is left exact, and so has right derived functors. SinceG′ is contravariant, i.e. it reverses the direction of maps, the right derived functorsRiG′ are computed using projective resolutions (so that after applying G′ we get acochain complex and consider its cohomology groups). The balancing theorem forExt says that the right derived functors of G′(A) = HomR(A,B) are RiG′(A) =ExtiR(A,B) = RiG(B). See [3, §2.7].

3.3 Tor and Ext

If A is a projective right R-module, then 0 → A is a projective resolution of A, soall higher derived functors are zero. Thus TorRn (A,B) = 0 for any left R-module Band any n ≥ 1.

By the balancing theorem, if B is a projective left R-module, then TorRn (A,B) = 0for any right R-module A and any n ≥ 1.

Now recall that projective modules are flat (Proposition 1.12), but not vice versa.If B is flat, then −⊗RB is an exact functor (i.e. it preserves short exact sequences).By cutting an exact complex into short exact sequence one shows that this impliesthat −⊗R B applied to a projective resolution P• → A gives an exact complex

. . . Pn ⊗R B → Pn−1 ⊗R B → . . .→ P1 ⊗R B → A⊗R B → 0.

Thus all higher derived functors are zero, i.e. TorRn (A,B) = 0 for any A and anyn ≥ 1.

Proposition 3.12 A left R-module B is flat if and only if TorR1 (A,B) = 0 for anyright R-module A (and then TorRn (A,B) = 0 for any A and any n ≥ 1).

Proof. It remains to prove that TorR1 (A,B) = 0 for any A implies that B is flat. Bythe definition of flatness we need to prove that the functor −⊗R B is exact. Let

0→ L→M → N → 0

24

be a short exact sequence of right R-modules. Then we have a long exact sequence

. . .→ TorR1 (N,B)→ L⊗R B →M ⊗R B → N ⊗R B → 0.

Since TorR1 (N,B) = 0 we see that −⊗R B is an exact functor. �

Example Let R = Z. Then for any n ≥ 1 we have

TorZn(Z, B) = TorZn(A,Z) = TorZn(Q, B) = TorZn(A,Q) = 0,

because Z is free and Q is flat.

Let G be a finitely generated abelian group. By basic group theory G is isomorphicto F × Zr, where r ≥ 0 and F is a finite abelian group, hence a product of finitecyclic groups of prime power order. Since TorZm(A,B) is an additive functor in eachargument, in order to compute it in the case when A and B are finitely generated,it remains to compute TorZm(Z/n,B).

We have a natural 2-term projective resolution of Z/n, namely 0→ Z→ Z, wherethe second map is the multiplication by n map [n]. By tensoring this resolution withB we get the complex

0 −→ B[n]−→ B −→ 0.

Thus for any abelian group B we have

TorZ0 (Z/n,B) = Z/n⊗Z B = B/nB, TorZ1 (Z/n,B) = B[n], TorZ≥2(Z/n,B) = 0.

For example, if0→ A→ B → C → 0

is a short exact sequence of abelian groups, then applying the derived functors ofZ/n⊗Z − we get the exact sequence

0→ A[n]→ B[n]→ C[n]→ A/n→ B/n→ C/n→ 0.

Theorem 3.13 For any abelian groups A and B we have TorZ>1(A,B) = 0 andExt>1

Z (A,B) = 0.

Proof. There exists a free abelian group F and a surjective homomorphism F → B,see the proof of Proposition 1.4. Let F ′ be the kernel of F → B. By Theorem 2.3the abelian group F ′ is also free. Hence 0 → F ′ → F is a projective resolution ofB. Since it has only two non-zero terms, all left derived functors vanish for m > 1.

Let I be an injective abelian group such that there is an injective homomorphismB → I. By Theorem 2.5 being injective and being divisible is the same thing forabelian groups. Thus I is divisible, but then I/B is divisible too. Hence I/B isinjective, and I → I/B → 0 is an injective resolution of B. Therefore, all rightderived functors vanish for m > 1. �

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It is a fact that for all abelian groups (not just finitely generated) the groupTorZ1 (A,B) is a torsion abelian group, and TorZ1 (Q/Z, B) is the torsion subgroup ofB. The proof is omitted, see [3, §3.1]. (This follows from the fact that Tor commuteswith direct limits so everything can be reduced to finitely generated abelian groups.)

3.4 Ext as the group of extensions

We start with an important supplement to the main theorem on derived functors(Theorem 3.8). Suppose that we have a commutative diagram of exact sequences ofR-modules

0 −→ L −→ M −→ N −→ 0f ↓ g ↓ h ↓

0 −→ L′ −→ M ′ −→ N ′ −→ 0(5)

If A is an R-module, then we have connecting homomorphisms

∂ : HomR(A,N) −→ Ext1R(A,L), ∂′ : HomR(A,N ′) −→ Ext1R(A,L′)

from the respective long exact sequences of Ext-groups. The important fact (usuallyrefered to as the ‘naturality’ of the long exact sequence of derived functors) is thatwe actually have a commutative square

HomR(A,N)∂−→ Ext1R(A,L)

h∗ ↓ f∗ ↓HomR(A,N ′)

∂′−→ Ext1R(A,L′)

(6)

where, as usual, h∗ sends a map φ : A → N to the composed map hφ : A → N ′.The map f∗ is induced by f in view of the functoriality of Ext1R(A,−). In fact, f∗,g∗ and h∗ induce maps between the long exact sequences of Ext-groups making theresulting diagram commutative. We only need to worry about connecting maps,because elsewhere this follows from the functoriality of Ext1R(A,−). The proof ofthis fact passes by replacing (5) by a commutative diagram of resolutions as in theHorseshoe Lemma; we omit this proof here, see [3, Thm. 2.4.6].

A short exact sequence of R-modules 0 → A → B → C → 0 is also called anextension of C by A.

Definition 3.14 Let us call two extensions of R-modules 0 → A → B → C → 0and 0→ A→ B′ → C → 0 equivalent if there is a commutative diagram

0 −→ A −→ B −→ C −→ 0|| ↓ ||

0 −→ A −→ B′ −→ C −→ 0

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It is clear that all maps in the above diagram are isomorphisms. It follows thatthis is indeed an equivalence relation.

Definition 3.15 Given two extensions of R-modules 0 → Aα→ B

β→ C → 0 and

0→ Aα′→ B′

β′→ C → 0 their Baer sum is the extension

0 −→ A −→ X −→ C −→ 0,

where X is the homology group of the complex

A(α,−α′)−→ B ⊕B′ β−β

′−→ C.

The map A→ X is obtained from (α, 0), and the map X → C is obtained from β.

Lemma 3.16 The Baer sum turns the set of equivalence class of extensions intoan abelian group; the zero element is the class of split extensions.

Proof. Do it as an exercise. Write down an extension whose equivalence class isinverse to the class of a given extension. (Hint: change the sign of one of the maps.)Show that the Baer sum of these extensions is split.

Definition 3.17 The class of an extension of R-modules

0 −→ A −→ B −→ C −→ 0

is the image of idC ∈ HomR(C,C) under the connecting homomorphism

∂ : HomR(C,C) −→ Ext1R(C,A).

Remark. Equivalent extensions have the same class. Indeed, if the extensions in(5) are such that f : L→ L′ and h : N → N ′ are the identity maps, then (6) showsthat the associated maps ∂ and ∂′ coinside.

Theorem 3.18 The class defines a bijection between the equivalent classes of ex-tensions of C by A and Ext1R(C,A). It is a group isomorphism under the Baer sumof extensions.

Proof. Let us realise an arbitrary element x ∈ Ext1R(C,A) as the class of someextension of C by A. Choose an injective R-module I with an injective map A→ I.Let M = I/A and call µ : I → M the natural surjective map. This gives an exactsequence of R-modules

0 −→ A −→ Iµ−→M −→ 0. (7)

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We have Ext1R(C, I) = 0 since I is injective, so we obtain an exact sequence

HomR(C, I)µ∗−→ HomR(C,M) −→ Ext1R(C,A) −→ 0.

Let φ : C → M be a map of R-modules such that φ ∈ HomR(C,M) goes tox ∈ Ext1R(C,A). The desired extension will be constructed as the ‘pull-back’ of (7)via φ. Namely, let Xφ ⊂ I ⊕ C be the kernel of the map µ − φ : I ⊕ C → M .The kernel of the natural projection Xφ → C is identified with A, so we have acommutative diagram of short exact sequences of R-modules

0 −→ A −→ Iµ−→ M −→ 0

|| ↑ φ ↑0 −→ A −→ Xφ −→ C −→ 0

The commutative diagram (6) in our case can be written as

φ ∈ HomR(C,M)∂−→ Ext1R(C,A)

↑ φ∗ ↑ ||id ∈ HomR(C,C)

∂′−→ Ext1R(C,A)

(8)

Thus the class of the bottom sequence ∂′(idC) = ∂(φ) = x, as desired. In particular,the constructed extension is split only if x = 0.

Any lifting φ′ of x has the form φ+ µρ, where ρ : C → I is a map of R-modules.The automorphism of I ⊕ C given by (a, b) 7→ (a + ρ(b), b) gives an isomorphismXφ−̃→Xφ′ compatible with maps Xφ → C and A → Xφ. So the extension of C byA constructed from φ′ is equivalent to the one constructed from φ.

To finish the proof that our construction gives a bijection between the equivalentclasses of extensions of C by A and Ext1R(C,A) we need to show that any extension0 → A → B → C → 0 is a pull-back of (7). For this it is enough to construct amap of R-modules B → I which extends A → I. Such a map exists because I isinjective. Then it automatically defines a map from from C = B/A to M = I/A.Now it is easy to see that B is constructed from I ⊕ C using the same recipe as Xabove.

It remains to check that the class of the Baer sum of extensions is the sum ofclasses. Suppose that x1, x2 ∈ Ext1R(C,A) come from φ1, φ2 ∈ HomR(C,M), respec-tively. Let 0 → A → B → C → 0 and 0 → A → B′ → C → 0 be the respectiveextensions. Consider the diagram

0 −→ A −→ Iµ−→ M −→ 0

↑ ↑ ↑0 −→ A⊕ A −→ B ⊕B′ −→ C ⊕ C −→ 0

where the map A⊕A→ A sends (x, y) to x+ y, the map B⊕B′ → I is the sum ofmaps B → I and B′ → I, and the map C ⊕ C → M sends (x, y) to φ1(x) + φ2(y).

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This diagram is commutative. The kernel of the sum map A ⊕ A → A is the‘antidiagonal’; call it A0. Let us pass to the quotients by A0 in A⊕A and in B⊕B′.This gives rise to the commutative diagram

0 −→ A −→ Iµ−→ M −→ 0

|| ↑ ↑0 −→ A −→ (B ⊕B′)/A0 −→ C ⊕ C −→ 0

Now consider the diagonal C ⊂ C ⊕C. Let X be its inverse image in (B ⊕B′)/A0.By comparing with Definition 3.15 we see that X is precisely the module appearingin the Baer sum of our two extensions. Thus by restricting to X and to the diagonalC ⊂ C ⊕ C we obtain a commutative diagram

0 −→ A −→ Iµ−→ M −→ 0

|| ↑ ↑0 −→ A −→ X −→ C −→ 0

where the bottom extension is the Baer sum of 0 → A → B → C → 0 and0 → A → B′ → C → 0. The restriction of the map C ⊕ C → M sending (x, y)to φ1(x) + φ2(y) to the diagonal is the map φ1 + φ2 : C → M . Hence the class of0→ A→ X → C → 0 is the sum of classes x1 + x2. �

Exercise 1. Classify the extensions of Z/2 by Z/2 in the category of abeliangroups (and then in the category of Z/4-modules). Relate extensions to elements ofExt1Z(Z/2,Z/2).

2. Similarly for the extensions of Z/2 by Z in the category of abelian groups.

3. Similarly for the extensions of Z/2 by Z/3 in the category of abelian groups.

End of the examinable part of notes

4 Group cohomology and the Brauer group

4.1 Group ring

Let G be a group.

Definition 4.1 The group ring of G is a free Z-module with generators g forg ∈ G, and multiplication defined by the condition that the product of generatorsattached to g and h is the generator attached to gh.

The group ring of G is denoted by Z[G]. By definition the elements of Z[G] arewritten as finite sums

∑g∈G agg, where ai ∈ Z. The generator attached to the

unit element e ∈ G is the multiplicative unit of Z[G], so we shall denote it by

29

1. The associativity in G implies that Z[G] is an associative ring. The ring Z[G] iscommutative if and only if the group G is commutative. We shall call the generatorsattached to g ∈ G the canonical generators or the canonical Z-basis of Z[G].

Examples 1. Let G = Cn be the cyclic group of order n with generator s. WriteCn = {e, s, s2, . . . , sn−1}. Then Z[Cn] = Z ⊕ Zs ⊕ Zs2 ⊕ . . . ⊕ Zsn−1 is the directsum of n copies of Z. The multiplication by s is the cyclic shift:

sn−1∑i=0

aisi = an−1 +

n−1∑i=1

ai−1si.

2. Let G = Z be the infinite cyclic group. Call its generator t. Then Z[Z] is thedirect sum of infinitely many summands Zti, where i ∈ Z, so that Z[Z] = ⊕i∈ZZti isthe ring of Laurent polynomials, since by definition the product of ti and tj is ti+j.

Definition 4.2 A Z[G]-module is called a G-module.

In other words, an abelian group M is a G-module if we are given a linear actionof G on M . We write HomG(A,B) for HomZ[G](A,B) and A⊗G B for A⊗Z[G] B.

For example, Z[G] is naturally a left G-module: g ∈ G acts on Z[G] by sendingthe generator attached to h ∈ G to the generator attached to gh. Thus G acts onthe canonical basis by permutations.

One denotes by MG the subgroup of G-invariant elements of M , that is, MG ={m ∈M |gm = m for any g ∈ G}. It is called the invariant subgroup of G.

Definition 4.3 A G-module M is called trivial if the action of G on M is trivial,i.e. every element of G acts as identity.

Thus any abelian group can be equipped with the structure of a trivial G-modulefor any group G. It is clear that MG is the maximal trivial G-submodule of M , soM is a trivial G-module if and only if M = MG. Note that Z[G] is not a trivialG-module if G has more than one element.

For a G-module M define MG as the quotient of M by the subgroup generated bym− gm for all m ∈M and all g ∈ G. It is clear that MG is the largest G-invariantquotient of M . It is called the coinvariant subgroup of G.

Proposition 4.4 (i) Define I ⊂ Z[G] as the zero sum subgroup, i.e. the kernel ofthe map of σ : Z[G] → Z sending

∑g∈G agg ∈ Z[G] to

∑g∈G ag ∈ Z. Then I is a

two-sided ideal of Z[G], called the augmentation ideal. The elements 1−g ∈ Z[G],where g ∈ G and g 6= e, form a Z-basis of I, so that Z[G]G = Z[G]/I ∼= Z.

(ii) Let G be finite. Define N =∑

g∈G g ∈ Z[G], called the norm element. ThenN is contained in the centre of Z[G] and Z[G]G = ZN.

30

Proof. (i) It is enough to notice that the augmentation map σ is a map of G-modules,or Z[G]-modules, which is the same, provided Z is given the structure of a trivialG-module. Indeed, the action of g ∈ G permutes the elements of the canonical basis,so the sum is invariant.

(ii) Since the action of g ∈ G permutes the elements of the canonical basis andN is the sum of these elements, we gee that gN = Ng = N. Hence N ∈ Z[G]G.Conversely, if gx = x for some x ∈ Z[G], then all coordiates of x must be equal.Finally, for any x ∈ Z[G] we have xN = Nx = σ(x)N, hence N is in the centre ofZ[G]. �

Note that the assumption that G is finite in (ii) is important. For example, ifG = Z then Z[G]G = 0.

Definition 4.5 Let M be a G-module. For n ≥ 0 define the n-th cohomology groupHn(G,M) = ExtnZ[G](Z,M) and the n-th homology group Hn(G,M) = TorZ[G]

n (Z,M).

Thus H0(G,M) = HomG(Z,M) = MG and H0(G,M) = Z ⊗G M = MG (checkthis!). In other terms, Hn(G,−) is the right derived functor RnF of the left exactfunctor F (M) = MG. Similarly, Hn(G,−) is the left derived functor RnF of theright exact functor F (M) = MG.

By the previous theory Hn(G,M) = ExtnZ[G](Z,M) can be calculated using a pro-jective resolution P• → Z→ 0 of the trivial G-module Z, by applying HomG(−,M)and taking the cohomology groups of the resulting cochain complex. Similarly, tocalculate Hn(G,M) = TorZ[G]

n (Z,M) we apply − ⊗G M to P• and then take thehomology groups.

4.2 Cohomology of cyclic groups

Projective (in fact, free) resolutions are especially easy to construct for cyclic groups.

Let us start with the infinite cyclic group G = Z, with t as a generator. Then

0 −→ Z[Z]1−t−→ Z[Z], which is the same as 0 −→ Z[t, t−1]

1−t−→ Z[t, t−1], is a projectiveresolution of the trivial Z-module Z. Recall that for any ring R we have R⊗RM =M , and that the multiplication by r ∈ R map R → R induces the multiplicationby r map M → M on the tensor products. Thus the homology groups Hn(Z,M)

are the homology groups of the chain complex 0→ M1−t−→ M → 0. It follows that

H1(Z,M) = MZ and Hn(Z,M) = 0 for n ≥ 2.

Similarly, the cohomology groups Hn(Z,M) are the cohomology groups of thecomplex

0 −→ HomG(Z[Z],M) −→ HomG(Z[Z],M) −→ 0.

Since HomR(R,M) = M , this complex is 0 → M1−t−→ M → 0. We deduce that

H1(Z,M) = M/(1− t)M = MZ and Hn(Z,M) = 0 for n ≥ 2.

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Theorem 4.6 Let G = Cm, the cyclic group of order m. Let s ∈ Cm be a generator.

(i) We have Hn(Cm,M) = Ker[MN−→M ]/(1−s)M for odd n ≥ 1 and Hn(Cm,M) =

M s/NM for even n ≥ 2 (and of course H0(Cm,M) = M s).

(ii) We have Hn(Cm,M) = Ker[MN−→ M ]/(1 − s)M for even n ≥ 2 and

Hn(Cm,M) = M s/NM for odd n ≥ 1 (and of course H0(Cm,M) = M/(1− s)M).

Proof. Consider the semi-infinite free resolution of the trivial Cm-module Z

. . .N−→ Z[Cm]

1−s−→ Z[Cm]N−→ Z[Cm]

1−s−→ Z[Cm]σ−→ Z −→ 0.

The exactness of this complex is due to the fact that Im(1 − s) = I = Ker(N) andIm(N) = ZN = Ker(1− s). �

Exercise 1. Show that if M is a trivial Cm-module, then H0(Cm,M) = M ,Hn(Cm,M) = M/m for even n ≥ 2, and Hn(Cm,M) = M [m] for odd n ≥ 1.

2. Write down explicitly the terms of the long exact sequence of Hn(Cm,−)attached to each of the following short exact sequences of trivial Cm-modules:

0 −→ Z −→ Q −→ Q/Z −→ 0,

0 −→ Z [m]−→ Z −→ Z/m −→ 0,

0 −→ Z/m −→ Q/Z [m]−→ Q/Z −→ 0.

4.3 Standard resolution

The aim of this section is to construct a projective resolution of an arbitrary group.This will allow us to calculate explicitly H1(G,M) and H2(G,M) in the general case.

Let Pn = Z[Gn+1] be the group ring of the (n + 1)-fold self-product of G. Thecanonical generators of Pn are ordered sequences (g0, g1, . . . , gn), where gi ∈ G. Weequip Pn with the structure of a G-module by making g ∈ G send (g0, g1, . . . , gn) to(gg0, gg1, . . . , ggn).

Lemma 4.7 For any n ≥ 1 the G-module Pn is free. We have Pn = Z[G] ⊗Z Qn,where Qn is a free abelian group generated by (e, a1, a1a2, . . . , a1a2 . . . an) for alla = (a1, a2, . . . , an) ∈ Gn.

Proof. For each a = (a1, a2, . . . , an) ∈ Gn define Pn(a) as a G-submodule of Pngenerated as an abelian group by (g, ga1, ga1a2, . . . , ga1a2 . . . an), where g ∈ G:

Pn(a) =⊕g∈G

Z(g, ga1, ga1a2, . . . , ga1a2 . . . an).

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It is clear that Pn is the direct sum of its submodules Pn(a), where a ∈ Gn. Wehave

Pn(a) =⊕a∈Gn

Z[G](e, a1, a1a2, . . . , a1a2 . . . an),

so Pn(a) ∼= Z[G]. Hence Pn is a free G-module and Pn = Z[G]⊗Z Qn. �

We organise the G-modules Pn into a chain complex by defining d : Pn → Pn−1by the formula

d(g0, g1, . . . , gn) =n∑i=0

(−1)i(g0, . . . , gi−1, gi+1, . . . , gn).

For example, d : P1 → P0 sends the canonical generator (g0, g1) ∈ Z[G2] to thedifference of generators g1 − g0 ∈ Z[G]. The differential d : P2 → P1 sends thecanonical generator (g0, g1, g2) ∈ Z[G3] to (g1, g2)− (g0, g2) + (g0, g1) ∈ Z[G2].

Lemma 4.8 Let σ : P0 = Z[G]→ Z be given by the sum of coordinates. Then

. . .d−→ Pn

d−→ Pn−1d−→ . . .

d−→ P1d−→ P0

σ−→ Z −→ 0

is an exact complex, hence P• is a projective resolution of the trivial G-module Z.

Proof. We note that (g0, . . . , gn) with removed gi and gj, where i < j, appears ind2(g0, g1, . . . , gn) with coefficients (−1)i+j−1 (when we first remove gi and then gj)and (−1)j+i (when we first remove gj and then gi). We conclude that d2 = 0, so oursequence is a complex.

To prove the exactness of this complex it is enough to show that the identity mapid : P• → P• is equal to the zero map up to homotopy. Indeed, by Lemma 3.5 mapsof chain complexes that are equal up to homotopy induce equal maps of homologygroups, but the identity map is zero only if the group is trivial. We now constructmaps sn : Pn−1 → Pn such that id = sn−1d+ dsn. Fix h ∈ G and let

sn−1(g0, . . . , gn−1) = (h, g0, . . . , gn−1).

It is easy to see that this works, so our proof is complete. �

We proceed to calculate the cohomology groups Hn(G,M). Apply HomG(−,M)to P• to obtain a cochain complex

0 −→ HomG(P0,M)d−→ HomG(P1,M)

d−→ . . .d−→ HomG(Pn,M)

d−→ . . .

For any ring R considered as a left R-module, any abelian group A, and any left R-module M we have HomR(R ⊗Z A,M) = HomZ(A,M). (Indeed, a homomorphismof R-modules R ⊗Z A → M is uniquely determined by its values at 1 ⊗ a, a ∈ A,

33

so it comes from a homomorphism A → M . Conversely, any homomorphism ofabelian groups f : A → M naturally extends to a homomorphism of R-modulesR⊗Z A→M sending g ⊗ a to gf(a).) In particular,

HomG(Pn,M) = HomZ[G](Z[G]⊗Z Qn,M) = HomZ(Qn,M).

Thus Hn(G,M) are the cohomology groups of the cochain complex

0 −→ HomZ(Q0,M) = Md−→ HomZ(Q1,M)

d−→ . . .d−→ HomZ(Qn,M)

d−→ . . .

A homomorphism Qn → M is determined by its values on the generators of Qn,that is, on (e, g1, g1g2, . . . , g1g2 . . . gn), where gi ∈ G. These values can be arbitraryelements of M . Therefore, such homomorphisms bijectively correspond to functionsf : Gn → M : the homomorphism defined by f sends (e, g1, g1g2, . . . , g1g2 . . . gn) tof(g1, . . . , gn). Thus Hn(G,M) are the cohomology groups of the cochain complex

0 −→Md−→ Fun(G,M)

d−→ . . .d−→ Fun(Gn,M)

d−→ Fun(Gn+1,M)d−→ . . .

All is now reduced to the calculation of the differentials between these groups offunctions.

Lemma 4.9 The differential d : Fun(Gn,M) → Fun(Gn+1,M) sends f : Gn → Mto the function df : Gn+1 →M whose value at g1, . . . , gn+1 is as follows.

n = 0 We have Fun(G0,M) = M , so f = m and (dm)(g) = gm−m;

n = 1: (df)(g1, g2) = g1f(g2)− f(g1g2) + f(g1);

n = 2: (df)(g1, g2, g3) = g1f(g2, g3)− f(g1g2, g3) + f(g1, g2g3)− f(g1, g2).

For the general n the formula for (df)(g1, . . . , gn+1) is

g1f(g2, . . . , gn+1)− f(g1g2, g3, . . . , gn) + f(g1, g2g3, . . . , gn)− . . .

+(−1)nf(g1, g2, . . . , gn−1gn) + (−1)n+1f(g1, . . . , gn).

Proof. The differential of the resolution P• sends (e, g) ∈ Q1 to g − 1 ∈ P0 = Z[G].So if a constant function from Fun(G0,M) = M sends 1 to m, then extending it ina manner compatible with the action of G we must have g 7→ gm. Thus m goes togm−m ∈ Fun(G,M).

Next, (e, g1, g1g2) ∈ Q2 goes to g1(e, g2) − (e, g1g2) + (e, g1) ∈ P1. Now for anyg ∈ G our function from Fun(G,M) sends g to f(g). So in terms of Q1 it sends(e, g) to f(g). Thus f goes to g1f(g2)− f(g1g2) + f(g1).

Finally, (e, g1, g1g2, g1g2g3) ∈ Q3 goes to

g1(e, g2, g2g3)− (e, g1g2, g1g2g3) + (e, g1, g1g2g3)− (e, g1, g1g2) ∈ P2.

Now for any g ∈ G our function from Fun(G2,M) sends (g1, g2) to f(g1, g2). So interms of Q2 it sends (e, g1, g1g2) to f(g1, g2). This gives the formula for n = 2.

The same arguments establish the formula in the general case. �

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Theorem 4.10 (i) The group H1(G,M) is the quotient of the group of functionsf : G → M satisfying f(g1g2) = g1f(g2) + f(g1) = 0, called 1-cocycles, by thesubgroup of functions of the form gm−m, where m ∈M , called trivial 1-cocycles.

(ii) The group H2(G,M) is the quotient of the group of functions f : G2 → Msatisfying g1f(g2, g3)− f(g1g2, g3) + f(g1, g2g3)− f(g1, g2) = 0, called 2-cocycles, bythe subgroup of functions of the form g1f(g2)−f(g1g2)+f(g1), where f ∈ Fun(G,M),called trivial 2-cocycles.

We note that if G and M are both finite, then Hn(G,M) is finite for all n ≥ 0.

For n = 1 the cocycle condition is usually written as

f(g1g2) = g1f(g2) + f(g1).

1-cocycles are also called crossed homomorphisms, because the cocycle conditioncan be seen as a ‘twisted’ version of linearity. Note that f(e) = 0. (Indeed, f(e) =f(e2) = ef(e) + f(e) = 2f(e), hence f(e) = 0.)

If M is a trivial G-module then a 1-cocycle G → M is just a homomorphismG → M . For example, if G is finite, then H1(G,Z) = 0 and H1(G,Q/Z) is thegroup of characters of G.

For n = 2 the cocycle condition is

g1f(g2, g3)− f(g1g2, g3) + f(g1, g2g3)− f(g1, g2) = 0.

2-cocycles are traditionally called factor sets or systems of factors.

4.4 Change of the group

The cohomology (and homology) groups are also functorial with respect to grouphomomorphisms f : H → G. Indeed, given f every G-module M naturally becomesan H-module. There are canonical homomorphisms

resGH : Hn(G,M) −→ Hn(H,M),

called the restriction maps. They are compatible with long exact sequences attachedto short exact sequences of G-modules. They are given by a general theory (whichwe haven’t covered) in view of the fact that the obvious functor from G-modules toH-modules is exact.

The situation becomes more explicit when H ⊂ G. Then Z[G] is free as a Z[H]-module. Thus free G-modules give rise to free H-modules, and hence projectiveG-modules are also projective as H-modules. So a projective resolution P• of thetrivial G-module Z is also a projective resolution of the trivial H-module Z, andcan be used to calculate Hn(H,M) as the cohomology groups of HomH(P•,M). In

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this case we can define the restriction maps resGH as the maps on cohomology groupsinduced by the obvious maps HomG(P•,M)→ HomH(P•,M).

If H is a normal subgroup of G, then for any G-module M the invariant subgroupMH is stable under the action of G. Indeed, if m ∈MH , then hgm = g(g−1hg)m =gm for any g ∈ G and h ∈ H. Hence gMH ⊆MH , so MH a G-module.

Definition 4.11 Let H be a normal subgroup of G. Define the inflation map asthe composition

inf : H1(G/H,MH) −→ H1(G,MH) −→ H1(G,M),

where the first map is induced by the homomorphism G → G/H, and the secondmap is induced by the inclusion of G-modules MH ↪→M .

In terms of cocycles the inflation map can be described as follows. Take a 1-cocycle ϕ : G/H → MH and consider it a function from G to M . This function isa 1-cocycle G→M whose class in H1(G,M) is the inflation of the class of ϕ.

Theorem 4.12 (inflation-restriction) Let H be a normal subgroup of G. Thesequence

0 −→ H1(G/H,MH)inf−→ H1(G,M)

res−→ H1(H,M)

is exact.

Proof. Exactness at H1(G/H,MH) means the injectivity of the inflation map. Weneed to show that if c : G/H →MH is a 1-cocycle, then considering c as a functionG → M we have the implication: if c is trivial as a 1-cocycle of G, then it isalready trivial as a 1-cocycle of G/H. So assume that there is an m ∈M such thatc(g) = gm−m for all g ∈ G. Now hm−m = 0 because the restriction of c to H isc(e) = 0. Then m ∈MH , so c is a trivial 1-cocycle of G/H.

Next we check the exactness at H1(G,M). Let c : G → M be a 1-cocycle suchthat there is an m ∈ M for which c(h) = hm − m, where h ∈ H. The 1-cocyclec′(g) = c(g)− (gm−m) is equivalent to c, so it defines the same cohomology class.Now c′(h) = 0 for any h ∈ H. We have c′(gh) = gc′(h)+c′(g) = c′(g), so c′ is constanton cosets gH. Thus we can consider c′ as a function G/H →M . But gH = Hg, sofor any h ∈ H and any g ∈ G we can write c′(g) = c′(hg) = hc′(g) + c′(h) = hc′(g).It follows that c′(g) ∈MH for any g ∈ G. Thus c′ is a 1-cocycle G/H →MH . �

Exercise. Use the inflation-restriction exact sequence to calculate H1(Sn,Z), whereSn is the symmetric group on n letters (n ≥ 2) and the action of π ∈ Sn on Z is viathe signature of a permutation: π(x) = sign(π)x.

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4.5 Central simple algebras

The material of this section is not examinable

Let R be a semisimple ring. Recall from Section 1.5 that R is semisimple if it issemisimple as a left R-module, i.e. R is a direct sum of non-zero left ideals whichare minimal with respect to inclusion (that is, are simple left R-modules).

Let Ms, s ∈ S, be representatives of isomorphism classes of simple left R-modules.

Define the isotypical component N (s) of a left R-module N as the submodule ofN generated by all submodules isomorphic to Ms.

Lemma 4.13 N =⊕

s∈S N(s)

Proof. It is clear that N is the sum of the N (s). We have

N (s) ∩⊕

t∈S, t6=s

N (t) = 0,

because the intersection is an R-module and so is a direct sum of simple modules,each of them isomorphic to Ms and to Mt, where t 6= s, which is impossible. Thisimplies that each element of N is uniquely written as a sum of elements of N (s),hence N is the direct sum of the N (s). �

By Theorem 1.15 all short exact sequences of R-modules are split. It follows thatfor any map of R-modules f : A→ B we have f(A(s)) ⊂ B(s).

Lemma 4.14 A submodule of A is stable under all R-module endomorphisms of Aif and only if it is a direct sum of some of the A(s).

Proof. Since for any endomorphism f : A→ A we have f(A(s)) ⊂ A(s), the conditionis clearly sufficient.

Now let B be a submodule of A stable under all R-module endomorphisms of A.Let Ms be a simple submodule of B. We need to show that A(s) ⊂ B. Let M ′

s be anysimple submodule of A isomorphic to Ms and let µ : Ms−̃→M ′

s be an isomorphism ofR-modules. Since Ms is a direct summand of A, the projection p : A→Ms is a mapof R-modules. The composition µp : A → M ′

s ↪→ A is an R-module endomorphismof A, thus µp(Ms) = M ′

s ⊂ B. Therefore, A(s) ⊂ B. �

Proposition 4.15 Let R be a semisimple ring. The following statements are equiv-alent:

(a) All simple left R-modules are isomorphic;

(b) the left R-module R has only one isotypical component;

(c) R has no two-sided ideals except 0 and R.

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Proof. (a)⇒(b) is trivial. Any simple R-module M is a quotient of the left R-moduleR. Indeed, pick m ∈ M , m 6= 0. Then Rm is a non-zero submodule of M , henceRm = M . This shows that (b)⇒(a).

The endomorphisms of the left R-module R are the right multiplications by theelements of R. Any two-sided ideal of R is stable under them. By Lemma 4.14 itmust be the sum of some of the isotypical components of R. Hence (b)⇒(c). Eachisotypical component of the left R-module R is stable under right multiplications,so is a two-sided ideal. Thus (c)⇒(b). �

Definition 4.16 A ring satisfying the equivalent conditions of Proposition 4.15 iscalled simple.

It is clear that any skew-field D is a simple ring. Recall that a skew-field is aring that satisfies all the axioms of a field except, possibly, commutativity, so skew-fields are also called division rings. The endomorphisms of the left D-module Dare right multiplications by the elements of D. Since the right multiplication by bfollowed by the right multiplication by a is the right multiplication by ba, we seethat EndD(D) = Dop is the opposite ring of D. Similarly, the endomorphisms ofthe left D-module Dn is the ring of matrices over Dop, that is,

EndD(Dn) = Matn(Dop).

If D is a field, there is no difference between D and Dop, so we obtain a usualidentification of the endomorphisms of an n-dimensional vector space over a fieldwith a given basis with the ring of n× n-matrices with entries in this field.

Theorem 4.17 (Wedderburn) Any simple ring is isomorphic to a ring of matri-ces over a skew-field.

Proof. Let D be a skew-field and let A = Dn. Let us prove that EndD(A) is a simplering. For any non-zero vectors u and v in A there is a D-endomorphism f : Dn → Dn

such that f(u) = v. (The vector u generates a submodule of A isomorphic to D, soit can be completed to a basis of Dn, so that u = e1 = (1, 0, . . . , 0). Now considerthe endomorphism of A = Dn sending e1 to v and ei, i 6= 1, to 0.) It follows that Ais a simple EndD(A)-module.

Let e1, . . . , en be the standard basis of A = Dn. The map

EndD(A) −→ An f 7→ (f(e1), . . . , f(en))

is a map of left EndD(A)-modules which is obviously injective. It is also surjectivebecause for any v1, . . . , vn ∈ Dn there is an endomorphism sending ei to vi, fori = 1, . . . , n. Thus EndD(A) is a simple ring by Proposition 4.15.

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Conversely, let R be a simple ring. There is a simple R-module M such that theleft R-module R is a direct sum of copies of M . Write 1 =

∑ni=1 µi, where µi 6= 0

belongs to the i-th summand of this direct sum. It follows that R = ⊕ni=1Rµi∼= Mn,

so the sum has n terms and the i-th summand is the left ideal Rµi ∼= M . We notethat Rµi is stable under right multiplication by its own elements.

By Schur’s lemma every element of EndR(M) is invertible, so this ring is a skew-field; call it D. The left action of R on M commutes with the left action of D on M(by the definition of D). This gives a homomorphism of rings ϕ : R → EndD(M).We want to show that ϕ is an isomorphism. It is clear that Ker(ϕ) is a two-sidedideal in R, so it can be 0 or R. Since R contains 1 which acts on M as identity, wesee that ϕ(1) = id, so the second case is not possible. Hence ϕ is injective.

Let us show that ϕ is surjective. It is clear that RµiR is a two-sided ideal of R,hence RµiR = R. Thus ϕ(Rµi)ϕ(R) = ϕ(R). We note that the right multiplicationby an element of Rµi on Rµi commutes with the left action of R on Rµi, so itdefines an element of D. Thus for any f ∈ EndD(M) and any x, y ∈ Rµi we havef(xy) = f(x)y ∈ Rµi. This shows that the left action of x ∈ Rµi on Rµi followedby any f ∈ EndD(M) is the left action of f(x) ∈ Rµi. It follows that ϕ(Rµi) is aleft ideal of EndD(M). Therefore,

ϕ(R) = ϕ(Rµi)ϕ(R) = EndD(M)ϕ(Rµi)ϕ(R) = EndD(M)ϕ(R) = EndD(M),

where the last equality is due to the fact that ϕ(1) = id.

Finally, one can show that M has finite dimension as a vector space over D, sothat M ∼= Dm. (Otherwise EndD(M) is neither Noetherian nor Artinian, but it hasto be because it is a finite sum of its left ideals.) We have seen that in this caseEndD(M) ∼= Matm(Dop), so we are done. (This also implies that n = m.) �

Examples. If k is a field, then Matn(k) is a simple ring. This is a k-algebra whosecentre is k. Hamilton’s quoternions H = R ⊕ Ri ⊕ Rj ⊕ Rij, where ij = −ji,i2 = j2 = −1, is another example of a k-algebra whose centre is k.

If D is a division ring, then it is easy to check that the centre of Matn(D) is thefield which is the centre of D.

Definition 4.18 Let k be a field. A simple ring with centre k is called a centralsimple algebra over k.

Recall that a k-algebra is an associative ring with 1 which is a vector space over kin such a way that multiplication is linear in each argument. We identify k with thesubalgebra k · 1 of A. For any a, b ∈ k and x, y ∈ A we have (ax)(by) = (ab)xy, so kis contained in the centre of A. If A and B are k-algebras, then A⊗k B is equippedwith multiplication by the rule (x⊗ y)(x′ ⊗ y′) = xy ⊗ yy′.

Let K/k be a finite Galois extension of fields with the Galois group G = Gal(K/k).Let n = [K : k]. Consider the K-algebra K[G] = Z[G] ⊗Z K, which is obtained

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from the group ring Z[G] by changing the coefficient domain from Z to K. Inparticular, K[G] is a vector space over K with basis g ∈ G, so K[G] consists oflinear combinations

∑g∈G agg with ag ∈ K.

Let f : G × G → K∗ be a 2-cocycle with coefficients in K∗. We now ‘twist’ themultiplication on K[G] using f , so that we define a new multiplication on the vectorspace K[G] as follows:

(∑g∈G

agg) · (∑h∈G

bhh) =∑g,h∈G

f(g, h) · ag · g(bh) · gh.

In a magical way the associativity of this product is equivalent to the cocycle conditionfor the 2-cocycle f . The resulting algebra is called the crossed product of K and Gwith respect to f . We note that this algebra contains K and has dimension n2 overk.

One can prove the following statements:

1. If A is a central simple algebra over k and K/k is a field extension, then A⊗kKis a central simple algebra over K. One says that K splits A if A⊗kK ∼= Matm(K)for some m ≥ 1.

2. If A and B are central simple algebras over k, then so is A⊗k B.

3. If D is a skew-field with centre k, then D ⊗k Dopp ∼= Matm(k), where m =dimk(D).

4. Every central simple k-algebra split by K is isomorphic to a crossed product ofK and G with respect to a 2-cocycle f : G×G→ K∗. The matrix algebra Matn(k)is obtained from f identically equal to 1 (so this is the ‘untwisted’ algebra K[G]).

5. 2-cocycles give rise to isomorphic crossed products if and only if they representthe same class in H2(G,K∗).

6. Let A and B be crossed products constructed from 2-cocycles representingcohomology classes x, y ∈ H2(G,K∗). Then A ⊗k B = Matn(C), where C is thecrossed product constructed from a 2-cocycle representing x+ y.

Definition 4.19 The relative Brauer group Br(K/k) is defined as the set of allcentral simple k-algebras A such that A⊗k K = Matn(K), where n = [K : k].

With the group structure defined in Property 6, Br(K/k) is a group canonicallyisomorphic to H2(G,K∗).

We note that Hm(G,K) = 0 for all m ≥ 1 (a consequence of the normal basistheorem and Shapiro’s lemma) and H1(G,K∗) = 0 (Hilbert’s theorem 90).

References

[1] S. Lang. Algebra. Addison-Wesley, 1965.

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[2] H. Cartan and S. Eilenberg. Homological algebra. Princeton University Press,1956. 2, 5, 7, 11, 13

[3] C. Weibel. An introduction to homological algebra. Cambridge University Press,1994. 24, 26

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