Algebra IV. Contents. Alexei Skorobogatov. December 12, 2017

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1 Algebra IV Alexei Skorobogatov December 12, 2017 Abstract This course is an introuction to homological algebra an group cohomology. Contents 1 Moules over a ring Definitions an examples Injective an projective moules Hom Tensor prouct Semisimple moules Moules over integral omains Torsion an ivisible submoules Moules over principal ieal omains Enough injectives Resolutions an erive functors Projective an injective resolutions Derive functors Tor an Ext Ext as the group of extensions Group cohomology an the Brauer group Group ring Cohomology of cyclic groups Stanar resolution Change of the group Central simple algebras

2 1 Moules over a ring 1.1 Definitions an examples [2, Ch.1, 1] Let R be an associative ring with 1. Recall that this means that R is an abelian group with respect to aition (the group operation is enote by +) that has an associative operation of multiplication (enote 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 o not require R to be commutative with regar to multiplication. R enotes the set of elements x R for which there exists y R such that xy = 1 an yx = 1. Such elements are calle invertible. The set R is a group. If R = R \ {0}, then R is calle a skew-fiel. If, moreover, R is commutative, then R is calle a fiel. Key examples of rings: Z; fiels Q, R, C, F q ; polynomial rings k[x 1,..., x n ], where k is a fiel; the ring of square matrices Mat n (k); the skew-fiel of quaternions H. The central concept of this course is that of a moule. Definition 1.1 A left R-moule M is an abelian group (written aitively) with a function : R M M (calle the left action of R) satisfying r (m 1 + m 2 ) = r m 1 + r m 2, (r 1 + r 2 ) m = r 1 m + r 2 m, for any r, r 1, r 2 R an m, m 1, m 2 M. 1 m = m, (r 1 r 2 ) m = r 1 (r 2 m), Unless explicitly mentione, all our moules are left moules. Examples M = R n with coorinate-wise left action of R, i.e. r R sens (r 1,..., r n ) to (rr 1,..., rr n ). These left R-moules are calle free of finite rank. Any left ieal I R (i.e. a subgroup of R close uner left multiplication by the elements of R) is a left R-moule. Given a linear transformation L : V V of a vector space V over a fiel k one turns V into a k[x]-moule by making x act by L. 2

3 Take R = Mat n (k) an M = k n ientifie with column vectors. The left action of R on M is efine by applying a square matrix to a column vector. If there are left moules, there shoul be also right moules! Their efinition is obtaine by replacing the property (r 1 r 2 ) m = r 1 (r 2 m) by (r 1 r 2 ) m = r 2 (r 1 m). An example of a right Mat n (k)-moule is the vector space of row vectors k n, on which square matrices act by right multiplication. From now on we shall enote the action of R on a left moule by rm an the action of R on a right moule by mr. Thus for the left moules we have (r 1 r 2 )m = r 1 (r 2 m), whereas for the right moules we have m(r 1 r 2 ) = (mr 1 )r 2. The opposite ring R op of R is obtaine by replacing the multiplication xy by yx. It is trivial that a right R-moule is the same as a left R op -moule. If R is commutative, then R op is the same as R, an there is no ifference between left an right moules. Some known classes of objects can be interprete as moules over a specific ring: Abelian groups = moules over the (commutative) ring Z. Vector spaces over a fiel k = moules over the (commutative) ring k. Representations of a group G over a fiel k = moules over the group algebra k[g]. This is the vector space over k with basis e g, g G, with multiplication given by e g e h = e gh. A homomorphism (or simply a map) of left R-moules f : L M is a homomorphism of groups that preserves the action of R: f(rx) = rf(x) for any r R an x L. An isomorphism (resp. an injective, a surjective map) of R-moules is a bijective (resp. an injective, a surjective) map of R-moules. A subgroup L M stable uner the action of R is cale a submoule of M. Since M is an abelian group uner +, we can form the quotient group M/L. But L is R-stable, so we have a well efine action of R on M/L, which is calle the quotient moule. For example, a left R-submoule of R is the same as a left ieal of R. Let S be a set. Suppose we are given a left R-moule M s for each s S. The irect prouct s S M s is efine as the set of vectors (m s ) s S, where m s M s. The aition is efine coorinate-wise. The action of r R multiplies each coorinate by r. For a fixe element s 0 S the canonical surjective map s S M s M s0 is a map of R-moules. If all M s are isomorphic to a given moule M, we enote the irect prouct by M S, because this is the same as the set of all functions S M. 3

4 The irect sum s S M s is the submoule of s S M s given by the conition that all but finitely many coorinates are zero. For a fixe element s 0 S the canonical injective map M s0 s S M s is a map of R-moules. If S is finite, then there is no ifference between the irect sum an the irect prouct. A free moule is a irect sum of copies of the left R-moule R. In the particular case when S is finite, a free moule is isomorphic to R n for some positive integer n. A sequence (finite or infinite) of moules an maps... f n 1 f n f n+1 f n+1 A n An+1 An+2... is calle a complex if f n+1 f n = 0 whenever this is efine. A complex is calle exact if Ker(f n+1 ) = Im(f n ) for all n. An exact complex 0 A α B β C 0 (1) is calle a short exact sequence. To give a short exact sequence is the same as to give an 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 usually ientify A with the submoule α(a) B, an C with the quotient moule B/α(A). An example of an exact sequence is 0 A α A C β C 0, where α(x) = (x, 0) an β((x, y)) = y. Such exact sequences are calle split. Not all sequences are split, for example this sequence of abelian groups (=Z-moules) is not: 0 Z/2 Z/4 Z/2 0. (2) Proposition 1.2 An exact sequence of left R-moules is split if an only if there is a map of R-moules σ : C B such that βσ = i C. (Such a map is calle a section of β.) Equivalently, if there is a map ρ : B A such that ρα = i A. (Such a map is calle a retraction of α.) Proof. If the sequence is split, then σ an ρ are the obvious maps. Now suppose that σ exists. We nee to ientify B with A C so that α an β are the natural injective an surjective maps, respectively. Let f : B A C be the map f(b) = (α 1 (b σβ(b)), β(b)). It is well efine because b σβ(b) Ker(β) = Im(α) an α inuces an isomorphism A Im(α). The kernel of f is containe in Ker(β) B. However, the restriction of f to Ker(β) = Im(α) is an isomorphism (α 1, 0), hence f is injective. 4

5 Let us prove that f is surjective. Let a A an c C. Since β is surjective, we can fin x B such that β(x) = c. For any a A we have β(x + a ) = c, so we can moify x by any element of A without changing c. For example, we can replace x by x a, where a = x σβ(x) a A. Then f(x a ) = (a, c). Exercises 1. Prove the secon claim of Proposition Show that (1) is split when C = R. 3. Show that (1) is split when C is a free R-moule. 4. Show that (2) is not split. 1.2 Injective an projective moules [2, Ch. 1, 2,3] An R-moule M is calle projective if for any exact sequence (1) an for any map of R-moules f : M C there is a map of R-moules 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 an so is split. Lemma 1.3 A irect sum is projective if an only if each summan is projective. Proof. Let M = s S M s an f : M C. If M s is projective for each s S, then the restriction of f to each M s (embee into M in the obvious way), call it f s : M s C, lifts to g s : M s B. By efinition each element of the irect sum has only finitely many non-zero coorinates, hence g := s S g s is a well efine function M B. It is clear that βg = f. Conversely, assume that M is projective. A map f s : M s C for one element s S extens trivially to a map f : M C whose restriction to any other summan is zero. It lifts to g : M B. Restricting to M s we get a lifting of f s. Proposition 1.4 (Criterion of projectivity) A moule is projective if an only if it is a irect summan of a free moule. Proof. Any moule M is a quotient of the free moule F freely generate as an R-moule by the elements of M. In other wors, F consists of (r m ) m M, where each r m R an only finitely many r m can be non-zero. There is a natural map of R-moules F M sening (r m ) m M F to m M r mm M. When M is projective, there is a section. By Proposition 1.2 M is a irect summan of F. The R-moule R is projective, because one element (namely, 1) can be lifte. By Lemma 1.3 all free moules are projective. Again by Lemma 1.3 a irect summan of a free moule is projective. The fact that every R-moule is a quotient of a projective R-moule is usually state as the category of left R-moules has enough projectives. 5

6 Example Projective moules are not always free. Let R = Z/6 an let M be the principal ieal of R generate by 3, so M = Z/2. The abelian group Z/6 is isomorphic to Z/2 Z/3. One checks that this gives an isomorphism of Z/6-moules Z/6 = Z/2 Z/3. Hence Z/2 is a projective Z/6-moules. Since 2 is not a power of 6, it is certainly not free. An R-moule M is calle injective if for any exact sequence (1) an for any map of R-moules f : A M there is a map of R-moules g : B M such that f = gα. The theory of injetive moules is in a certain sense ual to the theory of projective moels. If A is injective, then (1) has a retraction an so is split. Lemma 1.5 A irect prouct is injective if an only if each factor is injective. Proof. Let M = s S M s an f : A M. If M s is injective for each s S, then the s-coorinate of f, which is a map f s : A M s, is such that there exists g s : B M s with the property g s α = f s. The prouct (which can be infinite) g := s S g s is a well efine function B M. It is clear that gα = f. Conversely, assume that M is injective. A map f s : A M s for one element s S extens trivially to a map f : A M. Then there is g : B M such that gα = f. The s-coorinate g s of g has the esire property g s α = f s. Theorem 1.6 (Baer s criterion of injectivity) A moule M is injective if an only if for any map of left R-moules f : I M, where I is a left ieal of R, there is an element m M such that f has the form f(x) = xm. Equivalently, any map of left R-moules I M extens to a map of left R- moules R M. Proof. To see the equivalence of two properties we note that to give a map of left R-moules 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 extene to R by sening 1 to m. Conversely, any map I 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 extens to a map of R-moules R M, so one irection is clear. The proof of the converse uses Zorn s lemma. Let A B be left R-moules an let g : A M be a map of left R-moules. To show that M is injective we nee to show that g extens to a map of R-moules B M. All pairs (A, g ), where A is a submoule of B containing A an g : A M is a map of R-moules, form a partially orere set: a pair (A, g ) ominates (A, g ) if A A an the restriction of g to A is g. To apply Zorn s lemma we nee to check that each totally orer subset of this set of pairs has a maximal element, but this is clear as one takes the union of these sets. By Zorn s lemma this set has a maximal element. Call it (A 0, g 0 ). 6

7 Let us prove that A 0 = B. Otherwise there is an b B, b / A 0. To euce a contraiction we want to show that g 0 extens to A 0 + Rb A 0. We nee to use our assumption, so we nee an ieal. Consier the set of elements x R such that xb A 0. This is a left ieal of R. Let us call it I. For x I let f(x) = g 0 (xb); this is a map of R-moules. By the assumption there is an m M such that g 0 (xb) = xm. Now consier the map A 0 +Rb M which sens a+yb to g 0 (a)+ym, where a A 0 an y R. We nee to check that this map is well efine, i.e. if a + yb = a + y b for a A 0 an y R, then g 0 (a) + ym = g 0 (a ) + y m. But a a = (y y)b, hence y y I an in this case g 0 (a) g 0 (a ) = g 0 (a a ) = g 0 ((y y)b) = f(y y) = (y y)m, so our map is inee well efine. It is clearly a map of left R-moules, so we get the esire contraiction. 1.3 Hom [2, Ch. 2, 2,4] Let A an B be left R-moules. We enote by Hom R (A, B) the set of maps of R- moules A B. It is an abelian group uner aition of maps. We have canonical isomorphisms an Hom R (A 1 A 2, B) = Hom R (A 1, B) Hom R (A 2, B) Hom R (A, B 1 B 2 ) = Hom R (A, B 1 ) Hom R (A, B 2 ). This can be rephrase by saying that Hom R (A, B) can be viewe as an aitive bi-functor from the category of R-moules to the category of abelian groups that is covariant in the secon argument an contravariant in the first argument. We note a few examples. The group Hom R (R, M) is canonically isomorphic to M by the map that associates to f : R M the value f(1) M. If Z is given the trivial structure of an R-moule, that is, rn = n for every r R an n Z, then Hom R (Z, M) is the subgroup of M consisting of the elements on which R acts trivially. E.g., if R is the group ring Z[G], then Hom Z[G] (Z, M) = M G is the subgroup of G-invariant elements of M. Let 0 A 1 A 2 A 3 0 (3) be an exact sequence of left R-moules. Lemma 1.7 (i) The sequence of abelian groups 0 Hom R (A 3, B) Hom R (A 2, B) Hom R (A 1, B) 7

8 is exact. (ii) The R-moule B is injective if an only if 0 Hom R (A 3, B) Hom R (A 2, B) Hom R (A 1, B) 0 is exact for any short exact sequence (1.8). Proof. (i) is easy. (ii) is irect form the efinition of an injective moule. Lemma 1.8 (i) The sequence of abelian groups 0 Hom R (B, A 1 ) Hom R (B, A 2 ) Hom R (B, A 3 ) is exact. (ii) The R-moule B is projective if an only if 0 Hom R (B, A 1 ) Hom R (B, A 2 ) Hom R (B, A 3 ) 0 is exact for any short exact sequence (1.8). Proof. (i) is easy. (ii) is irect form the efinition of a projective moule. Now let R = Z an let M be an abelian group. The group Hom Z (Z/n, M) is the subgroup of M consisting of the elements x such that nx = 0. We enote this supgroup by M[n]. We get the exact sequence 0 A 1 [n] A 2 [n] A 3 [n]. The last map in this sequence is not always surjective. B = Z/2 an the exact sequence is A typical case is when Exercises 1. Show that 0 Z/2 Z/4 Z/2 0. (4) 0 Z Q Q/Z 0 is exact. Is it split? 2. Let M enote the quotient of M by its subgroup nm = {nx x M}. Use the snake lemma to prove that the following sequence is exact: 0 A 1 [n] A 2 [n] A 3 [n] A 1 /n A 2 /n A 3 /n 0. 8

9 1.4 Tensor prouct Now we woul like to efine the tensor prouct. This is an abelian group associate to two R-moules. More precisely, the general efinition of tensor prouct is for a right R-moule A an a left R-moule B. (This subtlety can be ignore when R is commutative an there is no ifference between left an right moules.) Let F be the free abelian group, i.e. the irect sum of copies of Z, generate by all pairs (a, b), where a A an b B. (So in general F will have infinitely many generators.) Let S F be the subgroup generate by all elements of the form an also by the elements (a + a, b) (a, b) (a, b), (a, b + b ) (a, b) (a, b ), (ar, b) (a, rb), for all r R. In particular, we have relations n(a, b) = (na, b) = (a, nb), where a A, b B, n Z. Define A R B = F/S. The image of the generator (a, b) F in A R B is enote by a b. From the efinition 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, an r R. Lemma 1.9 Let M be a left R-moule. The map m 1 m gives a canonical isomorphism of abelian groups M R R M. (Here R is a right moule over itself.) Proof. It is clear that our map is a homomorphism. The abelian group R R M is generate by r m = 1 rm, so M R R M is surjective. Let us prove that it is injective too. Recall that R R M is the quotient of a free abelian group F freely generate by all pairs (r, m). The function of sets R M M sening (r, m) to rm can be extene by linearity to a homomorphism F M. All the efining relations of the tensor prouct are containe in its kernel, so we obtain a homomorphism R R M M. The composition M R R M M is the ientity map. It follows that M R R M is injective. In particular, by taking R = Z we obtain the tensor prouct of abelian groups G 1 an G 2, enote by G 1 Z G 2. Note that this is completely ifferent from the usual prouct of two abelian groups G 1 G 2. For example, as a particular case of the canonical isomorphism R R M = M, we obtain a canonical isomorphism Z Z Z = Z. Similarly, Z/2 Z Z/2 = Z/2 Z/2 Z/2 = Z/2. Sometimes the result of tensor multiplication can be counter-intuitive. For example, Z/2 Z Z/3 = 0. This follows 9

10 from the relations 2(a b) = 2a b = 0 b = 0 an 3(a b) = a 3b = a 0 = 0, hence a b = 0 for all a A an b B. Exercise. Show that Q/Z Z Q/Z = 0. We note that tensor prouct is covariant in each argument. Both these functors are aitive, that is, there are obvious canonical isomorphisms (A 1 A 2 ) R B = (A 1 R B) (A 2 R B) an A R (B 1 B 2 ) = (A R B 1 ) (A R B 2 ). Let φ : A B A R B be the set-theoretic function mapping (a, b) to a b. The universal property of the tensor prouct states that if there is an abelian group C an a set-theoretic function f : A B C which is linear in each argument an satisfies f(ar, b) = f(a, rb), then f is uniquely written as f = gφ, where g : A R B C is a homomorphism of abelian groups. (For the proof note that φ is the restriction of the canonical surjective homomorphism F A R B to the basis A B of F. Similarly, f uniquely extens to a homomorphism F C. By the assumptions on f, the kernel of F A R B is containe in the kernel of F C. A stanar isomorphism theorem for abelian groups then prouces a homomorphism g.) Exercise. Let A be an abelian group. How that the map A A Z Z/n given by a a 1 efines an isomorphism A/n A Z Z/n. (Hint: Prove that the map is surjective, then construct a homomorphism A Z Z/n A/n such that the composition A/n A Z Z/n A/n is the ientity on A/n.) Lemma 1.10 Let 0 A 1 A 2 A 3 0 be an exact sequence of right R-moules. For any left R-moule B the sequence of abelian groups A 1 R B A 2 R B A 3 R B 0 is exact. Proof. Let Q be the quotient of A 2 R B by the image of A 1 R B. We have a natural map u : Q A 3 R B. It is enough to show that it is an isomorphism. We choose a set-theoretic section s of A 2 A 3, i.e. a function s : A 3 A 2 such that βs = i A3. Now efine a function A 3 B Q by sening a pair (a, b), where a A 3 an b B, to the element of Q which is the coset of s(a) b. This map oes not epen on the choise of s, because a ifferent set-theoretic section will change s(a) b by (x, b), 10

11 where x A 1. Similar calculations show that the function A 3 B Q is bilinear an sens (ar, b) an (a, rb) to the same element of Q. By the universal property of tensor prouct, the function A 3 B Q factors through a homomorphism of abelian groups v : A 3 R B Q. One checks that uv an vu are ientities, so u is injective an surjective, hence an isomorphism. Let A be an abelian group. With B = Z/n we obtain an exact sequence A 1 /n A 2 /n A 3 /n 0. The example of (4) shows that this sequence cannot in general be extene by 0 on the left. Definition 1.11 A left R-moule is cale flat if R B preserves exact sequences of right R-moules. We see that Z/n is not flat as an abelian group. Proposition 1.12 All projective moules are flat. Proof. There is a canonical isomorphism A R (B C) = (A R B) (A R C), an similarly for any number of summans. It follows that a irect sum of moules is flat if an only if each summan is flat. Thus all free moules are flat, an so all projective moules are flat by Proposition 1.4. We shall see that the converse is not true: Q is flat as an abelian group (see Therem 2.6 below) though visibly not a irect summan of a power of Z (since it is ivisible), hence not projective. Lemma 1.13 If M is a flat left R-moule, then for a non-zero ivisor a R an non-zero m M we have am 0. Proof. Consier the injective map R R of right R-moules sening 1 to a. If M is flat, then the inuce map R R M R R M is also injective. The ientification m 1 m of M with R R M allows us to rewrite this map as the map M M sening m to am. 1.5 Semisimple moules [2, Ch. 1, 4] (non-examinable) A non-zero left R-moule M is simple if M an 0 are the only submoules. A semisimple moule is a irect sum of simple moules. We will use the easy fact that if A, B M are submoules such that A + B = M an A B = 0, then the natural map A B M is an isomorphism. In this case we ientify M with A B. 11

12 Lemma 1.14 A moule is semisimple if an only if every submoule is a irect summan. Proof. Assume that M = M i, where each M i is a simple moule. Let A M be a submoule. Let I be the maximal (by inclusion) subset of inices such that A i I M i = 0. For each j / I the moule M j i I M i has non-zero intersection with A. We euce that A i I M i has non-zero intersection with M j. But M j is simple, hence M j A i I M i. Therefore, A i I M i = M, an we are one. Conversely, assume that the property hols for M. Then it hols for any nonzero submoule A M. Each such A contains simple submoules. Inee, let a A, a 0. Using Zorn s lemma we fin a maximal submoule B A that oes not contain a. By assumption we can write A = B C. We claim that C is simple. Inee, otherwise C = D E, where D an E are non-zero, so that A = B D E. If a B D an a B E, we get a contraiction with the fact that the sum of B, D, E is a irect sum (if a = b + = b + e, then since a / B the sum (b b ) + e = 0 is a zero sum of non-zero summans an this is not possible in a irect sum). To fix ieas, assume that a / B D. This is a contraiction with the choice of B. We conclue that A contains simple submoules. There exist sums of simple submoules of M which are irect (e.g. one simple submoule). Then any maximal such sum is M. Inee, otherwise the complement to it in M contains a simple submoule of M, so our irect sum is not actually maximal. Theorem 1.15 The following statements are equivalent: (a) The ring R is semisimple as a left R-moule; (b) each left ieal of R is a irect summan; (c) each left ieal of R is an injective moule; () all left R-moules are simisimple; (e) all short exact sequences of R-moules are split; (f) all left R-moules are projective; (g) all left R-moules are injective. Proof. (a) (b) follows from Lemma () (e) follows from Lemma If all moules are projective, then all short exact sequences are split, because there is a section. If all exact sequences are split, then every moule is a irect sum of a free moule 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 moule M can be embee into an injective moule, say M I. If all short exact sequences are split, then I = M M for some R-moule M. But all irect factors of injective moules are injective by 12

13 Lemma 1.5, hence M is injective. Conversely, if all moules are injective, then all short exact sequences are split, because then there is a retraction of the injective map in the sequence. This shows that (e) (g). It is clear that (g) (c). If I is a left ieal of R which is injective, then the inclusion of I in R has a retraction, so I is a irect summan. Thus (c) (b). If (b) hols, then any map of R-moules f : I M extens to R M. By Baer s criterion of injectivity (Theorem 1.6), M is injective, so we get (g). We have prove the equivalence of all the statements. A ring R satisfying the equivalent conitions of Theorem 1.15 is calle semisimple. Later we shall prove that the ring of n n-matrices Mat n (D) with entries in a skewfiel (=ivision algebra) D is a semisimple ring. 2 Moules over integral omains 2.1 Torsion an ivisible submoules [2, Ch. VII, 1] Let R be an integral omain, i.e. a commutative ring with 1 such that ab = 0 implies a = 0 or b = 0. An element m M is a torsion element if rm = 0 for some r R, r 0. Define M tors as the subset of torsion elements in M; this is a submoule. If M tors = 0, then M is calle torsion-free. If M = M tors, then M is calle a torsion moule. Example Abelian groups Z an Q are torsion-free, whereas Z/n, Q/Z an Z/n are torsion groups. Exercise Let M be the quotient of Z 2 by the subgroup generate by (m, n), where m, n Z. Determine M tors. What is (R/Z) tors? Lemma 2.1 Any projective moule is torsion-free. Proof. Free moules are torsion-free since R has no zero ivisors. Hence their submoules are torsion-free too. Now use Proposition 1.4. An element x M is infinitely ivisible if for any r R, r 0, we can write x = ry for some y M. Define M iv as the subset of infinitely ivisible elements in M; this is a submoule. If M = M iv, then M is calle an infinitely ivisible moule. Often the wor infinitely is omitte. Example Abelian groups Q/Z an Q are ivisible, whereas Z an Z/n are not. What is (R/Z) iv? Lemma 2.2 Any injective moule is ivisible. 13

14 Proof. Let m M an let r R, r 0. We nee to show that there is an s M such that m = rs. We note that the principal ieal I = rr an the ring R are isomorphic as R- moules: x rx is an isomorphism R I since R has no zero ivisors. The map of R-moules R M sening x to xm compose with the inverse of the isomorphism R I gives a map of R-moules f : I M such that f(rx) = xm for any x R. Consier the natural inclusion of the principal ieal I into R. Since M is injective, f : I M extens to a map of R-moules F : R M. We claim that s = F (1) oes the job. Inee, rs = rf (1) = F (r) = f(r) = m. 2.2 Moules over principal ieal omains A principal ieal omain (PID) is an integral omain where every ieal is principal. The crucial example for this section is the category of Z-moules, that is, abelian groups. Theorem 2.3 If R is a PID, then every submoule of a free moule is free. Every submoule of R n is isomorphic to R m for some m n. Proof. Accoring to Zermelo s theorem any set S can be well-orere. This means that there is a total orer a b on S such that every subset of S has a least element. We fix such an orer. In a well-orere set we can use transfinite inuction, which says that a statement is true for any s S if it is true for the least element of S an if it hols for all x < s, then it also hols for s. Let S be the inexing set of the free R-moule s S R (s), where each R (s) is a copy of R. Write 1 (s) for 1 consiere as an element of R (s). Let M be a submoule of s S R (s). Choose a total orer on S. For each t S we efine M t = M s t R (s). The projection to the t-th summan s S R (s) R restricte to M t is a map of R-moules f t : M t R. Since R-submoules of R are precisely the ieals of R, an all ieals are principal by assumption, we see that f t (M t ) = Ra t for some a t R. Whenever a t 0 choose m t M t such that f t (m t ) = a t. Define M t as the R-submoule of M t generate by the m s for s t. We claim that M t = M t for all t S. This is clear for the least element of S. By transfinite inuction it is enough to prove that if M s = M s for all s < t, then M t = M t. But any m M t can be written as m = rm t + (m rm t ), where m rm t is a finite linear combination of 1 (s) R (s) for s < t. Let q be the largest inex in this finite set. Then m rm t M q an, by inuctive assumption, m rm t M q. But then m M t. This proves that M t = M t for all t S, an hence M is generate by the m s for s S. We claim that (m s ) s S is a free basis of M; in other wors, every element of M is uniquely written as a finite linear combination of the m s with coefficients in R. 14

15 Otherwise 0 can be written as s T r sm s, where T S is finite an all r s m s 0 (which implies a s 0). If t is the largest element of T, then the projection to the t-th coorinate kills all terms except possibly r t m t which it sens to r t a t R. This is zero, hence r t = 0 since a t 0. This contraiction finishes the proof of the first statement. The secon statement is clear, because we prove that M has a basis whose elements are in a bijection with a subset of S. Corollary 2.4 If R is a PID, then projective moules an free moules are the same thing. Proof. Use Proposition 1.4. Theorem 2.5 If R is a PID, then injective moules an ivisible moules are the same thing. Proof. By Lemma 2.2 all injective moules are infinitely ivisible, so we nee to show that any ivisible moule M is injective. Let us use Baer s criterion (Theorem 1.6). So let f : I M be a map of R-moules, where I R is an ieal. Since R is a PID, we have I = Rr for some r R. As M is ivisible we can write f(r) = rs for some s M. Then f : I M extens to a map of R-moules R M which sens 1 to s. In particular, the abelian groups Q/Z, R/Z an Q are ivisible, hence injective. Theorem 2.6 If R is a PID, then flat moules an torsion-free moules are the same thing. Proof. We know from Lemma 1.13 that flat moules are torsion-free, so let us prove the converse. Let A B be a submoule an let M be a torsion-free R-moule. We nee to prove that A R M B R M is an injective map. Step 1: It is enough to prove our claim in the case when B is free. Inee, we know that any moule is a quotient of a free moule F, so there is an exact sequence 0 K F B 0. If we enote the inverse image of A in F by A, we obtain a commutative iagram of R-moules 0 K A A 0 0 K F B 0 This gives rise to a commutative iagram of abelian groups K R M A R M A R M 0 K R M F R M B R M 0 15

16 By assumption the mile vertical map is injective. Now a igram chase gives us that the right vertical map is injective too. Step 2: It is enough to prove our claim in the case when B = R. Inee, suppose that some element x A R M goes to a non-zero element of F R M. Any element of A R M is a finite sum of elements of the form a m, so x A 0 R M, where A 0 A is a submoule generate by the relevant a. The finitely generate submoule A 0 F is containe in some R n F, which is free of finite rank. Thus it is enough to show that A 0 R M R n R M is injective for any submoule A 0 R n. Write R n = R R n 1 an A 1 = A 0 R. Let A 2 be the image of A 0 in R n 1 uner the projection map. We obtain a commutative iagram an hence a commutative iagram 0 A 1 A 0 A R R n R n 1 0 A 1 R M A 0 R M A 2 R M 0 0 R R M R n R M R n 1 R M 0 because the bottom exact sequence of the previous iagram is split. By assumption the left vertical map is injective. Arguing by inuction we assume that the right vertical map is injective. Then a iagram chase (or the snake lemma) gives the injectivity of the mile vertical map. Step 3: en of proof. Now A is an ieal in R. Since R is a PID, we have A = ar. We nee to prove the injectivity of the natural map α : ar R M R R M. The map of R-moules R ar, r ar, gives rise to β : R R M ar R M. This map is clearly surjective. The composition αβ is the map R R M R R M, which is ientifie with the map M M given by m am. Since M is torsion-free, this map is injective. Now αβ is injective an β is surjective, an this implies that α is injective. Now we can justify the example of a moule that is flat but not projective given above: the abelian group Q is torsion-free, hence flat. But Q is ivisible, so it is not a irect summan of a free abelian group, hence Q is not projective. 2.3 Enough injectives We are now reay to show that the category of R-moules has enough injectives. Theorem 2.7 Let R be a (not necessarily commutative) ring. Every left R-moule is isomorphic to a submoule of an injective moule. 16

17 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 Hom Z (G, Q/Z) 0. Moreover, for any non-zero x G there is an f Hom Z (G, Q/Z) such that f(x) 0. Inee, let C G be the cyclic subgroup generate by x. It can be finite or infinite. If C is finite, then there is an injective homomorphism C Q/Z (e.g., sening x to 1/ C ). If C is infinite, then consier the homomorphism sening x to 1/2 Q/Z. In each case 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 extens to a non-zero homomorphism G Q/Z. Let I(G) = (Q/Z) HomZ(G,Q/Z) be the prouct of copies of Q/Z inexe by the elements of Hom Z (G, Q/Z). Since any prouct of injective moules is injective, I(G) is injective. There is a canonical map G I(G) that sens x G to the element of I(G) whose coorinate corresponing to f Hom Z (G, Q/Z) is f(x). By the previous paragraph this homomorphism is injective. Here is the proof for an arbitrary ring R. Step 1. Consier R as a right R-moule. The abelian group S = Hom Z (R, Q/Z) can be equippe with the structure of a left R-moule by making r R sen a homomorphism f(x) to f(xr). Inee, the action of b R sens f(x) to f(xb), an then the action of a R sens f(xb) to f(xab), which is the same as the action of ab. Step 2. Claim: the R-moule S is injective. Proof. For any R-moule M we have a canonical isomorphism of abelian groups Hom R (M, S) = Hom R (M, Hom Z (R, Q/Z)) = Hom Z (M, Q/Z). The map from left to right is obtaine by evaluating at 1 R. The map from right to left sens φ(x) : M Q/Z to the map which sens x M to φ(rx). Since Q/Z is injective as an abelian group, for any injective map of R-moules M N any map M Q/Z extens to a map N Q/Z. In other wors, the natural map Hom Z (N, Q/Z) Hom Z (M, Q/Z) (the restriction to M N) is surjective. The above isomorphism for N in place of M now gives that the natural map Hom R (N, S) Hom R (M, S) is surjective. This means that S is injective as a left R-moule. Step 3. For any M we have Hom R (M, S) 0; moreover, for any non-zero m M there exists an f Hom R (M, S) such that f(m) 0. This follows irectly from the previous steps. Now let I(M) be the prouct of copies of S inexe by the elements of Hom R (M, S). Since any prouct of injective moules is injective, I(M) is injective. There is a canonical map M I(M) that sens m M to the element of I(M) whose coorinate corresponing to f Hom R (M, S) is f(m). Then rm is sent to the element of I(M) whose f-coorinate is f(rm) = rf(m), so this is a map of left R-moules. 17

18 Step 4. Claim: this map is injective. Inee, let m M be non-zero. By Step 3 there exists an f Hom R (M, S) such that f(m) 0. Hence this m goes to a non-zero element of I(M). 3 Resolutions an erive functors 3.1 Projective an injective resolutions Let R be a ring (not necessarily commutative). A chain complex of R-moules is a (finite, infinite or semi-infinite) complex... A n+1 A n A n 1... The fact that it is a complex is equivalent to 2 = 0. A map of chain complexes f : A B is a family of maps of R-moules f n : A n B n commuting with the ifferentials, i.e. such that f n = f n 1 for all n. A cochain complex of R-moules is a complex... C n 1 C n C n+1... Maps of cochain complexes are efine as families of map commuting with. Let M be a left R-moule. We can associate to M chain an cochain complexes with goo properties, as follows. Definition 3.1 A left resolution of M is a complex of R-moules P 1 P 0 together with a map P 0 M such that the complex... P n P n 1... P 1 P 0 M 0 P n is exact. A left resolution is enote by P M. If each P n is projective, the resolution P M is calle projective. Definition 3.2 A right resolution of M is a complex of R-moules I 0... I n... together with a map M I 0 such that the complex 0 M I 0 I 1... I n... I 1 is exact. A right resolution is enote by M I. If each I n is injective, the resolution M I is calle injective. Lemma 3.3 Every moule has a projective an injective resolutions. Proof. We know that for any M there is a surjective map ɛ : P 0 M, where P 0 is free, hence projective. Define M 0 = Ker(ɛ). Take a surjective map P 1 M 0 with P 1 projective. Now let : P 1 P 0 be the composition P 1 M 0 P 0. It is easy to see that Im[P 1 P 0 ] = M 0 = Ker[P 0 M]. This gives the first bit of our resolution an proves the exactness at P 0. Now efine M 1 = Ker[P 1 P 0 ] an repeat the proceure to construct P 2, an so on. An injective resolution is built in exactly the same way using Theorem

19 Proposition 3.4 Suppose that we are given a projective resolution P M an a map M N. Then for any left resolution Q N there exist maps f n : P n Q n, n 0, making the following iagram commutative... P 2 P 1 P 0 M 0... Q 2 Q 1 Q 0 N 0 If g = (g n ) : P Q is another map of complexes making the iagram commutative, then there are maps s n : P n Q n+1 such that f n g n = s n 1 n + n+1 s n for n 1 an f 0 g 0 = 1 s 0. Proof. Consier the surjective map Q 0 N. By the efinition of a projective moule the composition P 0 M N can be lifte to f 0 : P 0 Q 0. This buils the first square of the iagram an shows that it is commutative. Now consier P 1 P 0 Q 0. The image of this map goes to 0 in N, so it can be viewe as a map P 1 Ker[Q 0 N]. Consier the surjective map Q 1 Ker[Q 0 N]. Since P 1 is projective, we get a map f 1 : P 1 Q 1 which buils the secon square an shows that it is commutative. Iterating this proceure constructs all vertical maps one by one an proves the commutativity of respective squares. To prove the secon statement, we consier the zero map M N an prove that for any map of complexes f : P Q making the iagram commutative there are maps s n : P n Q n+1 such that f n = s n 1 n + n+1 s n for n 1 an f 0 = 1 s 0. Inee, in this case f 0 sens P 0 to Ker[Q 0 N]. Since P 0 is projective, f 0 lifts to a map s 0 : P 0 Q 1. This gives f 0 = 1 s 0. Now f 1 s 0 1 sens P 1 to Ker[Q 1 Q 0 ], hence, by the projectivity of P 1, it lifts to s 1 : P 1 Q 2. We obtain f 1 s 0 1 = 2 s 1, which is f 1 = s s 1. Now continue by iterating this contruction. The secon statement of Proposition 3.4 is usually phrase as follows: a map of complexes f = (f n ) : P Q making the above iagram commutative is unique up to homotopy. Exercise State an prove the analogous statement for injective resolutions. 3.2 Derive functors Let A be a chain complex of R-moules. The n-th homology group H n (A ) is efine as the quotient of Ker[A n A n 1 ] by Im[A n+1 A n ]. One efines the cohomology groups H n (C ) of a cochain complex C similarly. A complex is exact if an only if all its (co-)homology groups are zero. We note that a map of chain complexes f = (f n ) : A B inuces maps on homology groups f = (f n ) : H n (A ) H n (B ). Lemma 3.5 Maps of chain complexes f, g : A B that are equal up to homotopy inuce equal maps of homology groups, i.e. f = g. 19

20 Proof. Easy exercise. Suppose that we are given maps of chain complexes A B C, that is, maps of R-moules A n B n C n for all n 0 that commute with the ifferentials. We call it a short exact sequence of compexes an write 0 A B C 0 if for each n 0 we have a short exact sequence of R-moules 0 A n B n C n 0. Lemma 3.6 A short exact sequence of complexes gives rise to a long exact sequence of homology groups... H n (A ) H n (B ) H n (C ) H n 1 (A ) H 1 (A ) H 1 (B ) H 1 (C ) H 0 (A ) H 0 (B ) H 0 (C ) 0. Proof. We shall use the following form of the snake lemma: a commutative iagram with exact rows gives rise to an exact sequence L M N 0 f g h 0 L M N Ker(f) Ker(g) Ker(h) Coker(f) Coker(g) Coker(h). The proof is left to the reaer as a (very goo an important) exercise. The iea is to apply the snake lemma to truncate pieces of the short exact sequence of complexes, as follows. We first apply it to the iagram an obtain A 1 /(A 2 ) B 1 /(B 2 ) C 1 /(C 2 ) 0 0 A 0 B 0 C 0 0 H 1 (A ) H 1 (B ) H 1 (C ) H 0 (A ) H 0 (B ) H 0 (C ) 0. Here we note that H 0 (B ) H 0 (C ) is surjective because B 0 C 0 is surjective. Next, we apply the snake lemma to A 2 /(A 3 ) B 2 /(B 3 ) C 2 /(C 3 ) 0 0 Ker[A 1 A 0 ] Ker[B 1 B 0 ] Ker[C 1 C 0 ] 20

21 an obtain H 2 (A ) H 2 (B ) H 2 (C ) H 1 (A ) H 1 (B ) H 1 (C ). We can glue this with the previous 6-term exact sequence because the maps between the first homology groups are the same as before. Iterating this proceure we buil a long exact sequence of homology groups. Let M be a left R-moule. For a right R-moule A we shall write F (A) = A R M. Then F is a right exact aitive functor from the category of right R-moules to the category of abelian groups. To say that F is a functor means that F (A) is an abelian group for every right R-moule A; moreover, to a map of right R-moules φ : A B it associates a homomorphism F (φ) : F (A) F (B) in such a way that F (i A ) = i F (A) an F (φ 1 φ 2 ) = F (φ 1 )F (φ 2 ). To say that F is aitive means that if φ, φ : A B, then F (φ + φ ) = F (φ) + F (φ ). Finally, to say that F is right exact means that we have the conclusion of Lemma We are going to efine left erive functors L i F from the category of right R- moules to the category of abelian groups, i 0. They will have these properties: L 0 F = F, an for any short exact sequence of right R-moules 0 A B C 0 we have a (semi-infinite) long exact sequence of erive functors... L n F (A) L n F (B) L n F (C) L n 1 F (A) L n 1 F (B) L n 1 F (C) L 1 F (A) L 1 F (B) L 1 F (C) F (A) F (B) F (C) 0. Definition 3.7 Define L n F (A) = H i (F (P )), where P A is a projective resolution of A. In other wors, L n F (A) is the n-th homology group of the complex... F (P n ) F (P n 1 )... F (P 1 ) F (P 0 ) 0. Theorem 3.8 (i) L 0 F = F ; (ii) the abelian group L i F (A) oes not epen on the choice of a projective resolution of A, for any i 0; (iii) L i F is an aitive functor, for any i 0; (iv) for any short exact sequence of R-moules there is an associate long exact sequence of erive functors. 21

22 Proof. (i) We have a short exact sequence 0 Im[P 1 P 0 ] P 0 A 0, which, since F is right exact (Lemma 1.10), gives an exact sequence F (Im[P 1 P 0 ]) F (P 0 ) F (A) 0. Since F (P 1 ) surjects onto F (Im[P 1 P 0 ]), we euce an exact sequence F (P 1 ) F (P 0 ) F (A) 0, whence F (A) = H 0 (F (P )). (ii) If Q A is another projective resolution, then Proposition 3.4 together with Lemma 3.5 gives well efine maps H n (F (P )) H n (F (Q )). Reversing the roles of P an Q prouces well efine maps in the opposite irection. Moreover, both compositions are the maps inuce by maps of complexes which up to homotopy are the ientity maps. Hence both compositions are the ientity maps. Therefore, the canonical maps H n (F (P )) H n (F (Q )) are isomorphisms. (iii) Proposition 3.4 together with Lemma 3.5 shows that for any map of R- moules f : A B we have associate well efine maps L i f : L i F (A) L i F (B). It is clear that L i (i A ) is the ientity map i : L i F (A) L i F (A). It is also clear that L i (fg) = (L i f)(l i g). We conclue that each L i F is a functor from the category of R-moules to the category of abelian groups; moreover, it is an aitive functor. (iv) Finally, the existence of a long exact sequence of erive functors follows from Lemma 3.6 an the following Horseshoe lemma. Its iea is that since the L i F o not epen on the choice of a projective resolution, we can construct a resolution of B from given resolutions of A an C, which will have certain aitional properties. Lemma 3.9 (Horseshoe lemma) Let P A an R C be projective resolutions. Define Q n = P n R n for all n 0. One can efine maps Q n Q n 1 for n 1 an a map Q 0 B to make Q a projective resolution of B in such a way that P A, Q B an R C form a short exact sequence of exact complexes. Remark The maps Q n Q n 1 that are going to be efine are not the irect sums of ifferentials : P n P n 1 an : R n R n 1. Nevertheless, each short exact sequence 0 P n Q n R n 0 is split because P n Q n = P n R n is the natural injective map an Q n = P n R n R n 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. 22

23 Proof. We nee to construct a map Q 0 = P 0 R 0 B. The first component P 0 A is the augmentation map of P A. Since R 0 is a projective R-moule, the map R 0 C can be lifte to a map R 0 B. It is easy to see that this efines a surjective map Q 0 B. The snake lemma now gives an exact sequence 0 Ker[P 0 A] Ker[Q 0 B] Ker[R 0 C] 0. We repeat the previous argument for Q 1 = P 1 R 1 to construct a map Q 1 Ker[Q 0 B] compatible with : P 1 P 0 an : R 1 R 0. Iterating the same proceure we prove the lemma. Lemma 3.10 (i) An R-moule M is a non-tivial irect sum if an only if we can write i M = p 1 + p 2, where p 1 an p 2 are non-zero enomorphisms of the R-moule M such that p 2 1 = p 1 an p 2 2 = p 2. (ii) If F is an aitive functor, then F (A B) is canonically isomorphic to the irect sum F (A) F (B). Proof. (i) If M = A B, then let p 1 be the projector to A an let p 2 be the projector to B. Let us prove the converse. Note that i M = p 1 + p 2, p 2 1 = p 1, p 2 2 = p 2 formally imply p 1 p 2 = p 2 p 1 = 0. Define A = p 1 (M) an B = p 2 (M). There is a natural map A B M sening a b to a + b. It is surjective since i M = p 1 + p 2. It is injective since A B = 0, because p 2 (A) = 0, p 1 (B) = 0, hence i M (A B) = 0. (ii) Write M = A B; this efines projectors p 1 an p 2. An aitive functor preserves ientities, sums an compositions, so i F (M) = F (i M ) = F (p 1 ) + F (p 2 ). Moreover, F (p 1 ) 2 = F (p 1 ) an F (p 2 ) 2 = F (p 2 ). By (i), F (M) is the irect sum of F (p 1 )F (M) = F (A) an F (p 2 )F (M) = F (B). Now we can complete the proof of part (iv) of Theorem 3.8. We choose any projective resolutions P A an R C an construct a projective resolution Q B as in the Horseshoe lemma. Apply our functor F to the split short exact sequence of resolutions. By Lemma 3.10 the fact that this sequence is split implies that the resulting sequence of complexes of abelian groups 0 F (P ) F (Q ) F (R ) 0 is exact. (Here F (P ) enotes the complex... F (P n ) F (P n 1 )...) Applying Lemma 3.6 to this exact sequence of complexes gives rise to a long exact sequence of erive functors. Exercise Let G be a left exact aitive functor from the category of left R-moules to the category of abelian groups. Using the symmetry between injective an projective moules (which reverses the irection of maps) efine the right erive functors R i G, i 0. Prove their properties analogous to the properties of left erive functors establishe above. 23

24 Definition 3.11 The left erive functors of F (A) = A R B are enote by L i F (A) = Tor R i (A, B). The right erive functors of G(A) = Hom R (B, A) are enote by R i G(A) = Ext i R(B, A). Remark Given a right R-moule A one can also consier the functor A R from the category of left R-moules to the category of abelian groups, call it F (B) = A R B. This functor is right exact, an so has left erive functors which are compute using projective resolutions. A balancing theorem (which we will not prove in this course) states that the left erive functors L i F give rise to the same abelian groups, i.e. L i F (B) = Tor R i (A, B) = L i F (A). Similarly, given a left R-moule B consier the contravariant functor Hom R (, B) from the category of left R-moules to the category of abelian groups, call it G (A) = Hom R (A, B). This functor is left exact, an so has right erive functors. Since G is contravariant, i.e. it reverses the irection of maps, the right erive functors R i G are compute using projective resolutions (so that after applying G we get a cochain complex an consier its cohomology groups). The balancing theorem for Ext says that the right erive functors of G (A) = Hom R (A, B) are R i G (A) = Ext i R(A, B) = R i G(B). See [3, 2.7]. 3.3 Tor an Ext If A is a projective right R-moule, then 0 A is a projective resolution of A, so all higher erive functors are zero. Thus Tor R n (A, B) = 0 for any left R-moule B an any n 1. By the balancing theorem, if B is a projective left R-moule, then Tor R n (A, B) = 0 for any right R-moule A an any n 1. Now recall that projective moules are flat (Proposition 1.12), but not vice versa. If B is flat, then R B is an exact functor (i.e. it preserves short exact sequences). By cutting an exact complex into short exact sequence one shows that this implies that R B applie to a projective resolution P A gives an exact complex... P n R B P n 1 R B... P 1 R B A R B 0. Thus all higher erive functors are zero, i.e. Tor R n (A, B) = 0 for any A an any n 1. Proposition 3.12 A left R-moule B is flat if an only if Tor R 1 (A, B) = 0 for any right R-moule A (an then Tor R n (A, B) = 0 for any A an any n 1). Proof. It remains to prove that Tor R 1 (A, B) = 0 for any A implies that B is flat. By the efinition of flatness we nee to prove that the functor R B is exact. Let 0 L M N 0 24

25 be a short exact sequence of right R-moules. Then we have a long exact sequence... Tor R 1 (N, B) L R B M R B N R B 0. Since Tor R 1 (N, B) = 0 we see that R B is an exact functor. Example Let R = Z. Then for any n 1 we have Tor Z n(z, B) = Tor Z n(a, Z) = Tor Z n(q, B) = Tor Z n(a, Q) = 0, because Z is free an Q is flat. Let G be a finitely generate abelian group. By basic group theory G is isomorphic to F Z r, where r 0 an F is a finite abelian group, hence a prouct of finite cyclic groups of prime power orer. Since Tor Z m(a, B) is an aitive functor in each argument, in orer to compute it in the case when A an B are finitely generate, it remains to compute Tor Z m(z/n, B). We have a natural 2-term projective resolution of Z/n, namely 0 Z Z, where the secon map is the multiplication by n map [n]. By tensoring this resolution with B we get the complex 0 B [n] B 0. Thus for any abelian group B we have Tor Z 0 (Z/n, B) = Z/n Z B = B/nB, Tor Z 1 (Z/n, B) = B[n], Tor Z 2(Z/n, B) = 0. For example, if 0 A B C 0 is a short exact sequence of abelian groups, then applying the erive functors of Z/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 an B we have Tor Z >1(A, B) = 0 an Ext >1 Z (A, B) = 0. Proof. There exists a free abelian group F an a surjective homomorphism F B, see the proof of Proposition 1.4. Let F be the kernel of F B. By Theorem 2.3 the abelian group F is also free. Hence 0 F F is a projective resolution of B. Since it has only two non-zero terms, all left erive functors vanish for m > 1. Let I be an injective abelian group such that there is an injective homomorphism B I. By Theorem 2.5 being injective an being ivisible is the same thing for abelian groups. Thus I is ivisible, but then I/B is ivisible too. Hence I/B is injective, an I I/B 0 is an injective resolution of B. Therefore, all right erive functors vanish for m > 1. 25