Algebra IV doc. Luk´aˇs Vokˇr´ınek, PhD. June 2, 2022 Contents Introduction iii Syllabus iii 1. Noetherian rings 1 2. Invariant theory 7 3. Localization 8 4. Primary decomposition 12 5. Chain complexes 15 6. Derived functors 24 7. Balancing Tor and Ext 28 8. Ext and extensions 33 9. Homological dimension 36 10. Group cohomology 38 11. Flatness is stalkwise 46 12. Simplicial resolutions 48 13. Representation theory 50 14. Characters of groups 54 i 15. Representations of symmetry groups Sn 57 16. Integrally closed rings, valuation rings, Dedekind domains 57 ii Introduction Introduction will be here at some point (or not). Luk´aˇs Vokˇr´ınek Syllabus Syllabus will be here at some point (or not). iii 1. Noetherian rings 1. Noetherian rings Definition 1.1. Let A be a ring. We say that an A-module M is noetherian, if it satisfies the ascending chain condition for submodules, i.e. if there exists no strictly increasing chain M0 ⊊ M1 ⊊ · · · ⊊ Mn ⊊ · · · of submodules of M. As a special case, we say that A is a noetherian ring, if it is noetherian as an A-module, i.e. if it satisfies the ascending chain condition for ideals in A. Theorem 1.2. An A-module M is noetherian iff every submodule of M is finitely generated. Proof. Assume that M is noetherian and let L ⊆ M be infinitely generated. We construct inductively a stricly increasing sequence of finitely generated submodules Ln ⊆ L in the following way: we start with L0 = 0 and then inductively Ln ⊊ L for otherwise L would be finitely generated and we set Ln+1 = Ln + Rxn+1 where xn+1 ∈ L ∖ Ln. For the opposite implication, assume that every submodule of M is finitely generated and that M0 ⊆ M1 ⊆ · · · is a sequence of submodules of M. Then M∞ = ∪nMn is a submodule. It is finitely generated by assumption, M∞ = R{x1, . . . , xk} and since each xi lies in some Mj, there exists n such that x1, . . . , xk ∈ Mn. Then Mn = Mn+1 = · · · . Theorem 1.3. Let 0 → M′ α −→ M β −→ M′′ → 0 be a short exact sequence of A-modules. Then M is noetherian iff both M′ and M′′ are noetherian. Proof. If M is noetherian then the lattice of submodules of both M′ and M′′ ∼= M/M′ are sublattices of the lattice of ideals of M and as such do not contain an infinite chain. Assume conversely that both M′, M′′ are noetherian and let M0 ⊆ M1 ⊆ · · · be a sequence of submodules. Then M′ n = α−1(Mn) is constant for n ≫ 0 and so is M′′ n = β(Mn). But then so must be Mn: for if x ∈ Mn+1, then β(x) ∈ M′′ n+1 = M′′ n and so β(x) = β(y) for some y ∈ Mn. Analogically, x − y = α(z) for some z ∈ M′ n, and thus x = y + α(z) ∈ Mn. (Alternatively: the inclusion Mn → Mn+1 is an extension of inclusions M′ n → M′ n+1 and M′′ n → M′′ n+1, which are isomorphisms for n ≫ 0 and 5-lemma gives the result.) Proof. Assume conversely that M′, M′′ are noetherian and let L ⊆ M be a submodule. Then for L′ = α−1(L), L′′ = β(L) we get a short exact sequence 0 → L′ → L → L′′ → 0. Since both L′ ⊆ M′ and L′′ ⊆ M′′ are finitely generated, so is L. Corollary 1.4. If A is a noetherian ring, the every finitely generated module M is noetherian. Proof. The sum of two modules can be expressed via a (split) short exact sequence 0 → M′ → M′ ⊕ M′′ → M′′ → 0, the previous theorem thus shows that every finitely generated free module An is noetherian and also every quotient of it, i.e. any finitely generated module. In the proceeding, the commutativity assumption is crucial. Definition 1.5. An A-algebra is a homomorphism of rings ρ: A → B. Mostly, it will be a mono and we will thus think of B as a supring of A. 1 1. Noetherian rings Example 1.6. A[x1, . . . , xn] is an A-algebra. Since B is canonically a B-module, by restricting scalars along ρ we may also treat it as an A-module. An alternative definition of an A-algebra is as an A-module B together with an A-bilinear mapping B × B → B (multiplication) that, together with the addition, makes B into a ring (this means that B is a monoid object in A−Mod). Definition 1.7. We say that an A-algebra B is finitely generated, when there exist b1, . . . , bn ∈ B that generate B as an A-algebra, i.e. via addition, multiplication and multiplication by scalars from A. We write B = A[b1, . . . , bn]. We say that an A-algebra B is finite, when B is a finitely generated A-module (i.e. there exist b1, . . . , bn ∈ B that generate B via addition and multiplication by scalars from B). We write B = A{b1, . . . , bn}. We remark that finite generation is equivalent to the existence of a surjective homomorphism of A-algebras A[x1, . . . , xn] → B (sending xi to the generators bi of B; this is so because A[x1, . . . , xn] is a free A-algebra on generators x1, . . . , xn). For any finite A-algebra there exists a surjective homomorphism of A-modules A{x1, . . . , xn} → B. Theorem 1.8. Let A be a noetherian ring and B a finite A-algebra. Then B is also a noetherian ring. Proof. By the corollary, B is a noetherian A-module, so every A-submodule of B is finitely generated as an A-module. This implies easily that every ideal of B (i.e. B-submodule ⇒ A-submodule) is a finite generated as an ideal (i.e. B-submodule). Example 1.9. The ring Z is noetherian. Then also Z[i] = {a + bi | a, b ∈ Z} is noetherian. Theorem 1.10. Let A be a noetherian ring, D ⊆ A a multiplicative subset. The also the localization D−1A is a noetherian ring. Proof. Again, the lattice of ideals of D−1A is a sublattice of the one for A. Theorem 1.11 (Hilbert basis theorem). If A is noetherian, then so is A[x]. Proof. Let I ⊆ A[x] be an ideal. We define an ideal J = {a ∈ A | ∃p ∈ I : p = axr + lot}, i.e. the ideal of leading coefficients of the polynomials from I. Let J = (a1, . . . , ak) and pick polynomials pi ∈ I with leading coefficients ai; we may assume that they all have the same degree r. The set A x β1 1 · · · x βn n = xβ , iff for some i ≥ 1 we have α1 = β1, . . . , αi−1 = βi−1, αi > βi. Since this is a linear order, we may speak of the leading term of a polynomial f ∈ k[x1, . . . , xn]: when f = aαxα + β<α aβxβ = aαxα + lot with aα ̸= 0, we call LC f = aα the leading coefficient, LM f = xα the leading monomial and LT f = aαxα the leading term. Somewhat more generally on monomial orders: we require that it should be a total order closed under the multiplication, i.e. xα ≤ xβ and xγ ≤ xδ ⇒ xα+γ ≤ xβ+δ and (often) also 1 ≤ xα. The second condition is equivalent to the monomial order being a well order. To prove this, we introduce a (useful) notion of a monomial ideal, i.e. an ideal generated by a collection of monomials fs. One can show that this is exactly {f | each monomial contained in f is divisible by some fs} (by observing that the right hand side is indeed an ideal, contains fs and is smallest among such). Now we are ready to prove that the monomial order is a well order: given a non-empty collection of monomials, consider a monomial ideal generated by them. By Hilbert basis theorem, it is finitely generated, so generated by a finite subcollection. Clearly any of the old generators (in fact any monomial in the ideal) is divisible by one of the new generators, thus sits above that generator in the order. We have thus reduced to a finite subset that has a smallest element since we assume the order to be total. 3 1. Noetherian rings Exercise 1.13. Show that an ideal J is monomial iff it satisfies f ∈ J ⇔ each monomial contained in f belongs to J. We will now give a definition that will allow us to describe a simple generalization of the membership algorithm f ? ∈ (g1, . . . , gs) from the univariate to the multivariate case. Definition 1.14. We say that elements g1, . . . , gs ∈ I form a Gr¨obner basis if the leading monomial of any g ∈ I is divisible by the leading monomial of some gi; in other words (LM g | g ∈ I) = (LM g1, . . . , LM gs) according to the characterization of monomial ideals above. With a Gr¨obner basis one can easily test membership f ∈ I: first we verify whether LM f is divisible by some LM gi. If not we get f /∈ I. If LM f = xα LM gi, we replace f by the polynomial f − LC f LC gi xα gi and we continue with testing (here we use that monomials are well ordered). Remark. We say that a Gr¨obner basis of an ideal I is reduced if all its members gi are monic and LM gi does not divide any term of any gj (this is in analogy with the reduced echelon form of a matrix, which is at the same time a special case). It holds that every ideal has a unique reduced Gr¨obner basis (we will not prove this). In the proceeding we explain how to compute such a basis. Then testing equality of two ideals becomes easy – we compute the reduced Gr¨obner bases and compare them. Even without reduced Gr¨obner bases, one can simply test whether each generator of one ideal belongs to the other and the other way around. 1.1. Buchberger algorithm The algorithm for finding a Gr¨obner basis of an ideal I = (f1, . . . , fl) proceeds in steps as follows: we compute for each fi, fj the so called S-polynomial S(fi, fj) by determining the least common multiple xα of the monomials LM fi, LM fj and we set S(fi, fj) = xα LT fi fi − xα LT fj fj. Then we use f1, . . . , fl to reduce the result, i.e. we subsequently subtract monomial multiples of the fk’s to exactly cancel the actual leading term (i.e. we use division with a remainder). If we get a nonzero polynomial whose leading term is no longer divisible by the leading term of any fk, we add the result to the set of generators, so that l increases by one and the bigger system of polynomials clearly generates I (the added polynomial may depend on the reduction, since we may choose multiple times the generator fk whose multiple gets subtracted). Since in each step the ideal (LM f1, . . . , LM fl) is enlarged and since k[x1, . . . , xn] is noetherian, after a finite number of steps we get to the situation where the reductions of all S-polynomials are zero. We will now show that the generating set then forms a Gr¨obner basis: Let f ∈ I = (f1, . . . , fl), so that f = p1f1 + · · · + plfl, and we assume that in this expression the monomial max{LM(pifi) | i = 1, . . . , l} 4 1. Noetherian rings is the smallest possible; let i0 be an index for which the maximum above is achieved. There are two possibilities: ˆ the leading terms do not cancel out, i.e. LM f = LM(pi0 fi0 ); then LM f ∈ (LM f1, . . . , LM fl); ˆ the leading terms do cancel out; then for indices i ̸= i0 with LM(pifi) maximal, we may replace pifi − LC(pifi) LC(pi0 fi0 ) pi0 fi0 = qiS(fi, fi0 ) + lot (the non-leading terms of pi and pi0 contribute to the lower order terms “lot”, the leading terms yield a multiple of the S-polynomial, since this was obtained as the smallest monomial combination in which the leading terms cancel out). By construction, every S-polynomial S(fi, fi0 ) can be replaced by a combination of the fj with smaller leading terms, and the terms in “lot” already have smaller leading terms; this gives a contradiction with minimality. Example 1.15. Compute the Gr¨obner basis of I = (f1, f2), where f1 = x3 − 2xy, f2 = x2y + x − 2y2. Solution. In the first step S(f1, f2) = yf1 − xf2 = −x2 f3 = x2 and no reduction is necessary. In the next step, the reduction of S(f1, f2) = −f3 is zero, further S(f1, f3) = f1 − xf3 = −2xy f4 = xy S(f2, f3) = f2 − yf3 = x − 2y2 f5 = x − 2y2 and again, no reductions are necessary. In fact, it is now possible to throw out f1, f2, since they lie in (f3, f4, f5). Let us compute S(f3, f4) = yf3 − xf4 = 0 S(f3, f5) = f3 − xf5 = 2xy2 ≡ 0 S(f4, f5) = f4 − yf5 = 2y3 f6 = y3 and now it is possible to leave out f3 = xf5 + 2yf4 a f4 = yf5 + 2f6. In the last step S(f5, f6) = y3 f5 − xf6 = −2y5 ≡ 0 Therefore (f5, f6) is a reduced Gr¨obner basis. ⋄ Example 1.16. Compute the Gr¨obner basis of I = (f1, f2, f3), where f1 = x2 + y2 + z2 − 1, f2 = x2 − y + z2, f3 = x − z. Solution. It will be convenient to write the subtraction of multiples of fi as reductions x2 ≡ −y2 − z2 + 1, x2 ≡ y − z2, x ≡ z, etc. In the first step we get S(f1, f2) = f1 − f2 = y2 + y − 1 f4 = y2 + y − 1 S(f1, f3) = f1 − xf3 = y2 + z2 − 1 + xz ≡ y2 + 2z2 − 1 f5 = y2 + 2z2 − 1 S(f2, f3) = f2 − xf3 = −y + z2 + xz ≡ −y + 2z2 f6 = y − 2z2 5 1. Noetherian rings It is now possible to throw out f1 = f2 + f4, f2 = (x + z)f3 − f4, f4 = f5 + f6 so that we have S(f3, f5) = y2 f3 − xf5 = −y2 z − 2xz2 + x ≡ −y2 z − 2z3 + x ≡ −y2 z − 2z3 + z ≡ −(1 − 2z2 )z − 2z3 + z = 0 S(f3, f6) = yf3 − xf6 = −yz + 2xz2 ≡ −yz + 2z3 ≡ −2z3 + 2z3 ≡ 0 S(f5, f6) = f5 − yf6 = 2z2 − 1 + 2yz2 ≡ 4z4 + 2z2 − 1 f7 = z4 + (1/2)z2 − 1/4 Again we can leave out f5 = (y + 2z2)f6 + 4f7, so the Gr¨obner basis is (f3, f6, f7). As an application, we can now solve the system of equations f1 = f2 = f3 = 0. This is equivalent to the system f3 = f6 = f7 = 0 and similarly to linear systems we may now compute the solution “from the back”: by solving f7 = 0 we get z = √ −1± √ 5 2 . Substituting into f6 = 0 we then obtain y = 2z2 = −2 ± 2 √ 5 an finally by substituting into f3 = 0 gives x = z = √ −1± √ 5 2 . ⋄ Example 1.17. Compute the Gr¨obner basis of I = (f1, f2), where f1 = x2 − y, f2 = x2 + (y − 1)2 − 1. Solution. In the first step S(f1, f2) = f1 − f2 = −y2 + y f3 = y2 − y and no reduction is necessary. In the next step we can leave out f2 = f1 − f3, further S(f2, f3) = y2 f2 − x2 f3 = x2 y + y4 − 2y3 ≡ y2 + y4 − 2y3 ≡ 0 (any power yk, k ≥ 1 reduces to y just using f3) and the reduced Gr¨obner basis is (f1, f3). The quotiend k[x, y]/I or perhaps rather k[x, y]/ √ I has a close connection to the solution set of f1 = 0, f2 = 0. It consists of three points [0, 0], [−1, 1], [1, 1] and therefore dim k[x, y]/ √ I = 3. At the same time dim k[x, y]/I = 4, since the point [0, 0] should be taken “twice”, concretely x(y − 1) /∈ I, but (x(y − 1))2 ∈ I, so that x(y − 1) ∈ √ I ∖ I (the function x(y − 1) vanishes on the three points, but not up to a sufficiently high order). ⋄ Lemma 1.18. If LM(f), LM(g) are coprime, then S(f, g) can be reduced to zero, using only f, g. Proof. For simplicity, we may assume f, g monic. By assumption S(f, g) = LM(g)f −LM(f)g and in each step we will subtract a multiple of the from tf where t is a term of g or adding a multiple of the form sg where s is a term of f, in such a way that in the end the S-polynomial will reduce to gf −fg = 0 (the point is that every term st turns up once with a plus sign and once with a minus sign and it can only be a leading term when s is a leading term of f or t is a leading term of g). HW 1. Solve the following system of polynomial equations using Gr¨obner bases x2 + y + z = 1 x + y2 + z = 1 x + y + z2 = 1 6 2. Invariant theory 2. Invariant theory This is a nice application of the Hilbert basis theorem. We assume here that k is a field of characteristic coprime to the order of a finite group G (this condition will also be important for the representation theory later in the course). We will consider an action of G on the polynomial ring k[x]. The invariants (the collection of invariant polynomials in this case) is the subset k[x]G = {f ∈ k[x] | ∀a ∈ G: a · f = f}. As an example, the symmetry group Sn acts on the variables and thus on the polynomials, e.g. (1 2)·x2 1x2 = x1x2 2. The main theorem in this respect is that for the elementary symmetric polynomials s1 = x1 + · · · + xn, s2 = x1x2 + · · · + xixj + · · · + xn−1xn, . . . , sn = x1 · · · xn the canonical map k[y] → k[x], yi → si is an isomorphism onto the invariants k[x]Sn . The action of σ ∈ Sn on k[x] is obtained from the action on variables in two steps: {x} Set //  {x}  k{x} Vect //  k{x}  k[x] Alg // k[x] The action on the set of variables induces a linear action on the vector space of linear forms (i.e. essentially on kn) and that induces an algebra action on the polynomial ring, i.e. it satisfies a · (f + g) = a · f + a · g, a · 1 = 1, a · (fg) = (a · f)(a · g). It follows that the G-invariants form a k-subalgebra. Theorem 2.1 (Hilbert’s on finite generation of invariants). The k-algebra k[x]G is finitely generated, i.e. there exists a surjection k[y] ↠ k[x]G. Proof. Denote by i: k[x]G → k[x] the inclusion. In this way, we can think of k[x] as an algebra over k[x]G. We will now construct a retraction p in the category of k[x]G-modules k[x] p "" k[x]G i << 1 // k[x]G by the formula p(f) = 1 |G| · a∈G a · f (the average of the elements in the orbit of f). The compatibility of the action with the algebra structure gives for f ∈ k[x]G: a · (fg) = (a · f)(a · g) = f(a · g) 7 3. Localization so that the action is indeed by k[x]G-linear maps and consequently so is p. Now im p ⊆ k[x]G since b · p(f) = b · ( 1 |G| · a∈G a · f) = 1 |G| · a∈G b · (a · f) = 1 |G| · a∈G (ba) · f = 1 |G| · a∈G a · f = p(f) since as a runs over elements of G, ba runs over the same set of elements (in a different order, i.e. a → ba is a bijection). Finally pi = 1 since, for f ∈ k[x]G, we have p(f) = 1 |G| · a∈G a · f = 1 |G| · a∈G f = f. Now let I ⊆ k[x] be the ideal generated by the homogeneous elements of k[x]G of positive degree, so that k(>0) [x]G ⊆ I ⊆ k[x]. By Hilbert basis theorem, we get I = (f1, . . . , fk) with fi some of the above generators, i.e. homogeneous elements of k[x]G of positive degree. We claim that k[x]G is generated as a k-algebra by the same set of elements f1, . . . , fk. Since the G-action respects the degrees of polynomials, a polynomial is G-invariant iff all its homogeneous components are G-invariant that leaves us to prove f ∈ k[x]G homogeneous ⇒ f ∈ k[f1, . . . , fk]. We prove this by induction on deg f. If deg f = 0, there is nothing to prove, so assume that deg f > 0. Now f ∈ I = (f1, . . . , fk) so we have an expression f = f1g1 + · · · + fkgk and we may assume that all the gi are homogeneous (replace the gi by their homogeneous components of the appropriate degrees and the equality will remain valid). Now apply the retraction p to get f = f1p(g1) + · · · + fkp(gk) (both the left hand side and the fi are G-invariant and p is k[x]G-linear). By induction, we may assume that all p(gi) already lie in k[f1, . . . , fk]. Thus, the same is true for f. 3. Localization Definition 3.1. A local ring is a ring (commutative with 1) with a unique maximal ideal. Theorem 3.2. A ring A is local iff its non-units form an ideal. In that case this ideal is the unique maximal ideal of A. 8 3. Localization Proof. The implication ⇒ follows from (a) ⊆ M for every non-unit a, the opposite implication is obvious. Definition 3.3. Let A be a ring and D ⊆ A a multiplicative subset, i.e. a subset satisfying 1 ∈ D and x, y ∈ D ⇒ xy ∈ D. The decomposition of A × D with respect to the equivalence relation (a1, d1) ∼ (a2, d2) ⇔ ∃d ∈ D: (a1d2 − a2d1)d = 0 will be denoted D−1A and called the localization of the ring A with respect to the multiplicative subset D. Its class will be denoted [a, d] = a d . A ring structure on D−1A is introduced by the formulas a1 d1 + a2 d2 = a1d2 + a2d1 d1d2 , a1 d1 · a2 d2 = a1a2 d1d2 . The mapping λ: A → D−1A, a → a 1 is a homomorphism of rings. The localization D−1A has the following universal property, saying that it is the universal supring where all the elements d ∈ D admit an inverse. Theorem 3.4. Let ρ: A → B be a ring homomorphism such that ρ(d) ∈ B× is a unit for all d ∈ D. Then there exists a unique ring homomorphism ρ: D−1A → B such that ρ = ρλ. A ρ // λ  B D−1A ρ << Proof. Since a d = λ(a)λ(d)−1, we are forced to set ρ(a d ) = ρ(a)ρ(d)−1. We will now show that this is a well defined mapping; we will leave to the reader to show that it is then a ring homomorphism. So let a1 d1 = a2 d2 , meaning that there exists d ∈ D such that (a1d2−a2d1)d = 0. Then also (ρ(a1)ρ(d2) − ρ(a2)ρ(d1))ρ(d) = 0. Since ρ(d) is a unit, we also have ρ(a1)ρ(d2)−ρ(a2)ρ(d1) = 0, implying easily that ρ(a1)ρ(d1)−1 = ρ(a2)ρ(d2)−1. Important special cases are: ˆ D = {1, a, a2, . . .}, then D−1A is obtained from A by adding an inverse to a, and we denote the result by A[a−1]. ˆ D = A ∖ p, where p ⊆ A is a prime ideal. Then D is indeed multiplicative and D−1A is denoted Ap – it is the so called localization of A at a prime ideal p. ˆ In particular, when A is an integral domain, 0 is a prime ideal and then A0 is the fraction field of A. HW 2. Prove the following isomorphisms: ˆ A[a−1] ∼= A[t]/(at − 1), ˆ (A/I)[t] ∼= A[t]/J and describe the ideal J, ˆ A/(I +J) ∼= (A/I)/J′ and describe the ideal J′ along the lines “it is essentially J, just. . . ”. Proposition 3.5. The localization D−1R is the trivial ring iff 0 ∈ D. 9 3. Localization Proof. The triviality means 1/1 = 0/1 and, by definition, this happens iff for some d ∈ D we have d = 0, i.e. iff 0 ∈ D. Theorem 3.6. The localization Ap at a prime ideal p is a local ring. Proof. It is easily seen that the complement of the ideal m = {a d | a ∈ p, d /∈ p} is composed of units. Definition 3.7. Let A be a ring. An A-module is an abelian group M together with an operation M × A → M, (x, a) → xa satisfying the axioms of a vector space, i.e. x1 = x, (xa)b = x(ab) x(a + b) = xa + xb, (x + y)a = xa + ya. An important example is an ideal – it is closed under addition and multiplication by elements of the ring. Theorem 3.8 (Nakayama lemma). Let A be a local ring with a maximal ideal m. Let N be a finitely generated A-module such that Nm = N. Then N = 0. Proof. Let x1, . . . , xn be a generating set of N. We may then write xj = x1a1j + · · · + xnanj for suitable aij ∈ m. Moving everything to the left, we obtain (x1, . . . , xn)(E −M) = 0, where M is a matrix composed of the elements aij. Multiplying by the adjoint matrix, we get (x1, . . . , xn) det(E − M) = 0, i.e. xj det(E − M) = 0. This means that the multiplication by det(E − M) gives on N the zero map. However det(E − M) ∈ 1 + m and it is thus a unit (it does not lie in m). Therefore N = 0. For a multiplicative subset D ⊆ A and the associated localization map λ: A → D−1A we study the relationship between the ideals of A and those of D−1A. We have maps between these sets that clearly preserve the ordering λ∗ : {ideals of A} //oo {ideals of D−1 A} : λ∗ with λ∗ (J) = λ−1 (J) = {a ∈ A | a 1 ∈ J} that clearly preserves primeness (e.g. A/λ−1(J) → B/J is clearly injective and a subring of a domain is itself a domain) and with λ∗(I) = D−1 A · λ(I) D−1I = {a d ∈ D−1 A | a ∈ I} (i.e. the ideal generated by the image λ(I). 10 3. Localization Clearly λ∗(λ∗(J)) = J and in the opposite direction λ∗ (λ∗(I)) = {a ∈ A | ∃d ∈ D: da ∈ I} We call this the D-saturation of I and also say that I is D-saturated if it equals its saturation, i.e. if da ∈ I ⇒ a ∈ I (division by d ∈ D). Obviously, by restriction, we get a bijection λ∗ : {D-saturated ideals of A} ∼= {ideals of D−1 A} : λ∗ Further, a prime ideal P is D-saturated iff it is disjoint from D (if saturated then d = d1 ∈ I ⇒ 1 ∈ I, i.e. nonsense, so that d /∈ I; if disjoint, one can divide by d showing D-saturatedness). λ∗ : {prime ideals of A disjoint from D} ∼= {prime ideals of D−1 A} : λ∗ Thus, if D = R ∖ P the left hand side consists of prime ideals contained in P and as such contains a maximal element P, implying that D−1A = AP has a unique maximal ideal, namely D−1 P = {a b | a ∈ P, b /∈ P} (alternatively, it consists exactly of the non-units of AP ). More generally, any ideal I that is maximal among those disjoint from D must be D-saturated, since its D-saturation is still a proper ideal disjoint from D, so that it is in fact a maximal D-saturated ideal and as such is a pullback of a maximal ideal of D−1A, hence prime. The point of the localization D−1A lies in having less ideals, in particular prime ideals, and this simplifies the structure theory of modules. We will see some examples of this. The localization of a module is defined similarly by universal property M ρ // λ  N D−1M ρ ;; where N is assumed to be an D−1A module, i.e. an A-module in which the multiplication map d · : N → N is an isomorphism (look at the action map D−1A → End(N) and employ the universal property of the localization D−1A). Straight from the definition we see that if the multiplication maps are isomorphisms on M then we can take λ = id, i.e. D−1M = M. In general, since HomA(M, N) ∼= HomA(M, HomD−1A(D−1 A, N)) ∼= HomD−1A(D−1 A ⊗A M, N) the so called extension of scalars gives a concrete construction D−1M = D−1A ⊗A M. It is then important that D−1A is a flat A-module (see below) and thus the localization functor is exact. We will now give a second construction D−1 M = {x d | x ∈ M, d ∈ D} where similarly to the case of A, it is imposed that x d = y e iff fex = fdy for some f ∈ D. To prove that this gives the previous localization, one has to prove that the maps D−1 A ⊗A M //oo D−1 M, 11 4. Primary decomposition given by a/d ⊗ x → (ax)/d and 1/d ⊗ x ← x/d, are well defined (the first is the extension of the canonical inclusion λ: M → D−1M) and inverse to each other. This implies easily that D−1A is flat since for f : M → N injective the induced D−1f : D−1M → D−1N satisfies D−1f(x/d) = f(x)/d = 0 iff ef(x) = 0, i.e. f(ex) = 0 and ex = 0 by injectivity of f; finally this gives x/1 = 0. Alternatively, one can express D−1A = d∈D d−1A = colimd∈D A where the maps in the diagram are exactly of the form e · : A → A from the copy of A with index d to the copy with index ed. It remains to show that the colimit indeed gives D−1A (easy) and that the diagram is filtered (very easy). Again, for D = R ∖ P we denote MP = D−1M. Theorem 3.9. For an A-module M we have: M = 0 ⇔ ∀P maximal: MP = 0. Before starting the proof we define the annihilator of x ∈ M to be the ideal Ann(x) = AnnM (x) = {a ∈ A | ax = 0}. Clearly x = 0 iff Ann(x) ∋ 1. The fraction a d ∈ D−1A then annihilates λ(x) = x 1 , i.e. ax d = 0 iff ∃e ∈ D: eax = 0 (i.e. ea ∈ Ann(x)) iff a d ∈ D−1 Ann(x), so that we finally get Ann(x 1 ) = D−1 Ann(x). (This implies, in particular, that x ∈ ker λ iff D ∩ Ann(x) ̸= ∅ since these are exactly ideals giving the trivial ideal in the localization D−1A.) Proof. The implication ⇒ is clear, so assume that 0 ̸= x ∈ M. Then Ann(x) ⫋ A is a proper ideal and there exists a maximal ideal P ⊇ Ann(x). Denoting D = A ∖ P as usual, we obtain D ∩Ann(x) = ∅ so that D−1 Ann(x) ̸∋ 1 is also proper. Since it equals Ann(x 1 ), we must have 0 ̸= x 1 ∈ MP and this module is thus also non-zero. Corollary 3.10. For an A-linear map f : M → N we have: f is mono/epi/iso ⇔ ∀P maximal : the localized map fP : MP → NP is such. Proof. This follows from the chain of equivalences: f mono iff ker f = 0 iff (ker f)P = 0 iff ker fP = 0 (since the localization, being exact, commutes with kernels) iff fP mono. 4. Primary decomposition Let R be a (possibly graded) noetherian ring and let M be an R-module. Let us investigate when multiplication by r ∈ R on the module M is non-injective – we may say that r is a zero divisor on M because this exactly means that there exists a nonzero x ∈ M such that rx = 0. We denote Ann(x) = AnnM (x) = {r ∈ R | rx = 0}, the so called annihilator of the element x; it is easy to see that this is an ideal. The zero divisors on M are thus exactly the elements of the union of all annihilators Ann(x) for x ̸= 0. Of course, it is enough to consider the maximal such and we will show that these are prime ideals. We say that a prime ideal p is an associated prime of the module M if p = Ann(x) for some x ∈ M. The set of all associated primes of M is denoted Ass(M). 12 4. Primary decomposition We will now explain a useful characterization of annihilators: an R-submodule generated by x is isomorphic to Rx ∼= R/ Ann(x) by the first isomorphism theorem applied to the Rlinear map R → M sending 1 → x, whose image is obviously Rx and whose kernel is Ann(x). Thus, equivalently, a prime ideal p is associated iff M contains a submodule isomorphic to the cyclic module R/p. Example 4.1. Ass(R/p) = {p} because R/p is an integral domain and thus the multiplication by any nonzero element is injective, i.e. Ann(x) = p for x ̸= 0. Lemma 4.2. Every maximal element of {Ann(x) | x ̸= 0} is an associated prime. In particular, for R noetherian, every annihilator Ann(x), for x ̸= 0, is contained in some associated prime. Remark. It is also true that, for a multiplicative subset D, a maximal element {Ann(x) | Ann(x) ∩ D = ∅} is an associated prime. This was proved in an earlier version and may be needed at some point... Proof. Let Ann(x) be maximal and let rs ∈ Ann(x). Then either sx = 0 and thus s ∈ Ann(x) or r ∈ Ann(sx) = Ann(x) by maximality. As a simple consequence, we obtain the following theorem: Theorem 4.3. Let R be a noetherian ring. Multiplication by r ∈ R on an R-module M is injective iff r does not lie in any associated prime of M. This theorem is useful especially because we will show that Ass(M) is finite for every noetherian (i.e. finitely generate) module M. The main tool here will be a so called primary decomposition. We say that a module M is p-primary if Ass(M) = {p}. We also say that M is primary if it is p-primary for some prime ideal p. In the case that M is not primary, it contains two submodules P ∼= R/p and Q ∼= R/q and in that case Ass(P ∩Q) ⊆ Ass(P)∩Ass(Q) = {p}∩{q} = ∅. Since every nonzero module has some associated prime, we get P ∩ Q = 0. Theorem 4.4. Let M be a finitely generated module over a noetherian ring R. Then there exists a finite collection of submodules Mi, for i = 1, . . . , n, such that 0 = i Mi and such that each M/Mi is pi-primary and the prime ideals pi are all distinct. If this expression is irredundant (i.e. no Mi can be removed) then Ass(M) = {p1, . . . , pn}. Proof. Let us call an expression M0 = Mi with all M/Mi primary a decomposition of the submodule M0. We will show that if M0 has no decomposition then there exists a strictly larger submodule without a decomposition and this would contradict M being noetherian. Since M0 admits no decomposition, M/M0 cannot be primary (for otherwise M0 = M0 would be a decomposition). As above, there exist two submodules M1/M0, M′ 1/M0 ⊆ M/M0 with zero intersection, i.e. with M1 ∩ M′ 1 = M0. If both M1 and M′ 1 had decompositions we would obtain a decomposition for M0 by intersecting these, so one of them does not admit a decomposition, as claimed. We will now show Ass(M) ⊆ {p1, . . . , pn}. First we prove Ann(x) = AnnM/M1 (x) ∩ · · · ∩ AnnM/Mn (x) (for a nonzero x ∈ M, we have r ∈ Ann(x) iff rx = 0 iff ∀i: rx ∈ Mi iff ∀i: r ∈ AnnM/Mi (x)). Assuming now that Ann(x) ⊊ AnnM/Mi (x) for all i, we pick si ∈ AnnM/M1 (x) ∖ Ann(x). 13 4. Primary decomposition Their product s1 · · · sn then lies in the intersection, hence in Ann(x), but si /∈ Ann(x), so Ann(x) is not prime. So for prime Ann(x) this must equal one of the AnnM/Mi (x), and the latter can only be prime if it equals pi. For the opposite inclusion we need the irredundancy: It gives i̸=j Mi ̸= 0 and this intersection thus contains some non-zero element, necessarily x /∈ Mj, that has AnnM (x) = AnnM/Mj (x) and, for some multiple y ∈ Rx ⊆ i̸=j Mi, we obtain AnnM/Mj (y) = Pj since M/Mj is Pj-primary. Finally, we will study the behaviour of primary decomposition under localization, so let D be a multiplicative subset. Lemma 4.5. Let x ∈ M. A maximal element of {Ann(dx) | d ∈ D} equals the D-saturation of Ann(x). In particular, for R noetherian, the D-saturation of every annihilator Ann(x) is an annihilator Ann(d0x). Proof. Let Ann(d0x) be maximal and let dr ∈ Ann(d0x). Then r ∈ Ann(dd0x) = Ann(d0x) and it is D-saturated. Clearly, it has the same D-saturation as Ann(x). For the localization map λ: M → D−1M we recall that Ann(x/1) = D−1 Ann(x) and since the localization gives a bijection {prime ideals of A disjoint from D} ∼= {prime ideals of D−1 A} (and those intersecting D give the full ring on the right hand side) we can determine the associated primes of D−1M: Ass(D−1 M) = {D−1 P | P ∈ Ass(M), D ∩ P = ∅}. This takes a particularly simple form for a P-primary module M over a noetherian ring (is this necessary?): then either D−1M is D−1P-primary when D ∩ P = ∅ or D−1M = 0 when D ∩ P ̸= ∅ (since then D−1M has no associated prime). Now apply this to a primary decomposition 0 = Mi with M/Mi being Pi-primary. We get 0 = D−1 Mi with D−1M/D−1Mi being D−1Pi-primary; when some D−1M/D−1Mi is zero, i.e. D−1Mi = D−1M, we may remove it from the decomposition. For a minimal associated prime Pj we then get only one non-zero submodule, namely 0 = D−1 Mj that together with the monomorphism (since the module M/Mj is Pj-primary, we have Ann(x/1) = D−1 Ann(x) ⊆ D−1Pj and is thus proper, showing that x/1 ̸= 0) M λj //  D−1M ∼=  M/Mj // // D−1M/D−1Mj 14 5. Chain complexes gives that Mj = ker λj and as such is unique. For completeness, still over a noetherian ring, we prove that for any prime P ⊇ Ann(x) there is an associated prime lying between these two: consider λ: M → MP and observe that Ann(x/1) = Ann(x)P is non-trivial. It is thus contained in some associated prime D−1Q ∈ Ass(MP ). As above, this means that Q ∈ Ass(M) (this is a bit circular, it seems that the general version of Lemma 4.2 is needed to conclude that there exists an annihilator maximal among those disjoint from D = R ∖ P and as such is the prime Q as above). This implies that any proper ideal I lies in prime that is minimal above it: since Ass(R/I) is finite, it contains a minimal element; by the above it must in fact be minimal among all primes containing I = Ann(1). As a final application of this, if I is P-primary then, in particular, P is the unique minimal prime above I (so smallest) and thus Proposition 4.6 gives √ I = I⊆Q prime Q = P. In particular, we obtain the following consequence: rs ∈ I ⇒ r ∈ √ I or s ∈ I (the first condition means that r belongs to the unique associated prime of R/I, the second that s = 0 in R/I). In the opposite direction, when I satisfies this condition we get that √ I is the union of all the associated primes by Theorem 4.3, and these include all minimal primes above I. Since it is also the intersection of all minimal primes above I, there must be only one associated prime and I is necessarily primary (with associated prime √ I). So for a primary ideal I, the radical √ I is a prime ideal. The converse is generally not true, since √ I being prime only means that there is a unique minimal prime over I, but some bigger prime may be associated as well. However, if √ I is a maximal ideal, then I is automatically primary. In particular, for a maximal ideal M, each power Mn is primary. It is interesting that Pn needs not be P-primary for a general prime P and its P-primary component P(n) = λ∗λ∗Pn is generally bigger and is called the n-th symbolic power of P. Proposition 4.6. The intersection of all prime ideals is the nilradical √ 0 = {r ∈ R nilpotent}. More generally P⊇I P = √ I. Proof. Clearly every nilpotent element lies in every prime. Thus let a ∈ R, denote D = {1, a, a2, . . .} the corresponding multiplicative subset and assume that a lies in every prime or, equivalently, every prime intersects D. Then the localization D−1R contains no prime and thus 1 = 0 in D−1R. By Proposition 3.5, this is equivalent to D containing zero, i.e. that some power of a is zero. The second point is obtained from the first by applying to the quotient ring R/I. 5. Chain complexes Definition 5.1. A sequence of R-modules A f −−→ B g −−→ C is said to be exact at B if im f = ker g. Similarly, one can define exactness of a longer sequence at any inner term. A sequence is exact, if it is exact at every inner term. Exercise 5.2. Characterize exactness of 0 → A → B, A → B → 0, 0 → A → B → 0, 0 → A → B → C, A → B → C → 0 and 0 → A → B → C → 0 (the last is referred to as a short exact sequence). In particular prove that any short exact sequence is isomorphic to an “extension” 0 → A → B ↠ B/A → 0. 15 5. Chain complexes In the condition im f = ker g, the inclusion ⊆ is equivalent to g ◦ f = 0, under which one may form the quotient ker g/ im f that measures the difference between the two submodules. One may thus express the exactness equivalently as g ◦ f = 0 and ker g/ im f = 0. These two parts are the main idea of the definition of a chain complex. Definition 5.3. A chain complex C is a diagram · · · → Cn+1 dn+1 −−−−→ Cn dn −−−→ Cn−1 → · · · in which dn ◦ dn+1 = 0 for all n. We often abbreviate all the maps to d and call them the boundary maps or differentials. The elements of Cn are called the n-chains. Denoting Bn = Bn(C) = im dn+1 ⊆ Cn the submodule of n-boundaries and Zn = Zn(C) = ker dn ⊆ Cn the submodule of n-cycles, the “square zero” condition d ◦ d = 0 is equivalent to Bn ⊆ Zn. The corresponding quotient module Hn = Hn(C) = Zn/Bn is called the n-th homology group, or just the n-th homology of C. As observed above, the corresponding sequence is exact at Cn iff Hn(C) = 0. If this happens for all n we say that the chain complex is acyclic or that the corresponding sequence is a long exact sequence. Example 5.4. Simplicial homology: Let K be a simplicial complex and choose a total ordering of its vertices (this can be avoided, see below). We write the n-simplices as ordered (n + 1)-tuples of its vertices, i.e. σ = [v0, . . . , vn] with v0 < · · · < vn. We define the operator di : Kn → Kn−1, di[v0, . . . , vn] = [v0, . . . , vi, . . . , vn]. The idea now is that the boundary of σ should be the collection d0σ, . . . , dnσ. This is algebraically achieved by considering Cn(K) = RKn the free module on the set of n-simplices and by defining d = (−1)idi. The n-th homology of the chain complex Cn(K) is the n-th simplicial homology of the simplicial complex K. The problem with ordering is solved when one defines Cn(K) = RKord n /∼, the free module on ordered n-simplices written again as [v0, . . . , vn], but this time without any restriction on the ordering of vertices of this n-simplex, modulo the relation [vσ(0), . . . , vσ(n)] = sign σ · [v0, . . . , vn]. Singular homology: Let X be a space and define similarly Cn(X) to be the free module on the set of all continuous maps ∆n → X, where ∆n denotes the standard n-simplex, i.e. the convex hull of any (n + 1)-tuple of affine independent points, e.g. E0, . . . , En ∈ Bn the standard point basis of the affine space x0 + · · · + xn = 1. The operators di are now given by restricting to the faces as above. The differential on C(X) is yet again d = (−1)idi. de Rham cohomology: ΩnM = {smooth n-forms on M}. Here the differential points in the opposite direction ΩnM → Ωn+1M. We will formalize this later as a cochain complex and de Rham cohomology is the cohomology of this cochain complex. 16 5. Chain complexes Definition 5.5. A chain map f : C → D between two chain complexes is a collection of homomorphisms fn for which the (ladder shaped) diagram · · · // Cn dn // fn  Cn−1 // fn−1  · · · · · · // Dn dn // Dn−1 // · · · commutes, i.e. df = fd. Every chain map induces maps Bn(C) → Bn(D) and Zn(C) → Zn(D) and thus also Hn(C) → Hn(D). We obtain Proposition 5.6. The n-th homology forms a functor Hn : Ch(ModR) → ModR. Definition 5.7. We say that a chain map f is a quasi-isomorphism if the induced map on homology is an isomorphism. As an example, a chain complex C is acyclic iff the unique map 0 → C is a quasiisomorphism iff the unique map C → 0 is a quasi-isomorphism. We will now present a special class of quasi-isomorphisms, analogous to homotopy equivalences in topology. First we need the corresponding algebraic notion of homotopy. Definition 5.8. Let f and g be two chain maps C → D. A chain homotopy from f to g is a collection of homomorphisms hn as in the (non-commutative!) diagram · · · // Cn+1 dn+1 //  Cn dn //  hn || Cn−1 // hn−1|| · · · · · · // Dn+1 dn+1 // Dn dn // Dn−1 // · · · such that dh + hd = g − f. We write h: f ∼ g or f ∼h g or simply f ∼ g if the homotopy is not important. A chain homotopy equivalence is a chain map f : C → D that admits a homotopy inverse, i.e. a chain map g: D → C together with homotopies gf ∼ 1, fg ∼ 1. Remark. Any continuous map between spaces induces a chain map between their singular chain complexes and any chain homotopy induces a chain homotopy (this is not completely straightforward). The simplicial situation is a bit more straightforward, but complicated enough to be explained at this point. We will give a nice interpretation (two in fact) of chain homotopy later. Proposition 5.9. Chain homotopic maps induce equal maps on homology. In particular, chain homotopy equivalences are quasi-isomorphisms. Proof. Let [z] ∈ Hn(C) be represented by a cycle z. Then g(x) − f(x) = dh(x) + h d(x) 0 so that [g(x)] = [f(x) + dh(x)] = [f(x)]. 17 5. Chain complexes Proposition 5.10. Chain homotopy equivalence is an equivalence relation that respects com- position. We may thus form the homotopy category of chain complexes and chain homotopy classes of maps where chain equivalences are exactly the isomorphisms. Proof. We prove transitivity: if f1 − f0 = dh + hd and f2 − f1 = dk + kd then f2 − f0 = d(h+k)+(h+k)d. Similarly if f1−f0 = dh+hd then gf1−gf0 = g(dh+hd) = d(gh)+(gh)d. We index the modules in our chain complexes by integers, but we will be using a lot chain complexes indexed by non-negative integers only. One can extend such a chain complex by zeros and thus think of it as a chain complex in the original sense. In doing so, the non-negatively graded chain complex · · · → C1 → C0 will also have the zero homology H0(C) = C0/B0(C) = coker(d1) since every 0-chain is a cycle. Another variation, briefly mentioned above with connection to de Rham cohomology is that of a cochain complex. Another situation where cochain complexes arise naturally is upon applying contravariant functors to chain complexes – the direction of homomorphisms changes. We will distingiush notationally by using upper indices. Definition 5.11. A cochain complex C is a diagram · · · → Cn−1 dn−1 −−−−→ Cn dn −−−→ Cn+1 → · · · in which dn ◦ dn−1 = 0 for all n. We get notions of cochains, cocycles, coboundaries and cohomology, cochain maps and cochain homotopy in an obvious way. Again, non-negatively graded cochain complexes will play an important role and they will look C0 → C1 → · · · so that the zeroth cohomology will be H0(C) = Z0(C)/0 = ker d0. Proposition 5.12. In a pullback square B g //  C  B′ g′ // C′ the induced map ker g → ker g′ is an iso. In addition, if g′ is epi then so is g. Proof. The first point follows from simple properties of pullbacks (tutorial). The second point is verified on elements and was done in the tutorial (more formally, it follows from ModR being abelian, but that I have to understand first). 18 5. Chain complexes Theorem 5.13 (snake lemma). Given a commutative diagram with exact rows A i // α  B p // β  C // γ  0 0 // A′ i′ // B′ p′ // C′ there exists a natural exact sequence ker α → ker β → ker γ δ −−→ coker α → coker β → coker γ. If, in addition the map i is injective (i.e. the top exact sequence can be prolonged to the left by zero to a short exact sequence), so is the map ker α → ker β and similarly for the surjectivity of the map B′ → C′. Proof. One first proves the version with short exact sequences in both rows. Since limits commute with limits, starting with the square B //  C  B′ // C′ and applying kernels first in the horizontal and then in the vertical direction yields the same result as applying them in the opposite order, i.e. ker α is indeed the kernel of the map ker β → ker γ and this proves exactness at ker α and ker β. By the dual argument, we are left to construct the “connecting homomorphism” δ and to prove exactness at its domain and codomain. We define δ(c) = (i′ )−1 βp−1 (c) where we need to verify that the preimage of βp−1(c) indeed exists. By exactness, this amounts to showing that 0 = p′βp−1(c) = γpp−1(c) = γ(c) and this holds since we assume c ∈ ker γ. Now the preimage p−1(c) is not unique and we have to show that the result does not depend on the choice. However, the choice is unique up to im i that is mapped by β to im(i′α) and further by (i′)−1 to im α that is zero in coker α. Clearly, if c = p(b) for some b ∈ ker β then the above prescription yields zero, so the composition ker β → ker γ δ −−→ coker α is indeed zero. Let now c ∈ ker γ be such that δ(c) = 0, i.e. (i′)−1βp−1(c) = α(a). Now p−1(c) + i(a) is still a preimage of c, and lies in ker β, by an easy inspection. Finally, if i is not mono, replace A by im i and apply the mono case. A α !! // // im i // // α′  B p // β  C // γ  0 0 // A′ i′ // B′ p′ // C′ The mono case also easily gets that the map ker α → ker α′ is epi and thus upon replacing ker α′ by ker α, the sequence remains exact everywhere except at ker α, as claimed. 19 5. Chain complexes Remark. One can construct from a chain A α −−→ B β −−→ C a diagram (coming from certain map of double complexes) 0 // A // α  A ⊕ B // βα⊕1  B // β  0 0 // B // C ⊕ B // C // 0 (the rows are not comprised of the inclusions and projections, they have to be twisted slightly), which gives the exact sequence relating kernels and cokernels of the maps α, βα and β. Proposition 5.14 (5-lemma). In a commutative diagram with exact rows A // α  B // β  C // γ  D // δ  E ε  A′ // B′ // C′ // D′ // E′ if α, β, δ, ε are iso, then so is γ. More precisely, α is only required to be epi and ε to be mono. Proof. Apply Lemma 5.20; denoting the image of the map B → C for simplicity by BC etc., we obtain short exact sequences 0 // BC // βγ  C // γ  CD // γδ  0 0 // B′C′ // C′ // C′D′ // 0 and the snake lemma gives that γ is mono provided that βγ and γδ are mono. The second is easier, just apply the snake lemma in 0 // CD // γδ  D // δ  DE // δε  0 0 // C′D′ // D′ // D′E′ // 0 to get γδ mono if δ is. The other condition is more complicated and involves the application of the snake lemma to 0 // AB // αβ  B // β  BC // βγ  0 0 // A′B′ // B′ // B′C′ // 0 to obtain βγ mono provided that β is mono and αβ is epi. Finally, αβ epi follows from α epi by the last application of snake lemma where one needs to prolong the sequence one step to the left, say by kernels of the maps A → B and A′ → B′. Altogether, γ is mono if β and δ are mono and α is epi. Dually, γ epi follows from β and δ epi and ε mono. The following is a converse to Proposition 5.12. I left it as an exercise, I think. 20 5. Chain complexes Proposition 5.15. In a commutative square B g //  C  B′ g′ // C′ if both g and g′ are epi and the induced map ker g → ker g′ is an iso then it is a pullback square. Exercise 5.16. This is about self-duality of homology. Show that there exists a factorization coker dn+1 // ker dn−1%% %% Cn+1 dn+1 // Cn dn // 99 99 Cn−1 dn−1 // $$ $$ Cn−2 im dn+1 ;; ;; im dn−1 and that the map on the top has kernel Hn(C) and cokernel Hn−1(C). The diagram is selfdual, so starting with a cochain complex, interpreting it as a chain complex in the opposite category and taking homology there yields exactly the cohomology of the original cochain complex. Theorem 5.17 (long exact sequence of homology). A short exact sequence 0 → A → B → C → 0 of chain complexes induces a natural long exact sequence of homology · · · → Hn+1(C) δ −−→ Hn(A) → Hn(B) → Hn(C) δ −−→ Hn−1(A) → · · · . Proof. Applying the previous exercise, we will consider the map coker dn+1 → ker dn−1 for the involved chain complexes and write them as Cn(C)/Bn(C) → Zn−1(C) etc. so that we obtain a diagram Cn(A)/Bn(A) //  Cn(B)/Bn(B) //  Cn(C)/Bn(C) //  0 0 // Zn−1(A) // Zn−1(B) // Zn−1(C) with exact rows (coker commutes with coker, similarly ker commutes with ker). Snake lemma gives a portion of the claimed long exact sequence, as required. Corollary 5.18. In a short exact sequence of chain complexes as above, A is acyclic iff B → C is a q-iso. Dually, C is acyclic iff A → B is a q-iso. Corollary 5.19. In a commutative diagram of chain complexes with exact rows 0 // A i // α  B p // β  C // γ  0 0 // A′ i′ // B′ p′ // C′ // 0 if two of α, β, γ are q-iso’s, so is the third. 21 5. Chain complexes Lemma 5.20. A long exact sequence · · · → Cn+1 → Cn → Cn−1 → · · · can be split into short exact sequences 0 → Bn → Cn → Bn−1 → 0. Conversely, any collection of short exact sequences as above can be spliced into a long exact sequence. Proof. Bn+1## ## Bn−1## ## · · · // Cn+1 // "" "" Cn // << << Cn−1 // ## ## · · · Bn >> >> Bn−2 In a general chain complex, one has to replace the short exact sequences by 0 → Zn → Cn → Bn−1 → 0 and add to these the short exact sequences 0 → Bn → Zn → Hn → 0 that define the homology as the quotient Zn/Bn. Again, one can splice such short exact sequences into a chain complex C with homology H. Definition 5.21. A resolution of a module A is a non-negatively graded chain complex C together with an “augmentation” map ε: C0 → A such that · · · → C1 → C0 ε −−→ A is an acyclic chain complex (the “augmented” chain complex). We say that C is a projective resolution if, in addition, C consists of projective modules. There is a nice “global” characterization of this, using the chain map ε: C → A[0] where A[0] denotes the chain complex whose only nonzero chains are in dimension zero and are A. Thus, the map is precisely · · · // C1 //  C0 ε  · · · // 0 // A Now the homology of the augmented chain complex agrees with the homology of C except in dimensions 0 and −1, where it is ker ε/B0(C) and coker ε. The first can be rewritten as ker(C0/B0(C) ε −−→ A) = ker(H0(C) ε −−→ A) while the second can be rewritten as the cokernel of the same map H0(C) ε −−→ A. Since this is the induced map on homology, we are finished with the equivalence. We will give a different, more conceptual proof later. 22 5. Chain complexes Definition 5.22. Dually, a (co)resolution is a cochain complex C together with a coaugmentation map A → C0 such that the coaugmented cochain complex A → C0 → C1 → · · · is acyclic. An injective (co)resolution has all objects Cn injective. Definition 5.23. A functor F is additive if its action on morphisms C(c′, c) → D(Fc′, Fc) is a homomorphism of groups. Any additive functor F has an extension to a functor between chain complexes since it preserves composition and zero. We denote this extension again by F. Example 5.24. Hom functors HomR(A, −), HomR(−, A) (the second contravariant though), tensor product functors − ⊗R A, A ⊗R −. We will now discuss basic properties of functors related to exactness or, equivalently, homology. It is easy to see that F0 = 0 from the characterization of 0 as an object where 1 = 0 (one may also view this as a nullary case of Lemma 5.29). We now make a couple of definitions: Definition 5.25. An additive functor F is said to be right exact if the image of an exact sequence A → B → C → 0 is an exact sequence FA → FB → FC → 0 Equivalently, F pereserves cokernels. (By Lemma 5.29, it preserves finite coproducts and this is thus equivalent to preservation of finite colimits.) A left exact functor is an additive functor that preserves kernels. A functor is exact if it preserves short exact sequences. Exercise 5.26. Show that a functor is exact iff it preserves all exact sequences iff it is left exact and right exact. In particular, an exact functor preserves acyclic chain complexes. A generalization of this is the following. Lemma 5.27. An exact functor F commutes with homology, i.e. Hn(FC) = FHn(C). In particular, F preserves q-iso’s. Proof. This is so since homology is defined using kernels and cokernels. Example 5.28. The tensor product functor − ⊗R A is right exact; it is exact iff A is flat. Similarly for the other tensor product functor. The hom functor HomR(A, −) is left exact; it is exact iff A is projective. Similarly for the other hom functor (here A should be injective for exactness), note however that this depends on writing the contravariant one as Modop R → Ab (and not as ModR → Abop ). Lemma 5.29. Additive functors preserve biproducts. Equivalently, additive functors preserve exactness of split short exact sequences. 23 6. Derived functors Proof. A biproduct is a diagram A i // C q // p oo B j oo satisfying i j · p q = 1, p q · i j = 1 0 0 1 . Since F preserves composition, addition, identities and zeros, the same is true for the image under F. Thus, any additive functor is exact on certain exact sequences. Over a field, any additive functor is exact. As we saw, this should mean that it preserves certain q-iso’s. Here is a precise claim. Lemma 5.30. Additive functors preserve chain homotopies. Proof. The proof is practically the same as for the previous lemma: A chain homotopy is given by some formulas and these are presereved by additive functors. Example 5.31. Consider a short exact sequence 0 → Z 2 −−→ Z → Z/2 → 0 and apply − ⊗ Z/2; this yields 0 → Z/2 0 −−→ Z/2 1 −−→ Z/2 → 0 that is clearly not exact. Thus, − ⊗ Z/2 is not exact and does not preserve q-iso’s (since C → 0 is a q-iso while FC → 0 is not). In the next sections, our main aim will be to measure the non-exactness of an additive functor. We will see that in the second short exact sequence the zero on the left can be replaced by a continuation – a long exact sequence of derived functors. 6. Derived functors Derived functors of F at A are defined by taking a projective resolution P → A[0] then applying F and taking homology, i.e. LnF(A) = Hn(FP). The main technical problem to solve is showing the independence of the choice of a projective resolution (and then obviously proving basice properties). Classically, one shows that between any two projective resolutions, there exists a chain map P // Q ~~ A[0] and that any two chain maps are chain homotopic (i.e. the map is unique up to chain homotopy). Applycation of F preserves this and taking homology makes the comparison map 24 6. Derived functors unique. There are, however, situation where projective resolutions do not exist, only some weaker version. In such situations, the existence of maps directly from P to Q cannot be expected. There is a weaker version of uniqueness (very much in the modern higher categorical sense), namely the category of (weakly) projective resolutions P → A[0], i.e. Ch(ModR)proj/A[0] is contractible (i.e. its classifying space is) from which we will only need that any two objects can be connected by a zig-zag of morphisms and any two such zig-zags can be connected by a zig-zag of zig-zags (this is just one dimensional triviality). The proof is not more difficult and one can find the classical approach in all books on homological algebra that I know of (I might later add a short summary). We will be working with a collection of objects P, called P-projective objects or Pprojectives, that is required to satisfy ˆ every object admits an epi from a P-projective and ˆ it is closed under kernels of epis, i.e. for a ses 0 → A → B → C → 0 with P-projective B and C, the same is true of A. Typically, P is also assumed to be closed under finite biproducts, but we will not need this assumption. We say that P is adopted to F if in addition to the above assumptions F is exact on ses’s of P-projectives. By splicing, it is then exact on bounded below les’s of P-projectives as well. We can then construct a P-projective resolution of any object A in the following way. Construct inductively ses’s 0 → Kn → Pn → Kn−1 → 0, starting from K−2 = 0 and P−1 = A, so that K−1 = A as well, in such a way that Pn ∈ P for each n ≥ 0 (it exists by the first point). Then splice these ses’s to get a les · · · → P1 → P0 ε −−→ A → 0. We will denote this P-projective resolution ε: P → A[0] (with the above augmented chain complex the mapping cone of this augmentation map ε). We will now show (only partially1, as required for our exposition) that the category of P-projective resolutions of a fixed A is weakly contractible. First we present a relative version of the above construction: Let f : A → B be a map and ε: Q → B[0] a resolution, not necessarily P-projective. Then we 1 In general, let Pi → A[0] be an I-diagram of P-projective resolutions and replace it by a fibrant diagram Pf i in the model structure with pointwise cofibrations and weak equivalences (here the fibrations of chain complexes are not necessarily surjective, but acycclic fibrations are and that should be enough); take the limit of this diagram and pullback P //  lim Pf i ∼  A[0] // lim A[0] This then forms a resolution P → A[0] and one can then replace it by a P-resolution with the map P → lim Pf i corresponding to a cone P ⇒ Pf i ⇐ Pi that together with the natural transformation shows that the category of P-resolutions is weakly contractible – thus, one should in fact assume that Pf i is itself a P-resolution, i.e. that P should be closed under products (finite should be sufficient as we can assume I to be finite in the sense that NI is (locally) finite). 25 6. Derived functors can complete the following diagram P ε // φ  A[0] f[0]  Q ε // B[0] in such a way that if f is epi then so is φ (I guess that φ is epi from dimension 1 onwards regardless of f!!! In fact, take the pullback of ε along f[0], which is an epi q-iso – see below – and thus we only need to consider the case f = 1). One proceeds exactly as above but using a pullback square 0 // Kn //  Pn //  Kn−1 // 0 0 // Ln // Pn //  Kn−1 //  0 0 // Ln // Qn // Ln−1 // 0 (this requires observation that kernels and pullbacks commute and also that a pullback of epi is epi). This applies in particular to the following situation: given two resolutions P′ → A[0] and P′′ → A[0] we can form their pullback P //  P′′  P′ // A[0] and since epi q-iso’s are closed under pullbacks (the epi part we know, then one takes kernels, which agree and are acyclic) we see that all the maps in the square are such. Replacing P by a P-projective resolution P as above, we thus get a span of epis between P-projective resolutions P′ ← P → P′′. We will need a further level of dimension: Given two spans P and P between P′ and P′′ as above (i.e. two epis P → P and P → P) we may form their pullback over P and get P //  P  P // P //  P′′  P′ // A[0] 26 6. Derived functors and finally resolve P by a P-projective P to get a span between spans: P′ P >> P OO  oo // P `` ~~ P′′ Now we finally utilize the above to the definition and properties of derived functors. We assume that P is adopted to F. Given an epi φ: P → P′ between P-projective resolutions of A, the kernel ker φ is then an acyclic chain complex of P-projectives and as such remains acyclic upon applying F. Thus, the map FP → FP′ is still an (epi) q-iso. Applying homology then yields an iso H∗FP → H∗FP′. We thus get a diagram of isomorphisms H∗FP′ H∗FP ∼= 99 ∼= %% H∗FP ∼= OO ∼=  ∼=oo ∼= // H∗FP ∼= ee ∼= yy H∗FP′′ so that we get a well defined (i.e. unique) comparison isomorphism H∗FP′ ∼= H∗FP′′. We may thus define L∗F(A) = H∗FP where P → A[0] is any P-projective resolution and we get that any two possible such definitions are isomorphic in a canonical way, allowing us to talk e.g. about individual elements of L∗F(A). Now given a ses 0 → A → B → C → 0 we take an arbitrary P-projective resolution R → C[0], then construct a P-projective resolution Q → B[0] together with an epi Q → R and finally take kernels to get 0 // P //  Q // ≃  R // ≃  0 0 // A[0] // B[0] // C[0] // 0 By the properties of P we conclude that P consists of P-projectives and, by 5-lemma, the left vertical map is a q-iso, making it a P-projective resolution. Thus, upon applying H∗F to the top row we get a les consisting of the left derived functors L∗F(A), L∗F(B), L∗F(C). Finally, if A is itself P-projective then the augmented chain complex P → A[0] remains exact upon applying F and, thus, L0F(A) = FA and LnF(A) = 0 for n > 0. In particular, we get that P is contained in the collection P = {A | LnF(A) = 0 for all n > 0}. Since this class satisfies the properties, it is thus the maximal such class for a given functor F. 27 7. Balancing Tor and Ext Remark. We will now show independence of LnF of the class P. Thus, let Q be another class and consider a Q-resolution Q → A[0], further a P-resolution P → A[0] etc. as in P′ // ∼  A[0] Q′ // ∼  A[0] P // ∼  A[0] Q // A[0] Since both composites P′ → P and Q′ → Q are epi q-iso’s between complexes of P-projectives or Q-projectives, they remain q-iso’s upon applying F so that the middle map FQ′ → FP is a q-iso by the 2-out-of-6 property, proving LQ ∗ F ∼= LP ∗ F. The uniqueness of this isomorphism follows by comparing to the maximal class above P ⊆ P = Q ⊇ Q that is independent by the mere existence of an isomorphism showing that the above comparison maps can be thought of as comparison maps for the class P = Q and are thus unique. Remark. One should also prove that each LnF is additive and I thought that this would require the class P to be closed under finite biproducts, and it seems so. Proposition 6.1. A right exact functor F is exact iff ∀n > 0: LnF = 0 iff L1F = 0. 7. Balancing Tor and Ext We define TorR n (A, B) = Ln(−⊗RB)(A). There is a second candidate, namely Ln(A⊗R−)(B). One we show that these are the same, we will know that these can be defined using flat resolutions (since flat modules are acyclic). A similar situation arises for Extn R(A, B) = Rn(HomR(A, −))(B) and the symmetric version obtained from the contravariant hom functor. We will show that in both cases, the two derived functors are canonically isomorphic. We will concentrate on the tensor products since these are both covariant and thus easier. The two derived functors are obtained as homology of chain complexes P ⊗R B and A ⊗R Q where P → A[0] and Q → B[0] are projective resolutions. It thus seems more than logical to compare these using a span P ⊗R B ← P ⊗R Q → A ⊗ Q. The question is what this P ⊗R Q should be. We can draw a diagram where we write only ⊗ for simplicity ...  ...  · · · Pp−1 ⊗ Qq oo 1⊗d  Pp ⊗ Qq d⊗1oo 1⊗d  · · ·oo · · · Pp−1 ⊗ Qq−1 oo  Pp ⊗ Qq−1 d⊗1oo  · · ·oo ... ... 28 7. Balancing Tor and Ext where dh = d ⊗ 1 and dv = 1 ⊗ d will not be exactly one’s first guess, resulting in the squares anti-commuting rather than commuting, i.e. dhdv = −dvdh. We will now make this kind of structure formal. Definition 7.1. A double (chain) complex is a diagram of modules Dp,q and homomorphisms dh : Dp,q → Dp−1,q, dv : Dp,q → Dp,q−1 that satisfy (dh + dv)2 = 0, i.e. dh dh = 0, dv dv = 0, dh dv + dv dh = 0. This way of presenting the axioms suggests the following definition. Definition 7.2. For a double complex D, the total complex Tot+ D is a chain complex with (Tot+ D)n = p+q=n Dp,q and with differential d = dh + dv. There is another version of the total complex Tot× D where the sum is replaced by the product. The two versions are useful for different applications, the first is related to left derived functors and the second for right derived functors. Since we concentrate on the tensor product case, we will stick to the sum version and will denote it simply by Tot D. In order to make the tensor product into an example, we have to introduce signs. We define (f ⊗ g)(x ⊗ y) = (−1)|g|·|x| fx ⊗ gy where |g| = |gx| − |x| is the degree of g (we will treat this more formally later). Thus the identity has degree |1| = 0, while the differential has degree |d| = −1. This gives as particular cases dh (x ⊗ y) = (d ⊗ 1)(x ⊗ y) = dx ⊗ y, dv (x ⊗ y) = (1 ⊗ d)(x ⊗ y) = (−1)|x| x ⊗ dy and the squares clearly anti-commute with this notation (one can also prove this formally by first verifying (f⊗g)(h⊗k) = (−1)|g|·|h|fh⊗gk and then using this to compute (d⊗1+1⊗d)2 = 0). We speak of Koszul sign convention. Later, when we will deal with chain complexes, we will be using P ⊗Q to denote the total complex Tot P ⊗ Q of this double complex. We will now introduce two very useful special cases of the tensor product construction. The motivation comes from topology, where C(X × Y ) = C(X) ⊗ C(Y ) (also for Koszul sign convention), at least when one deals with cellular complexes where products of cells are cells. In this way one obtains the cylinder of C by tensoring with the chain complex of the interval: cyl C = Tot(cyl R[0] ⊗ C) since R[0] is interpreted as a point and then the cylinder on the point is the interval; it remains to specify this interval: cyl R[0] = · · · → 0 → R d −−→ R ⊕ R Denoting the 1-dimensional generator by e (edge) and the 0-dimensional generators by v−, v+ (initial and terminal vertices), we define de = v+ − v−. 29 7. Balancing Tor and Ext Exercise 7.3. Prove that chain maps cyl C → D are in bijection with triples (f, g, h) where f and g are chain maps C → D and h is a chain homotopy f ∼ g. In topology, one can recover the two involved maps from a homotopy (as restrictions to the two ends of the cylinder), while in homological algebra, this is not the case – the best one can get is the difference g − f = dh + hd. Another example that we will need is the cone. We define similarly cone C = Tot(cone R[0] ⊗ C) where again the chain complex cone R[0] = · · · → 0 → R d −−→ R has differential de = v, i.e. d = 1. We will now draw a picture of the double complex cone R[0] ⊗ C and a simpler realization thereof: ...  ...  ...  ...  Rv ⊗ C1 1⊗d  Re ⊗ C1 d⊗1oo 1⊗d  C1 d  C1 1oo −d  Rv ⊗ C0  Re ⊗ C0 d⊗1 oo  C0  C01 oo  ... ... ... ... where the minus sign comes from (1 ⊗ d)(e ⊗ x) = (−1)|e|e ⊗ dx = e ⊗ (−dx), since |e| = 1. Clearly the zeroth column form a subcomplex (since R[0] ⊆ cone R[0] and then one applies the tensor product), so C → cone C. The quotient is the first column, i.e. the chain complex C[1] – called the suspension of C – is just C shifted by one dimension up and with opposite differential (again, the quotient of R[0] → cone R[0] is R[1] and C[1] = Tot R[1] ⊗ C). What comes now is a concrete description of the pushout C // f  cone C  D // cone f (with horizontal cokernels C[1], see below) that results from replacing the subcomplex C ⊆ 30 7. Balancing Tor and Ext cone C by D via f. It is the total complex of the double complex ...  ...  D1 d  C1 f oo −d  D0  C0 f oo  ... ... Similarly to the case of cone C, we get a subcomplex and a quotient, forming a short exact sequence 0 → D → cone f → C[1] → 0 Exercise 7.4. Verify that the connecting homomorphism in the homology long exact sequence is the map Hn+1(C[1]) = Hn(C) → Hn(D) induced by f. Conclude that f is a q-iso iff cone f is acyclic. Apply to the augmentation map. Proposition 7.5. Let D be a first quadrant double complex, i.e. such that Dp,q = 0 whenever p < 0 or q < 0. If D has exact columns, i.e. if for each p the chain complex (Dp,•, dv) is acyclic, then Tot D is acyclic. Dually, the same conclusion holds for first quadrant double complexes with exact rows. Remark. This version works for right halfplane double complexes (or upper halfplane complexes in the second case), but the Tot×-version requires this stronger assumption, I think. Proof. Denote D(0) = D. As above, the zeroth column forms a subcomplex D0,• with quotient D(1), obtained by removing the zeroth column. Continuing this way, we obtain short exact sequences 0 → Dp,• → Tot D(p) → Tot D(p+1) → 0 which shows that the natural projection maps Tot D = Tot D(0) → Tot D(1) → · · · are all q-iso’s. Since Tot D(n+1) is concentrated in dimensions ≥ n + 1, it has zero Hn and thus the same is true for Tot D. Now consider the double complex obtained as a tensor product of the augmented chain complex P and the chain complex Q, i.e. ...  ...  ...  A ⊗ Q1  P0 ⊗ Q1 oo  P1 ⊗ Q1 oo  · · ·oo A ⊗ Q0 P0 ⊗ Q0 oo P1 ⊗ Q0 oo · · ·oo 31 7. Balancing Tor and Ext This has exact rows since these are obtained by tensoring the augmented chain complex P with a projective Qq. Thus, the total complex is acyclic. Since it is (up to suspension) the cone of the map ε ⊗ 1: P ⊗ Q → A ⊗ Q, this map is a q-iso. Symetrically, the map 1 ⊗ ε: P ⊗ Q → P ⊗ B is also a q-iso and we obtain the following theorem: Theorem 7.6 (balancing of Tor). There exists a natural isomorphism Ln(− ⊗ B)(A) ∼= Ln(A ⊗ −)(B) between the derived functors of the tensor product functor. In fact, one can see easily that for a right exact bifunctor F, we only need that F(P, −) should be exact for any projective P as well as F(−, Q) for any projective Q and we obtain LnF(A, −)(B) ∼= LnF(−, B)(A) We will now shortly comment on the derived functors of the hom functor. The covariant one is easier, so we start with this: Rn (Hom(A, −))(B) = Hn (Hom(A, I)) where B[0] → I is an injective resolution. The situation of the other hom functor is exactly the same when interpreted in the opposite category; translating to the ordinary category of modules, we get Rn (Hom(−, B))(A) = Hn (Hom(P, B)) where P → A[0] is a projective resolution. Again, we can form a double cochain complex Hom(P, I) and its total complex Tot× Hom(P, I) that admits a cospan Hom(P, B) → Tot× Hom(P, I) ← Hom(A, I) with both maps q-iso’s by an analogous argument. Theorem 7.7 (balancing of Ext). There exists a natural isomorphism Rn (Hom(A, −)(B) ∼= Rn (Hom(−, B)(A) between the derived functors of the hom functor. We will now study hom complexes from a different perspective – related, but it may be easier to forget about what we did up to now. So let C and D be chain complexes and construct a chain complex Hom(C, D) with the aim of giving the category of chain complexes the closed symmetric monoidal structure. Symmetry is perhaps worth mentioning first, since it is given by a (not so much now) surprising isomorphism B ⊗ C ∼= −−→ C ⊗ B, x ⊗ y → (−1)|x|·|y| y ⊗ x. Now our goal is the adjointness B ⊗ C → D B → Hom(C, D) 32 8. Ext and extensions We will first study this on the level of the underlying graded modules (i.e. ignore the differentials). This becomes n+k=ℓ Bn ⊗ Ck → Dℓ Bn → k Hom(Ck, Dn+k) so we want to endow the graded module Hom(C, D)n = k Hom(Ck, Dn+k) with a differential so that the map at the top is a chain map iff the map at the bottom is. The differential will be derived from the requirement that the counit is a chain map, by observing that the counit is (as usual) the evaluation map ev: Hom(C, D) ⊗ C → D, f ⊗ c → fc. We will denote the differential on the hom complex by D and we thus require d ev = ev(D ⊗ 1 + 1 ⊗ d), that by applying to f ⊗ c amounts to d(fc) = (Df)c + (−1)|f| f(dc). We can thus write Df = df − (−1)|f|fd = [d, f] (the graded commutator). It is rather straightforward that this indeed makes Ch(ModR) into a closed symmetric monoidal category. The 0-chains of Hom(C, D) are by the construction not-necessarily-chain maps f : C → D. The 0-cycles are those that satisfy Df = 0, i.e. df − fd = 0 and these are exactly the chain maps.2 (Chain maps of degree n are defined as n-cycles, i.e. they are required to satisfy df = (−1)nfd.) For two chain maps f, g, i.e. 0-cycles, we have [f] = [g] in H0(Hom(C, D)) iff g − f ∈ B0(Hom(C, D)) iff there exists h ∈ Hom(C, D)1 with Dh = g − f; this means dh + hd = g − f and this h is a chain homotopy from f to g. As a result H0(Hom(C, D)) = [C, D] the group of chain homotopy classes of chain maps. This is another explanation of chain homotopy (we had definition, then as a map from the cylinder, now as a homology relation in the hom complex). 8. Ext and extensions We will now apply the derived functor Ext1 to study extensions of modules, i.e. short exact sequences ξ : 0 → B → X → A → 0 We start with a simple question: When does the sequence split? By applying Hom(A, −) we obtain an exact sequence 0 → Hom(A, B) → Hom(A, X) → Hom(A, A) ∂ −−→ Ext1 (A, B) → Ext1 (A, X) → · · · 2 This corresponds to the fact that the unit is R[0] and maps out of R[0] are exactly the 0-cycles. 33 8. Ext and extensions Clearly, ξ admits a splitting iff 1 ∈ Hom(A, A) lies in the image (more precisely, any preimage is such a splitting A → X) iff ∂(1) = 0. We define θ(ξ) = ∂(1) ∈ Ext1 (A, B) and we just observed that this is the (unique) obstruction to the existence of a splitting. Lemma 8.1. ξ splits iff θ(ξ) = 0. In particular, it splits when Ext1 (A, B) = 0. Naturality of the class θ with respect to maps of ses’s should be rather clear: we need ξ : 0 // B // X //  A // 0 ζ : 0 // B // Y // A // 0 (the map X → Y is then necessarily an iso by 5-lemma) so that the portions of long exact sequences of derive functors Hom(A, A) ∂ // 1  Ext1 (A, B) 1  Hom(A, A) ∂ // Ext1 (A, B) have both vertical maps identities – they are induced by maps in the transformation above so we require these to be identities. We will then say that the extensions ξ and ζ are isomorphic and we see that then θ(ξ) = θ(ζ). We have just defined, for fixed A and B, a map θ: {extensions 0 → B → X → A → 0}/iso → Ext1 (A, B). Theorem 8.2. The above map θ is bijective. Proof. We first produce a map in the opposite direction. Let B → I be an embedding of B into an injective module and let C be the cokernel so that we have a ses ζ : 0 → B i −→ I p −−→ C → 0. The les of Ext(A, −) then gives · · · → Hom(A, I) p∗ −→ Hom(A, C) → Ext1 (A, B) → Ext1 (A, I) 0 → · · · (since I is injective). Thus, for any α ∈ Ext1 (A, B) there exists a preimage f : A → C and any other preimage is of the form f + pg for g: A → I. We form a pullback of the ses above along f and obtain ξf : 0 // B // Xf //  A // f  0 ζ : 0 // B i // I p // C // 0 34 8. Ext and extensions Concretely Xf = {(x, a) | p(x) = f(a)} and we thus have an isomorphism Xf ∼= −−→ Xf+pg, (x, a) → (x + g(a), a) that respects the inclusion of B and projection onto A so this is in fact an isomorphism of extensions ξf ∼= ξf+pg. This finishes the construction of the inverse mapping, we need to verify that these are indeed inverse to each other. We thus study the obstruction θ(ξf ) of the obstruction above. Again, the transformation of ses’s gives a transformation between the sequences of derived functors Hom(A, A) ∂ // f∗  Ext1 (A, B) 1  Hom(A, C) ∂ // Ext1 (A, B) and this means precisely θ(ξf ) = ∂(1) = ∂f∗(1) = ∂(f) = α by construction (f was chosen as a preimage of α). It remains to show that the constructed inverse mapping is surjective, i.e. that every extension is obtained as a pullback from ζ: ξ : 0 // B // X //  A //  0 ζ : 0 // B i // I p // C // 0 Start by extending the map i into the injective I along the inclusion B → X, as suggested in the diagram. We then complete the diagram by a map f : A → C that is just the induced maps on cokernels. Since both X → A and p are epi and the induced map on kernels is an iso, Proposition 5.15 yields that the square is indeed a pullback. Example 8.3. Since Ext1 (Z/m, Z/n) = 0 if gcd(m, n) = 1, every ses 0 → Z/n → X → Z/m → 0 splits. We will prove later a more general result for non-commutative groups. Since Ext1 as a derived functor is additive in both variables, we get Ext1 (B, A) = 0 for any finite abelian groups of coprime orders (split into a direct sum and apply the above). Interestingly, since Ext1 (A, B) is an abelian group, the same must be true for the set of extensions up to isomorphism. It is instructive to transport the addition along the above isomorphism. The result looks as follows. Take two extensions and consider their biproduct 0 → B ⊕ B → X ⊕ Y → A ⊕ A → 0. Now take the pullback as in the above proof along the diagonal A → A ⊕ A to obtain an extension of A by B ⊕ B. Perform the dual construction, i.e. form the pushout along the codiagonal B ⊕ B → B (i.e. the addition) and finally obtain an extension of A by B; this is the sum of the original extensions. 35 9. Homological dimension Remark. The higher groups Extn (A, B) are in bijection with classes of longer extensions 0 → B → Xn → · · · → X1 → A → 0 modulo an equivalence relation generated by not necessarily invertible transformations 0 // B // Xn //  · · · // X1 //  A // 0 0 // B // Yn // · · · // Y1 // A // 0 There is a way of explaining this in a more natural way as follows. Consider the middle part as a chain complex X concentrated in dimension 1 through n and rewrite the exact sequence as 0 → B[n] → X → A[1] → 0, a ses of chain complexes. The natural transformation above becomes 0 // B[n] // X //  A[1] // 0 0 // B[n] // Y // A[1] // 0 and the middle map is a q-iso by 5-lemma. To make this comparison complete, one should show that chain complexes that are not concentrated in dimensions 1 through n can be truncated to the latter (easy). 9. Homological dimension We have just shown that Ext1 (−, −) = 0 iff every ses splits and we know that vanishing of Ext1 (A, −) is equivalent to A being projective so Ext1 (−, −) = 0 is also equivalent to every module being projective and dually also to every module being injective. We will now study higher dimensional analogues of such statements. Definition 9.1. A projective dimension of a module A, denoted pd(A), is defined to be the length of the shortest projective resolution of A, i.e. pd(A) ≤ n iff A admits a projective resolution · · · // 0 // Pn // · · · // P0  A There are similar notions of a flat dimension and injective dimension (the second using injective coresolutions): A  I0 // · · · // In // 0 // · · · Lemma 9.2. TFAE 36 9. Homological dimension 1. pd(A) ≤ n, 2. in any exact sequence 0 → Mn → Pn−1 → · · · → P0 → A → 0 with Pi projective, also Mn is projective, 3. Extn+1 (A, −) = 0. Proof. The implications 2 ⇒ 1 ⇒ 3 are trivial. Thus, let Extn+1 (A, −) = 0 and consider an exact sequence 0 → Mn → Pn−1 → · · · → P0 → A → 0. Denoting M0 = A, we split it into ses’s 0 → Mk+1 → Pk → Mk → 0. Applying Ext(−, B) yields, by projectivity of Pk the following isomorphisms for i ≥ 1: Exti+1 (Pk, B) 0 ← Exti+1 (Mk, B) ∼= ←−− Exti (Mk+1, B) ← Exti (Pk, B) 0 We thus obtain Ext1 (Mn, B) = · · · = Extn (M1, B) = Extn+1 (M0, B) = 0 and since this holds for any B, the module Mn is projective. A dual statement then shows that the injective dimension id(B) ≤ n is also equivalent to Extn+1 (−, B) = 0. Together, these results yield. Corollary 9.3. sup{pd(A) | A ∈ ModR} = inf{n | Extn+1 = 0} = sup{id(A) | A ∈ ModR}. Definition 9.4. The number from the previous corollary is called the global dimension of R and denoted gl. dim(R). Similarly, one can prove that the supremum of flat dimensions of modules does not depend on the side and equals the smallest n for which Torn+1 = 0, called Tor. dim(R). Example 9.5. A ring R has global dimension 0 iff Ext1 = 0 iff every module is projective iff every module is injective. A ring R has global dimension 1 iff Ext2 = 0 iff in every ses 0 → M → P → A → 0, the module M is projective; since A could be arbitrary, e.g. the cokernel of an arbitrary inclusion M → P, this is equivalent to a submodule of a projective module being projective. Dually, this is equivalent to a quotient of an injective module being injective. Any PID has global dimension 1: It can be proved by induction on n that a submodule of Rn is free of rank ≤ n, starting from n = 1 where this is just the definition of a PID. Theorem 9.6 (Hilbert on szyzygies). gl. dim k[x1, . . . , xn] = n. More generally, if gl. dim R = d then gl. dim R[x] = d + 1. 37 10. Group cohomology The full strength of the Hilbert theorem on szyzygies gives part 2 of Lemma 9.2 with projective replaced by free. Theorem 9.7 (K¨unneth). Assume that Tor. dim R ≤ 1, i.e. that every submodule of a flat module is flat. Let C be a chain complex of flat modules and A and module. Then there exists a natural ses (unnaturally split) 0 → Hn(C) ⊗ A → Hn(C ⊗ A) → Tor1(Hn−1(C), A) → 0. Proof. Apply Tor(−, A) to the ses 0 → Zn → Cn → Bn−1 → 0 of flat modules to obtain a ses of chain complexes 0 → Z ⊗ A → C ⊗ A → B[1] ⊗ A → 0 where the outer chain complexes are endowed with zero differential. Now apply the les of homology: · · · → (B[1] ⊗ A)n+1 ∂n −−−→ (Z ⊗ A)n → Hn(C ⊗ A) → (B[1] ⊗ A)n ∂n−1 −−−−→ (Z ⊗ A)n−1 → · · · . The connecting homomorphism ∂n is the canonical map i ⊗ 1: Bn ⊗ A → Zn ⊗ A that fits into a les of Tor(−, A) applied to 0 → Bn → Zn → Hn → 0, yielding (recalling that Zn must be flat) 0 → Tor(Hn, A) → Bn ⊗ A ∂n −−−→ Zn ⊗ A → Hn ⊗ A → 0. In other words, coker ∂n = Hn ⊗ A and ker ∂n−1 = Tor(Hn−1, A). Thus, one may replace the les above by a ses with Hn(C ⊗ A) in the middle, surrounded by the cokernel and the kernel. 10. Group cohomology This is a particular derived functor for modules over the group ring ZG for a group G. It is a free abelian group on the set G, i.e. its elements are formal Z-linear combinations of elements of the group G, say ag · g with only finitely many nonzero coefficients ag ∈ Z. The multiplication is extended Z-linearly from the multiplication in G, i.e. ( ah · h) · ( bk · k) = (ahbk) · hk = ( hk=g ahbk) · g. More abstractly, the free abelian group functor turns finite products into finite tensor products, Z(X × Y ) ∼= ZX ⊗ ZY , i.e. it is strongly monoidal. Thus, the multiplication in G induces ZG ⊗ ZG ∼= Z(G × G) −→ ZG 38 10. Group cohomology and similarly for the unit (which is then just the element 1 ∈ G interpreted as an element of ZG). A ZG-module is then equivalently an abelian group M together with an action of G via homomorphisms of groups, i.e. a · (x + y) = a · x + a · y. This is easily seen to be so by interpreting the module structure as a ring homomorphism ZG → EndZ(M) and by the freeness of ZG, this is induced uniquely by a group homomorphism G → AutZ(M). Another point of view is that this is a functor G → Ab hitting M (or in the first interpretation an Ab-enriched functor ZG → Ab). Example 10.1. The symmetry group Sn acts on V ⊗n by permuting the vectors in the tensor product. An important construction is that of the invariants (V ⊗n)Sn , i.e. the submodule of tensors that are invariant under the action, i.e. such that t · σ = t (since it is naturally a right action, i.e. a right ZG-module). Another related construction are the coinvariants (V ⊗n)Sn , i.e. the quotient by the congruence generated by t · σ ∼ t. When char k = 0, these are two equivalent definitions of the n-th symmetric power SnV . Definition 10.2. The invariants of a ZG-module M is the submodule MG = {x ∈ M | ∀a ∈ G: a · x = x}. The coinvariants of a ZG-module M is the quotient module MG = M/(a · x ∼ x | a ∈ G, x ∈ M). The first is the limit of the diagram G M (( the second is the colimit of the same diagram. There is another intrepretation of the same, using the trivial ZG-module Z. In general any abelian group admits a trivial action where a · x = x for any a ∈ G. Lemma 10.3. MG = HomZG(Z, M) and MG = M ⊗ZG Z (for a right ZG-module M). Proof. The point is that Z = (ZG)G, since the congruence identifies exactly the generators of ZG. Thus, f : Z → M f : (ZG)/(a · x ∼ x) → M f : ZG → M such that f(a · x) = f(x) m ∈ M such that a · m = m m ∈ MG and similarly M ⊗ZG Z ∼= M ⊗ZG ZG/(a·x ∼ x) ∼= (M ⊗ZG ZG)/(m⊗a·x ∼ m⊗x) ∼= M/(a·m ∼ m) = MG. Perhaps, it is better to relate it to HomZ and ⊗Z. Definition 10.4. The n-th group homology with coefficients in a ZG-module M is Hn(G; M) = Ln(−)G(M) = TorZG n (M, Z) or TorZG n (Z, M). The n-th group cohomology with coefficients in a ZG-module M is Hn (G; M) = Rn (−)G (M) = Extn ZG(Z, M). 39 10. Group cohomology We will study these via a projective resolution of Z ∈ ModZG. Example 10.5. Denote by Ck the cyclic group of order k written multiplicatively (i.e. it is Z/k but that is usually written additively), with elements 1, t, · · · , tk−1 and with tk = 1. A projective resolution was constructed in the tutorial, where the norm N = tk−1 + · · · + 1 denotes the sum of all the elements of the group · · · t−1 // ZCk N // ZCk t−1 // ZCk ev1  Z Now compute the homology of Ck with coefficients Z, i.e. apply − ⊗ZCk Z · · · 0 // Z k // Z 0 // Z and take homology to obtain Hn(Ck; Z) =    Z n = 0 Z/k n = 1, 3, 5, . . . 0 n = 2, 4, 6, . . . Example 10.6. Denote by C∞ the infinite cyclic group with elements powers tk of the generator t. The group ring ZC∞ is then the ring of Laurent polynomials. A projective resolution was constructed in the tutorial · · · // 0 // ZC∞ t−1 // ZC∞ ev1  Z Now compute the homology of C∞ with coefficients Z, i.e. apply − ⊗ZC∞ Z · · · // 0 // Z 0 // Z and take homology to obtain Hn(C∞; Z) = Z n = 0, 1 0 n = 2, 3, 4, . . . Remark. It can be shown that Hn(G; Z) = Hn(BG; Z) equals the singular homology of the classifying space BG = K(G, 1). Thus, despite G = Ck being finite, the classifying space BCk is infinite dimensional (and also ZCk has infinite global dimension). On the other hand BC∞ ≃ S1 is homotopy equivalent to a circle. We will now construct a general projective resolution of Z ∈ ModZG, the so called bar resolution. It has two versions – reduced and unreduced. We start with the second. 40 10. Group cohomology Definition 10.7. The unreduced bar resolution is the chain complex Bu with chains Bu n = ZG(G × · · · × G) = Z(G × (G × · · · × G)) where we denote the ZG-generators as [g1⊗· · ·⊗gn] and thus the Z-generators as g[g1⊗· · ·⊗gn]. The differential in this complex is d = (−1)idi for ZG-linear operators d0[g1 ⊗ · · · ⊗ gn] = g1 · [g2 ⊗ · · · ⊗ gn] di[g1 ⊗ · · · ⊗ gn] = [g1 ⊗ · · · ⊗ gigi+1 ⊗ · · · ⊗ gn] dn[g1 ⊗ · · · ⊗ gn] = [g1 ⊗ · · · ⊗ gn−1] and with augmentation ε: Bu 0 → Z, ε[] = 1. The (unreduced) bar resolution is the quotient B of Bu by the subcomplex spanned by [g1 ⊗ · · · ⊗ 1 ⊗ · · · ⊗ gn], where 1 denotes the unit of the group G. The classes are denoted [g1 | · · · | gn] and it is thus understood that thys symbol is zero when some gi = 1. In the formula for dn, one could imagine the rhs multiplied from the right by gn to get a more symmetrical version that will also be correct if we equip Bu with trivial right ZG-module structure, see Hochschild (co)homology. Theorem 10.8. Both Bu and B are free resolutions of Z ∈ ModZG. Proof. The proof that Bu → Z[0] is indeed an augmented chain complex was done in the tutorial. We will now show that the generators that contain 1 somewhere span a subcomplex. In the expression for d[g1 ⊗ · · · ⊗ 1 ⊗ · · · ⊗ gn] with 1 at position i, all terms contain this very same 1 except for the contributions di−1 and di that give (−1)i−1 · [g1 ⊗ · · · ⊗ gi−11 ⊗ · · · ⊗ gn] + (−1)i · [g1 ⊗ · · · ⊗ 1gi ⊗ · · · ⊗ gn] = 0 (the same generator with opposite signs). We will now show that Bu → Z[0] is a q-iso in Ch(ModZG) or equivalently in Ch(Ab), since the homology is computed the same way in ModZG and Ab. We will prove this by showing that the augmented chain complex is chain homotopy equivalent to the zero complex in Ch(Ab), i.e. that it admits a contraction h: 0 ∼ 1, dh + hd = 1. Since the chain homotopy will only be Z-linear, we define it on the Z-generators by setting h(g · [g1 ⊗ · · · ⊗ gn]) = [g ⊗ g1 ⊗ · · · ⊗ gn]. Easily di+1h = hdi and, thus, all terms in dh + hd cancel out with the exception of d0h = 1. We need to treat separately the cases involving the augmentation (dh + hε)(g · []) = d[g] + h1 = g · [] − [] + h1 so that we need to set h1 = []. Finally (εh + h0)1 = ε[] = 1. The same formula works for the reduced version. 41 10. Group cohomology Now we study examples. Obviously H0(G; Z) = ZG = Z and this corresponds to H0(BG; Z) = Z since BG is always connected. We proceed to H1 so we write out explicitly the lower dimensions of the unreduced bar resolution Bu = · · · → ZG{[g ⊗ h]} → ZG{[g]} → ZG{[]}. The coinvariants (−)G = Z ⊗ZG − then replace the free ZG-modules by the corresponding free Z-modules, i.e. Z ⊗ZG Bu = · · · → Z{[g ⊗ h]} → Z{[g]} → Z{[]}. We will now compute the first homology of this complex, i.e. H1(G; Z). The differential in the original bar resolution takes d[g] = g[] − [] and after quotienting out the action this becomes zero. Going up by one dimension d[g ⊗ h] = g[h] − [gh] + [g] becomes on coinvariants · · · // Z{[g ⊗ h]} // Z{[g]} 0 // Z{[]} [g ⊗ h]  // [h] − [gh] + [g] Altogether we get H1(G; Z) = Z{[g]}/([gh] ∼ [g] + [h]) the free abelian group generated by the elements of the group with addition forced to equal the original group multiplication. This is easily seen to give the abelianization Gab of G, e.g. by its universal property g_  G //  A [g] Gab >> This corresponds to the fact that for a path connected space X we have H1(X; Z) = π1(X)ab and π1(BG) = π1(K(G, 1)) = G. Exercise 10.9. Show that H1(G; M) = Der(ZG; M)/ PDer(ZG; M) using the concrete description of the cochain complex HomZG(Z, M) below. Here a derivation of a ring G with coefficients in an R-R-bimodule M is a group homomorphism D: R → M satisfying the Leibniz rule D(r · s) = Dr · s + r · Ds. A principal derivation is one of the form Dx(r) = rx − xr for some x ∈ M. In the case R = ZG and a left ZG-module M made into a right ZG-module trivially (as above), the formulas become D(g · h) = Dg + g · Dh, Dx(g) = gx − x. We will now study certain extensions of groups very closely related to H2(G; M). We will restrict to certain extensions (to be specified in a minute via a certain action) 1 → M i −→ X p −−→ G → 1 42 10. Group cohomology where we write all groups multiplicatively and assume M commutative (later on, we will rewrite M additively, but at this point it would seem rather confusing). There is an action of X on M by conjugation (since it is the kernel of p and thus a normal subgroup: X → Aut(M), x → (m → xmx−1 = x m) and by commutativity, the restriction to M is trivial and thus this action factors through X/M ∼= G and we denote the action by the power on the left as above, i.e. am. Definition 10.10. Let M be a ZG-module. An extension 1 → M i −→ X p −−→ G → 1 is to be understood as an extension of groups with M commutative and such that the given G-action agrees with the conjugation action coming from the extension. Our aim will be to classify the extensions for a fixed G ∈ Grp and M ∈ ZGMod. We will now choose a based section of p, i.e. a mapping σ: G → X satisfying p(σ(a)) = a and p(1) = 1 (i.e. thinking of G = X/M it is a mapping that picks a representative in each class and picks 1 ∈ 1M). Now we may rewrite the conjugation action as a m = σ(a) · m · σ(a)−1 . If σ happens to be a homomorphism then the extension is split and we will see that it is then isomorphic to the so called semidirect product M ⋊ G. We will now construct the so called factor set that is an obstruction to σ being a homomorphism: [a, b] = σ(a) · σ(b) · σ(ab)−1 . By the based property, we get [a, 1] = 1 = [1, b] and we say again that the factor set is based. We will now explain the importance of the factor set: Given an extension and a based section, the mapping M × G → X, (m, a) → m · σ(a) is a bijection with inverse x → (x · σ(p(x))−1, p(x)). We may thus transport the group structure from X to M × G and obtain an isomorphic group in a somewhat “canonical form” that will allow us to compare two extensions: We first compute the product of the images of (m, a) and (n, b) inside X: m · σ(a) · n · σ(b) = m · a n · σ(a) · σ(b) = m · a n · [a, b] ∈M · σ(ab) ∈σ(G) . Thus this corresponds to the pair with components as indicated and the transported group structure is (m, a) · (n, b) = (m · a n · [a, b], ab). Assuming now that φ: G × G → M is a based mapping, we observe that the identity for this product is always (1, 1). We will now study when this product is associative (it will then have inverses as well), in which case we denote the resulting group M ×φ G with multiplication (m, a) · (n, b) = (m · a n · φ(a, b), ab). By construction, M ×[−,−] G is always a group. 43 10. Group cohomology Lemma 10.11. The product is associative iff aφ(b, c) · φ(a, bc) = φ(a, b) · φ(ab, c). If this is the case, inverses exist. Proof. This is a straightforward computation: (m, a) · ((n, b) · (p, c)) = (m, a) · (n · b p · φ(b, c), bc) = (m · a (n · b p · φ(b, c)) · φ(a, bc), abc) = (m · a n · ab p · a φ(b, c) · φ(a, bc), abc) ((m, a) · (n, b)) · (p, c) = (m · a n · φ(a, b), ab) · (p, c) = (m · a n · φ(a, b) · ab p · φ(ab, c), abc) The first claim thus follows from commutativity of M. The second claim follows from the observation that the equation (m, a) · (n, b) = (m · a n · φ(a, b), ab) = (1, 1) with parameter (m, a) has a unique solution b = a−1 and then one can solve for n from the first component, giving a right inverse. Symmetrically, the unique solution has a = b−1 and one can solve for m, giving a left inverse. When the product is associative, these have to be equal. We will now show that the factor set [−, −]: G×G → M can be interpreted as a 2-cocycle in the cochain complex HomZG(B, M). We write out the terms of this cochain complex in low dimensions B = · · · → ZG{[a | b | c]} → ZG{[a | b]} → ZG{[a]} → ZG{[]} HomZG(B, M) = · · · ← Mapb(G3 , M) ← Mapb(G2 , M) ← Mapb(G1 , M) ← Mapb(G0 , M) where Mapb denotes the set of all “based” mappings, i.e. those whose value is 1 whenever one of the arguments is 1. Thus, the factor set is a 2-cochain. The differential of a 2-cochain is (remember that we write the group M multiplicatively and the action as a power) (δφ)(a, b, c) = a φ(b, c) · φ(ab, c)−1 · φ(a, bc) · φ(a, b)−1 . The equation from the lemma claims exactly this. Any other based section differs by σ′(a) = β(a) · σ(a) for some based mapping β : G → M and we compute the corresponding factor set: Lemma 10.12. [a, b]′ = [a, b] · (δβ)(a, b), i.e. the two factor sets differ by a 2-coboundary. Proof. This is again a simple computation: [a, b]′ = σ′ (a)σ′ (b)σ′ (ab)−1 = β(a)σ(a)β(b)σ(b)σ(ab)−1 β(ab)−1 = β(a) · a β(b) · [a, b] · β(ab)−1 with all factors in the commutative group M and with (δβ)(a, b) = aβ(b) · β(ab)−1 · β(a). We may thus summarize this technical part by stating that there is a well defined 2cohomology class associated with an extension, called the factor set [−, −] ∈ H2(G; M). 44 10. Group cohomology Theorem 10.13. The mapping {extensions 0 → M → X → A → 0}/iso → H2 (G; M), associating to an extension its factor set, is bijective. Proof. We first show that the mapping is injective. Given two extensions X and X′ with factor sets [−, −] and [−, −]′ that are cohomologous, i.e. differ by a coboundary δβ, one can change the based section of X by β to obtain a new based section with corresponding factor set [−, −]′. Now the construction above gives isomorphisms X ∼= M ×[−,−]′ G ∼= X′ . To prove surjectivity, let φ be a 2-cocycle and consider the group M ×φ G. If we equip it with the obvious based section σ(a) = (1, a) then the corresponding factor set will be [a, b] = σ(a)σ(b)σ(ab)−1 = (1, a) · (1, b) · (1, ab)−1 = (φ(a, b), ab) · (1, ab)−1 = (φ(a, b), 1) · (1, ab) · (1, ab)−1 = (φ(a, b), 1) that equals φ(a, b) as an element of M ⊆ M ×φ G, as required. Theorem 10.14. Let G be a finite group of order k. Then the multiplication by k is zero on Hn(G; M) and Hn(G; M) for n > 0, i.e. the order of any element divides k. Proof. We will show that the multiplication by k map on B is homotopic to the map that is zero in all dimensions except dimension 0 where it is multiplication by N = g∈G g. B2 // 0  k  B1 // 0  k  B0 N  k  B2 // B1 // B0 We define h[g1 | · · · | gn] = (−1)n+1 · g∈G [g1 | · · · | gn | g]. Clearly dih = −hdi (thanks to the above alternating sign) so that everything in dh + hd cancels out except dn+1h[g1 | · · · | gn] = g∈G [g1 | · · · | gn] = k · [g1 | · · · | gn] so that dh + hd = k as claimed except in dimension 0 where (dh + hd)[] = d(− g∈G [g]) = g∈G [] − g[] = k[] − N[]. Now apply either M ⊗ZG − or HomZG(−, M) to obtain a chain homotopy between the corresponding maps on the resulting chain complexes. In (co)homology the maps become equal. 45 11. Flatness is stalkwise Corollary 10.15. Let G and M be finite with gcd(|G|, |M|) = 1. Then Hn(G; M) = 0 and Hn(G; M) = 0 for n > 0. Consequently, any extension 0 → M → X → G → 0 splits, i.e. is isomorphic to the semidirect product M ⋊ G. Proof. The multiplication by k = |G| is both zero by the previous theorem and an isomorphism, since it is induced by an isomorphism M k −−→ M (let l = |M| and ak + bl = 1; then the inverse is clearly the multiplication by a). Remark. This is a generalization of Example 8.3 to the case of nonabelian G. The theorem holds even for nonabelian M and is proved from the above abelian case by “group theoretic induction” (descreasing order be quotienting out the centre, I think). 11. Flatness is stalkwise We use this opportunity to talk about various special instances of flat modules. The main goal is to prove the theorem Theorem 11.1. Let R be a commutative ring. An R-module A is flat iff for every maximal ideal P ⊆ R the localization AP is a flat RP -module. The main ingredient of the proof is the so called flat base change for Tor. Let S be an R-algerba that is flat as an R-module. Then for an R-module A and S-module B we have TorR n (A, B) ∼= TorS n(A ⊗R S, B) Proof of the claim. Consider a projective resolution P → A[0]. Extending the scalars via the exact functor S ⊗R − we thus obtain a resolution P ⊗R S → A ⊗R S[0] that is easily seen to be projective again (the extension takes R → S, thus free to free and thus projective to projective). We may thus use this to compute TorS n(A ⊗R S, B) = Hn(P ⊗R S ⊗S B) = Hn(P ⊗R B) = TorR n (A, B). Now we are ready to prove the theorem. Proof of the theorem. Assuming A flat, we get TorRP n (AP , B) = TorRP n (A ⊗R RP , B) = TorR n (A, B) = 0 and AP is flat. In the opposite direction, assuming AP flat over RP , we need to show TorR n (A, B) = 0 and this is equivalent to TorR n (A, B)P = 0 for all P maximal. Since Tor is some homology group and localization is exact, we may view this as Hn(Q ⊗R B)P ∼= Hn((Q ⊗R B)P ) ∼= Hn(QP ⊗RP BP ) ∼= TorRP n (AP , BP ) = 0. We will now show that over local rings, flat modules are very close to free modules. First a general result. 46 11. Flatness is stalkwise Definition 11.2. An R-module A is finitely presentable if there exists a ses Rs → Rt → A → 0 for some finite s and t. Exercise 11.3. Show that for a finitely presentable A and any ses 0 → K → L → A → 0 with L finitely generated, also K is finitely generated. Consider the following map B ⊗R A∗ ∼= HomR(R, B) ⊗R HomR(A, R) ◦ −−→ HomR(A, B). Proposition 11.4. If A is finitely presentable and B is flat then this map is an isomorphism. Proof. We proved this at the tutorial. The idea is to prove this for A = R, then for finite coproducts, then for cokernels (using B flat). Corollary 11.5. If A is finitely presentable and flat, it is projective. Proof. In the map A ⊗ A∗ → HomR(A, A) the simple tensor ai ⊗ ηi maps to the composition A ηi −−→ R ai −−→ A and a finite sum of such clearly maps to A      η1 ... ηn      −−−−−→ Rn a1 · · · an −−−−−−−−−−−−→ A. If this happens to be the preimage of the identity then A is a direct summand of Rn and is thus projective. Over a local ring, projective modules are exactly free modules (Kaplansky theorem). We will prove a simpler version. Theorem 11.6. A finitely generated projective module over a commutative local ring is free. Proof. Write Rn = A ⊕ B. We will find a new basis of Rn whose first k elements lie in A and the last l elements lie in B. Then for the projection p: Rn → A, we get A = p(Rn ) = p(R{v1, . . . , vn}) = R{p(v1), . . . , p(vn)} = R{v1, . . . , vk} and thus v1, . . . , vk form a basis of A, showing that it is free. Denoting by k = R/M the residue field and by applying k ⊗R − to the above biproduct we obtain another biproduct kn = A/MA ⊕ B/MB. Since these are now k-vector space, we may find the basis (v1, . . . , vn) with the corresponding properties. Take arbitrary preimages v1, . . . , vk ∈ A and vk+1, . . . , vn ∈ B. They will form a basis of Rn since it maps Matn×n R ∋ (v1, . . . , vn) → (v1, . . . , vn) ∈ Matn×n k to an invertible matrix over k, thus its determinant must be nonzero in k and thus it is a unit of R. 47 12. Simplicial resolutions 12. Simplicial resolutions Let C be a category. A monad on C is a monoid in the strictly monoidal category ([C, C], ◦) of endofunctors of C. Thus, it is an endofunctor T equipped with two natural transformations µ: T ◦ T → T, η: 1 → T, the multiplication and the unit, satisfying the associativity and unitality axioms µ ◦ (µ ◦ 1) = µ ◦ (1 ◦ µ), µ ◦ (1 ◦ η) = 1 = µ ◦ (η ◦ 1). More concisely, one requires natural transformations µk : Tk → T that are closed under compositions. This means that 1 = µ1 (as a composition of zero µk’s) and e.g. µ2 µ2 µ1 = µ3 = µ2 µ2 µ1 and similarly for the unary µ0 µ2 µ1 = µ1 = µ0 µ2 µ1 There is a universal strictly monoidal category with a monoid. We will give a concrete description and then, instead of showing the universal property, give the unique instance of it that we are interested in, i.e. to monads. It is the category ∆ of all finite ordinals (topologists would only consider non-empty ordinals; this would correspond to a non-unital version) [n] = {0 < 1 < · · · < n} with [0] = {0} and [−1] = ∅. Morphisms in ∆ are the order preserving maps and the monoidal product is the “join” [m] ∗ [n] = [m + 1 + n] or more intuitively {0 < · · · < m} ∗ {0 < · · · < n} = {0 < · · · m 0<··· 0} = (t). Every nonzero ideal is of the form Mn = (tn) (so that R is in fact a PID and noetherian). Conversely, if R has nonzero ideals exactly Mn = (tn), it is a DVR. ˆ The prime ideals of a DVR R are exactly 0 and M so that R has Krull dimension 1, i.e. the longest chain of primes consists of one inclusion – in this case 0 ⊆ M. Proof. This is all fairly straightforward. For the first point, use v(u−1) = −v(u); further, since r/tn ∈ K has valuation 0, it is a unit of the ring. For the second point, the implication ⇒ is exactly the first point. For the implication ⇐, observe that every nonzero element of the fraction field K can be written uniquely as k = u · tn with t ∈ Z and we may thus introduce v(k) = n. For the third point, observe that r | s iff v(r) ≤ v(s) so that every nonzero ideal I is generated by any nonzero element of minimal valuation. By the first part, we may write it as r = u · tn and thus I = (r) = (tn). In the opposite direction, R must then be a PID, 58 16. Integrally closed rings, valuation rings, Dedekind domains hence UFD. Since irreducible elements of a PID, up to associatedness, correspond precisely to nonzero prime ideals and the only such is (t), there is a unique irreducible and the second point applies. The last point is clear. Theorem 16.6. For a domain R, the following conditions are equivalent ˆ R is a DVR, ˆ R is a noetherian local ring whose unique maximal ideal is nonzero and principal, ˆ R is a noetherian local ring of Krull dimension 1 that is also integrally closed. Proof. We have proved that the first point implies the other (except we did not mention explicitly DVR ⇒ UFD ⇒ integrally closed). It remains to prove that any of the other conditions imply that R is a DVR. Start with the second point. Let M = (t) be the maximal ideal of R. We will show that all nonzero proper ideals are of the form Mn. Clearly I ⊆ M and we claim that there exists the largest n for which I ⊆ Mn. Otherwise, I would lie in the intersection M∞ def = n Mn that we will show to be zero by Nakayama lemma: M∞ is finitely generated since R is noetherian and clearly satisfies M · M∞ = M, thus M∞ = 0. Thus let a ∈ I ⊆ Mn = (tn) with a /∈ Mn+1. We may write a = u · tn and by assumption u /∈ M ⇒ u ∈ R×. Thus a is associate to tn and thus already a alone generates Mn; we get I = Mn = (tn), as claimed. Now assume the third set of conditions. We will prove all conditions in the second point, where M = 0 would imply that R has Krull dimension 0, so it remains to show that M is principal. By Nakayama lemma M2 ⊊ M, for equality would give M = 0. Let t ∈ M ∖ M2. We claim that M = (t). Clearly I = (t) is a proper nonzero ideal, thus contained in a unique prime ideal M. Proposition 4.6 gives √ I = M and this implies Mn ⊆ I for some n by finite generation of M. Starting from this, we will show inductively Mn ⊆ I ⇒ Mn−1 ⊆ I, finishing with M ⊆ I ⊆ M, as claimed. Thus let x ∈ Mn−1. Since we want to show that x ∈ I = (t), we consider the element x/t ∈ K of the fraction field and we want x/t ∈ R, which we prove by exploiting the fact that R is integrally closed. Consider the multiplication by x/t: x/t · : M → R Clearly x/t · M ⊆ 1/t · Mn ⊆ R. Now the image must be a submodule, i.e. an ideal, and we claim that it cannot be the trivial ideal R: for otherwise there would exist m ∈ M such that x/t · m = 1, i.e. t = xm ∈ Mn and we obviously assume n ≥ 2 and t /∈ M2. Thus the image of the multiplication map must be contained in M: x/t · : M → M Now the Cayley-Hamilton-Nakayama-like argument below gives a monic polynomial F ∈ R[x], such that the multiplication by F(x/t) is a zero map. Since K is a field and M ̸= 0, this implies that F(x/t) = 0 in K, as required. Theorem 16.7. Let S be a (commutative) R-algebra and let M be an S-module that is finitely generated over R. Then for every s ∈ S there exists a monic polynomial F ∈ R[x] such that F(s) · x = 0 for all x ∈ M, i.e. F(s) lies in the kernel of S → End(M). Remark. Keeping R commutative, we may replace a non-commutative S by its commutative subalgebra R[s] ⊆ S and apply the theorem to it, getting the same conclusion even for S non-commutative. 59 16. Integrally closed rings, valuation rings, Dedekind domains Proof. Write M = R{x1, . . . , xn} and express the action of s ∈ S on M in two ways with respect to this generating set: (x1, . . . , xn) · sE = (s · x1, . . . , s · xn) = (x1, . . . , xn) ·    r11 · · · r1n ... ... rn1 · · · rnn    = (x1, . . . , xn) · A where A denotes the n-by-n matrix in the formula, with elements in R. One can write this concisely as (x1, . . . , xn) · (sE − A) = (0, . . . , 0). Multiplying by the adjoint matrix gives (x1, . . . , xn) · det(sE − A) = (0, . . . , 0), i.e. the multiplication by det(sE − A) annihilates the generators x1, . . . , xn and thus M. We may set F(x) = det(xE − A) ∈ R[x]. We may now apply this characterization of DVRs to introduce Dedekind domains. Importantly, the last condition localizes well, so we define: Definition 16.8. A Dedekind domain is a noetherian domain of Krull dimension 1 that is integrally closed. Theorem 16.9. Let R be a domain. TFAE ˆ R is a Dedekind domain, ˆ R is noetherian and for all nonzero primes P, the localization RP is a DVR. Proof. We have proved that R is integrally closed iff RP is integrally closed and it remains to show the same for the Krull dimension, but this is easy, since localization at P picks out of the prime ideals of R those that are contained in P. The point is that the primes in a DD and in a DVR form the following posets: · · · Mi · · · Mj · · · M 0 0 These clearly correspond to one another. Now we want to show an interpretation of a DD in terms of fractional ideals. Definition 16.10. Let R be a domain with a fraction field K. A fractional ideal is an R-submodule A ⊆ K of the form 1/d · I for an ideal I ⊆ R. Remark. Over a noetherian domain, this is equivalent to A being a finitely generated R-submodule of K. We introduce a product of fractional ideals similarly to that of ideals, i.e. AB is the ideal generated by the products ab, for a ∈ A and b ∈ B. Clearly (1/d · I)(1/e · J) = 1/(de) · IJ so that this product is indeed a fractional ideal. Clearly the unit is R so that we get an induced notion of an invertible fractional ideal A as that for which there exists a fractional ideal B such that AB = R. 60 16. Integrally closed rings, valuation rings, Dedekind domains Example 16.11. A principal fractional ideal is one of the form (k) = (r/d) = 1/d · (r) for k = r/d ∈ K. Clearly, this has inverse (k−1). In a principal ideal domain, these are all examples. Consider, for a nonzero fractional ideal A, the following fractional ideal A′ = {k ∈ K | kA ⊆ R} (since A contains some element d ∈ R, we have A′d ⊆ R so that A′ = 1/d · I for the ideal I = A′d). By definition, A′A ⊆ R and we will prove that the equality holds iff A is invertible, in which case A−1 = A′. The implication ⇒ is obvious, so assume that A is invertible. Then A−1 ⊆ A′ and consequently R = A−1 A ⊆ A′ A ⊆ R and so we must get equality everywhere and A′ is also an inverse. But inverses are unique in monoids. Theorem 16.12. In a Dedekind domain, every fractional ideal is invertible. In addition, every nonzero proper ideal I ⊆ R admits a unique decomposition I = P1 · · · Pr into a product of prime ideals. Proof. Let A be a fractional ideal and consider the fractional ideal A′ as above. We need to show that A′A = R. This means that the inclusion A′A → R is an isomorphism and we know that this may be checked on localizations. These are (AP )′ AP = (A′ )P AP = (A′ A)P → RP (the localization AP is clearly the RP -submodule generated by A). These will be isomorphisms provided that AP is invertible. This follows from RP being a PID. Now let I ⊆ R be an ideal. The primary decomposition of I is I = I1 ∩ · · · ∩ Is with each Ii primary, say Ass R/Ii = {Pi}. Thus, Ii is contained in a unique maximal ideal Pi and consequently these are pairwise comaximal, i.e. Ii + Ij = R, giving I = I1 · · · Is by the following proposition. Now the Pi-primary component Ii of I is uniquely determined since Pi is minimal over I and in fact Ii/I is the kernel of the localization map R/I → RPi /IPi or, slightly better, Ii is the preimage of IPi under the localization map λi : R → RPi . Now since RPi is a DVR, the ideal IPi is a power Mki i of the maximal ideal and thus pulls back to the corresponding power Pki i , since this power is Pi-primary Pki i , thus (R ∖ Pi)-saturated, and clearly maps to Mki i . The uniqueness follows easily from all primes being invertible: for if P1 · · · Pr = Q1 · · · Qs then for any prime Qj we have P1 · · · Pr ⊆ Qj so Pi ⊆ Qj. Symmetrically Qj′ ⊆ Pi ⊆ Qj and appying this for Qj minimal must give equality. We may this common prime in the group of invertible fractional ideals and proceed by induction. 61 16. Integrally closed rings, valuation rings, Dedekind domains In fact, these are both equivalent conditions, i.e. a domain where every nonzero fractional ideal is invertible is a Dedekind domain (or a field). Also, a domain where every nonzero proper ideal factors uniquely into a product of prime ideals is a Dedekind domain (or a field). The first claim is not difficult: One shows that all invertible fractional ideals must be finitely generated (AB = R implies AB = R for some finitely generated fractional subideal ˆA ⊆ A by looking at 1 ∈ R; by uniqueness of inverses A = A), hence R is noetherian. Every localization RP at a nonzero prime P will also have all nonzero fractional ideals invertible (the fractional ideals are of the form AP and as such admit an inverse A′ P ). Thus, it remains to show that every noetherian local ring with all fractional ideals invertible must be a DVR. Let t ∈ M ∖ M2 and consider the fractional ideal t−1M with inverse (t)M−1 ⊆ MM−1 = R. Now (t)M−1 ̸⊆ M since otherwise t ∈ (t) ⊆ M2, so (t)M−1 = R giving (t) = M as required. The second claim is much more complicated. Proposition 16.13. Assume that I + J = R. Then IJ = I ∩ J. More generally if Ii are pairwise comaximal then I1 · · · Ir = I1 ∩ · · · ∩ Ir. Proof. The containment IJ ⊆ I ∩ J holds always, so let z ∈ I ∩ J. Write x + y = 1, giving z = (x + y)z = xz + zy with both terms in IJ. The general case is obtained by application to I1 · · · Ir−1 and Ir once we show that these are comaximal which is a bit tricky. So let xi + yi = 1 with xi ∈ Ii and yi ∈ Ir. Then we have 1 = (x1 + y1) · · · (xr−1 + yr−1) = x1 · · · xr−1 + terms containing some yi ∈Ir ∈ I1 · · · Ir−1 + Ir. Remark. I would say that Ii are comaximal if Ij + i̸=j Ii = R and the second part shows that pairwise comaximal implies comaximal, which I find a bit surprising. 62