# Injective Modules and Matlis Duality

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1 Appendix A Injective Modules and Matlis Duality Notes on 24 Hours of Local Cohomology William D. Taylor We take R to be a commutative ring, and will discuss the theory of injective R-modules. The following facts are commonly used results: (1) For any R-module M, there is a unique smallest injective module containing M, denoted E R (M) and called the injective hull. It has the property that any injective module containing M has E R (M) as a submodule. (2) Any injective module over a noetherian ring has a unique direct sum decomposition into indecomposable injective modules. (3) The indecomposable injective modules of a noetherian ring are of the form E R (R/p) for p Spec R. (4) Matlis Duality Suppose (R, m) is a complete local ring, E = E R (R/m), and ( ) is the R-module functor Hom R (, E). Then (a) If M is noetherian, then M is artinian. (b) If M is artinian, then M is noetherian. (c) If M is noetherian or artinian, then M = M. Note: We say a module M is noetherian if it satisfies the ascending chain condition on submodules, and that it is artinian if it satisfies the descending chain condition on submodules. 1

2 2 APPENDIX A. INJECTIVE MODULES AND MATLIS DUALITY A.1 Essential Extensions Definition A.1. An R-module E is injective if the functor Hom R (, E) is exact; i.e. if for every injective map of R-modules f : M N and R-module homomorphism g : M E, there exists h : N E such that g = h f. Theorem A.2 (Baer s criterion). An R-module E is injective if and only if for every ideal a R and R-module homomorphism f : a E, there exists an R-module homomorphism g : R E such that g a = f. Note that since the zero homomorphism can always be extended (trivially), we may consider only the nonzero ideals a and nonzero homomorphsims a E when confirming the injectivity of an R-module using Baer s Criterion. Exercise A.3. If E and E are R-modules, then E E is injective if and only if E and E are injective. Solution. Suppose E E is injective. Let a R be an ideal and f : a E be an R-module homomorphism. Then f 0 : a E E is an R-module homomorphism, and therefore there exists g : R E E extending f 0. Therefore π 1 g : R E is a homomorphism, and for x a, (π 1 g)(x) = π 1 (g(x)) = π 1 (f(x), 0) = f(x), so π 1 g extends f. Therefore E is injective by Baer s criterion. By a symmetric argument, so is E. Suppose E and E are injective, a R is an ideal, and f : a E E is an R-module homomorphism. Then π 1 f : a E and π 2 f : a E are R-module homomorphisms, hence they extend to g : R E and g : R E. Therefore g g : R E E is an R-module homomorphism, and for x a, (g g )(x) = (g(x), g (x)) = (π 1 (f(x)), π 2 (f(x))) = f(x). Hence g g extends f, and so E E is injective by Baer s Criterion. If R is a domain, we say that an R-module M is divisible if for all r R\{0}, r M = M, i.e. for every m M, there exists m M such that m = r m. Exercise A.4 (Part 1). If R is a domain, then all injective R-modules are divisible, and if in addition R is a P.I.D., then all divisible R-modules are injective. Solution. Suppose R is a domain and E is an injective R-module. For any r R \ {0} and e E we can define a map f : rr E by setting f(r) = e and extending linearly. Since E is injective there exists an R-module homomorphism g : R E extending f, and r g(1) = g(r) = e. Therefore E is divisible. Suppose that R is a P.I.D. and M is a divisible module. Let a = ar R be a nonzero ideal of R and let f : ar M be an R-module homomorphism. Since M is divisible we can find m M such that a m = f(a). Let g : R M be defined by g(r) = r m. Then g is an R-module homomorphism and we have that g(a) = a m = f(a). Therefore g extends f, and so M is injective. As an immediate application of the above exercise, we have that Q/dZ is an injective Z-module for d Z.

3 A.1. ESSENTIAL EXTENSIONS 3 Exercise A.4 (Part 2). Every nonzero abelian group has a nonzero homomorphism to Q/Z, and if we let ( ) = Hom Z (, Q/Z), then for each Z-module M, the natural map M M is injective. Solution. Let G be a nonzero abelian group, i.e. a nonzero Z-module. Let x G and let H = Z x G be a submodule of G. Define f : H Q/Z as follows: if n x 0 for n > 0, then f(x) = Z Q/Z. Otherwise, let n be the smallest positive integer such that n x = 0, and set f(x) = 1 n + Z Q/Z. Then f is a well-defined Z-module homomorphism from H to Q/Z. But since we have an injective map (namely inclusion) from H to G and Q/Z is injective (by the above exercise), there exists an R-module homomorphism g : G Q/Z extending f. Therefore g is nonzero, and the first statement is proved. In particular we have proved more: given any nonzero x G we can find a Z-module homomorphism g : G Q/Z such that g(z) 0. Now let M be any Z-module. The natural map M M = Hom Z (Hom Z (M, Q/Z), Q/Z) takes m (f f(m)). Suppose 0 m M. Then by the last sentence of the previous paragraph, we can find f Hom R (M, Q/Z) such that f(m) 0. Therefore m is not in the kernel of the map M M. Therefore the kernel of this map is 0, i.e. the map is injective. Exercise A.5. Let R be an A-algebra. Then (1) If E is an injective A-module and F is a flat R-module, then Hom A (F, E) is an injective R-module; and (2) Every R-module embeds in an injective R-module. The solution is below is modified from Mel Hochster s local cohomology notes, page 6. Solution. By definition, Hom A (F, E) is an injective R-module if the functor Hom R (, Hom A (F, E)) is exact. For any R-module M, Hom R (M, Hom A (F, E)) = Hom A (M R F, E) by the adjointness of Hom and. Since F is a flat R-module, the functor R F is exact, and since E is an injective A-module, the functor Hom A (, E) is exact. Therefore Hom R (, Hom A (F, E)) = Hom A ( R F, E) is the composition of two exact functors, hence is exact. This proves part (1). Let M be an R-module. Letting ( ) be the functor Hom Z (, Q/Z), we have that M M is injective by exercise A.4(4). Let F be a flat (e.g. free) R- module and f an R-module homomorphism such that f : F Hom Z (M, Q/Z) = M is surjective. Since ( ) is a left-exact contravariant functor on R-modules, f : M F is injective. Therefore M embeds in F. But since Q/Z is an injective Z-module and R is a Z-algebra (since any ring is), F = Hom Z (F, Q/Z) is an injective R-module by part (1). Proposition A.6. Let θ : M N be an injective map of R-modules. The following are equivalent:

4 4 APPENDIX A. INJECTIVE MODULES AND MATLIS DUALITY (1) For any homomorphism ɛ : N Q, if ɛ θ is injective, then so is ɛ. (2) Every nonzero submodule of N has nonzero intersection with θ(m). (3) Every nonzero element of N has a nonzero multiple in θ(m). Proof. Since θ is injective, we can identify M with θ(m) and write M N. In this context, ɛ θ = ɛ M. (1) (2): Let N be a nonzero submodule of N. Then the quotient map ɛ : N N/N is not injective, and so we must have that ɛ M is not injective. Therefore there exists 0 m M such that ɛ(m) = 0 + N, i.e. m N. (2) (3): Let 0 n N and consider the submodule R n N. By our hypothesis, there exists 0 m M R n, i.e. n has a nonzero multiple in M. (3) (1): Let ɛ : N Q be an R-module homomorphism such that ɛ M is injective. Let 0 n N, and take r R such that 0 r n M. Then ɛ(r n) = ɛ M (r n) 0. Therefore r n 0, i.e. n 0. Therefore ɛ is injective. For good measure we prove some additional implications (2) (1) Let ɛ : N Q be an R-module homomorphism such that ɛ M is injective. If ker ɛ 0, then there exists 0 m ker ɛ M = ker ɛ M = 0, a contradiction. Therefore ɛ is injective. (3) (2) Suppose N N is a nonzero submodule. Then there exists 0 n N, and so there exists r R such that r n M. Since r n N also, we have that M N 0. Definition A.7. If θ : M N satisfies any of the conditions of the above proposition, we say that N is an essential extension of M. Example A.8. If U R is a multiplicative system of nonzerodivisors of R then U 1 R is an essential extension of R. That U contain only zerodivisors of R is necessary for the natural map R U 1 R to be injective. If 0 r u U 1 R then u r u = r 1 and r 1 is in the image of R, so U 1 R is an essential extension of R. Example A.9. Let (R, m) be a local ring and N an m-torsion R-module (this means that every element of N is killed by some power of m). Then the socle of N is denoted and defined as soc(n) = (0 : N m). The extension soc(n) N is essential: If 0 y N, let t be the smallest integer such that m t y = 0. Then m t 1 y = 0, i.e. y has a nonzero multiple in soc(n). Exercise A.10. Let I be an indexing set, and for each i I let M i and N i be R-modules. Then i I M i i I N i is essential if and only if M i N i is essential for each i I. Solution. Suppose that i I M i i I N i is essential, and fix j I. Let 0 L j N j be an R-module. Then L = L j i I\{j} 0 is a nonzero submodule of i I N i. Therefore 0 L i I M i = (L j M j ) i I\{j} 0,

5 A.1. ESSENTIAL EXTENSIONS 5 and so L j M j 0. Hence M j N j is essential. Now suppose that M i N i is essential for each i I. Let 0 L i I N i. Then L = i I L i, where L i = L N i. Choose i I such that L i 0. Then since M i N i is essential, M i L i 0, hence i I M i L 0. Therefore i I M i i I N i is essential. Exercise A.11. Let k be a field, R = k[[x]], and N = R x /R. Then soc(n) N is essential and N soc(n) N N is not essential. Solution. The ring R is local with maximal ideal m = (x) and the elements of N are of the form f + R for some k N and f k[x] with degree less than k. x k For any such element, m k = (x k ) kills that element. Therefore N is m-torsion, an so by example A.9, soc(n) N is essential. Now consider the extension N soc(n) N N. Let y N N be given by y i = 1 x + R. Let 0 i i N g i R be any nonzero element of R. Let j N such that g (x j )\(x j+1 g ). Then the j+2th coordinate ( of g y is x +R = gj+gj+1x j+2 ) x + 2 gj+g R. But this element is not in soc(n) since x j+1x x + R = gj 2 x + R 0. Therefore g y / N soc(n), and so N soc(n) N N is not essential. Proposition A.12. Let L M N be nonzero R-modules. (1) L N is essential if and only if L M and M N are essential. (2) Let I be an index set and let N i be R-modules such that for each i, M N i N and i N i = N. Then M N is essential if and only if each M N i is essential. (3) There exists a unique module N N with M N N maximal with respect to the property that M N is essential. Proof. (1) Suppose L N is essential. Let 0 M M N and 0 N N. Then M L 0 since L N is essential and N M N L 0 since L N is essential. Therefore L M and M N are essential. Now suppose L M and M N are essential and let 0 N N. Then M N M 0 since M N is essential, and therefore L (M N ) 0 since L M is essential. But since L N L (M N ), we re done. (2) Suppose M N is essential. Fix i I and let 0 N i N i. Then N i N, and since M N is essential, M N i 0. Therefore M N i is essential. Now suppose M N i is essential for all i I. Let 0 N N. Since N = i N i, there exists i I such that N N i 0. Since M N i is essential, M N M (N N i ) 0. Therefore M N is essential. (3) Let S = {S N M S is essential} and order S by inclusion. Let S 1 S 2 be an increasing sequence of elements of S. Then since each S i is essential over M, so is S = i N S i, and we have that S S i for all i N. Therefore S is a maximal element for the increasing sequence, and by Zorn s Lemma, there exist maximal elements of S. Let S, S be two maximal elements of S. Then since M S and M S are both essential extensions, so

6 6 APPENDIX A. INJECTIVE MODULES AND MATLIS DUALITY is M S S. So S S S and S S S, S, and so by the maximality of S and S we must have that S = S S = S. Therefore the maximal element of S is unique. Definition A.13. The module N in Proposition A.12(3) is called the maximal essential extension of M in N. If M N is essential and N has no proper essential extensions, then N is called a maximal essential extension of M. Proposition A.14. Let M be an R-module. The following are equivalent: (1) M is injective; (2) every injective homomorphism M N splits; (3) M has no proper essential extensions Proof. (1) (2): Let f : M N be an injective homomorphism. Since M is injective and id M : M M is a homomorphism, id M extends to a map g : N M. Therefore g splits f. (2) (3): Suppose M N is an essential extension. Then there is a splitting f : N M of the inclusion map ι. Since f ι is the identity map, it is injective, and so since M N is essential, f is injective. Therefore f is bijective and so M = N. (3) (1): By Exercise A.5 we can embed M in an injective module E. If N 1 N 2 is an increasing sequence of submodules of E such that N i M = 0 for all i, then M i N N i = 0. Therefore there exists a submodule N of E maximal with respect to the property of having zero intersection with M by Zorn s Lemma. So now f : M E/N is an essential extension, hence M = E/N. Therefore E = M + N and since M N = 0, we have that E = M N. Since M is a direct summand of an injective module, M is itself injective. Note in particular that this means that if E is an injective module and E M for some R-module M, then M = E F for some R-module F. Proposition A.15. Let M be an R-module. If M E with E injective, then the maximal essential extension of M in E is an injective module, hence a direct summand of E. Maximal essential extensions of M are isomorphic. Definition A.16. The maximal essential extension of M is called the injective hull of M and is denoted E R (M) (or E(M) if the ring is understood). Definition A.17. Let M be an R-module. An injective resolution of M is a complex of injective R-modules 0 E 0 0 E 1 1 E 2 2 with H 0 (E ) = M and H i (E ) = 0 for i 1; it is called minimal if E 0 = E R (M), E 1 = E R (E 0 /M), and E i+1 is the injective hull of coker i for each i i.

7 A.2. NOETHERIAN RINGS 7 A.2 Noetherian rings We can characterize noetherian rings using injective modules Proposition A.18. A ring R is noetherian if and only if every direct sum of injective R-modules is injective. We have already seen that the direct sum of two injective modules (and hence of finitely many) is injective; the content of the above proposition lies in the infinite direct sum case. The only part of the proof that we need to fill in the details of is the following: Given a chain of ideals a 1 a 2 of R and setting a = i N a i, we have natural homomorphisms a R R/a i E R (R/a i ). From these we have a homomorphism a i N E R(R/a i ). We need to show that this map factors through i N E R(R/a i ). Indeed, let x a. Then there exists j N such that x a i for all i j. So for any i j, the map R R/a i takes x to 0. Therefore the image of x in i N E R(R/a i ) is zero for indices greater than or equal to j, i.e. x i N E R(R/a i ). This completes the last part of the proof of the proposition. Definition A.19. An R-module M is called a-torsion for an ideal a of R if every element of M is killed by some power of a. Theorem A.20. Let p be a prime ideal of a noetherian ring R and set κ = R p /pr p, the fraction field of R p. Let E = E R (R/p). (1) If x R \ p then E x E is an isomorphism, hence E = E p ; (2) (0 : E p ) = κ; (3) κ E is an essential extension of R p -modules and E = E Rp (κ); (4) E is p-torsion and Ass(E) = {p}; (5) Hom Rp (κ, E) = κ and Hom Rp (κ, E R (R/q) p ) = 0 for primes q p. Injective modules, for all their foibles, have a structure theorem when R is noetherian. Theorem A.21. Let R be an injective module over a noetherian ring R. There exists a direct sum decomposition E = E R (R/p) µp p Spec R and the numbers µ p are independent of the decomposition. Proposition A.22. Let U R be a multiplicative set. (1) If R is an injective R-module, the U 1 R-module U 1 E is injective (injective modules localize).

8 8 APPENDIX A. INJECTIVE MODULES AND MATLIS DUALITY (2) If M N is a (maximal) essential extension of R-modules, then U 1 M U 1 N is a (maximal) essential extension of U 1 R-modules (essential extensions localize). (3) The indecomposable injectives over U 1 R are the modules E R (R/p) for p Spec R with p U =. Definition A.23. Let M be an R-module and E its minimal injective resolution. For each i, we have a decomposition E i = E R (R/p) µi(p,m). p Spec R The number µ i (p, M) is called the ith Bass number of M with respect to p. It is well-defined according to the following theorem: Theorem A.24. Let R be a noetherian ring and M an R-module. Let p be a prime ideal, and set κ = R p /pr p. Then µ i (p, M) = rank κ Ext i R p (κ, M p ). We call a homomorphism of local rings ϕ : (R, m) (S, n) a local homomorphism if ϕ(m) n. Theorem A.25. Let ϕ : (R, m, k) (S, n, l) be a local homomorphism of local rings. If S is module-finite over R, then Hom R (S, E R (k)) = E S (l). Remark A.26. Let (R, m, k) be a local ring. Then the above theorem states that for any ideal a of R we have that E R/a (k) = (0 : ER (k) a), and since E R (k) is m-torsion, we have E R (k) = t N (0 : ER (k) m t ) = t N E R/m t(k). A.3 Artinian rings We define the length of a module (denoted l(m)) as the supremum of the lengths of a composition series 0 = M 0 M 1 M 2 M. If 0 M 1 M 2 M 3 0 is a short exact sequence of R-modules then l(m 2 ) = l(m 1 ) + l(m 2 ). If R is a local ring, then the residue field is the only simple module of R and so the subquotients in a composition series for M are all isomorphic to it. Lemma A.27. Let R be a local ring with residue field k, set ( ) = Hom R (, E R (k)), and let M be an R-module. Then (1) The natural map M M is injective; (2) l(m ) = l(m)

9 A.4. MATLIS DUALITY 9 In particular, point (2) implies that M = 0 if and only if M = 0, i.e. that ( ) is a faithful functor. Corollary A.28. l(e R (k)) = l(r). In particular, E R (k) has finite length if and only if R is artinian. Theorem A.29. Let R be a local ring. Then R is an injective R-module if and only if it is artinian and rank k soc(r) = 1. A.4 Matlis duality Remark A.30. Let a R be an ideal and let M be an R-module. Then there are natural surjections M/a 3 M M/a 2 M M/aM 0. The a-adic completion of M, denoted M, is the inverse limit of this system: { } lim i (M/a i M) = (..., m 2, m 1 ) i M/a i M m i+1 m i a i M There is a canonical homomorphism of R-modules M M. The salient properties of this construction are summarized below: (1) ker(m M) = i N a i M; (2) R is a ring and R R is a ring homomorphism; (3) M is an R-module and M M is compatible with these structures. When R is noetherian, then also (4) The ring R is noetherian; (5) If (R, m) is local, then R is local with maximal ideal m R; (6) If M is finitely generated, then M = R R M, and so M is a finitely generated R-module; (7) One has R/a i R = R/a i for each i. If M is a-torsion, then it has a natural R-module structure and the map M R R M is an isomorphism; (8) If M and N are a-torsion, then Hom R(M, N) = Hom R (M, N). Theorem A.31. Let (R, m, k) be a local ring, R ists m-adic completion, and set E = E R (k). Then E R(k) = E, and the map R Hom R (E, E) taking r to the homomorphism e r e ( multiplication by r ) is an isomorphsim.

10 10 APPENDIX A. INJECTIVE MODULES AND MATLIS DUALITY Corollary A.32. For a local ring (R, m, k), the module E R (k) is artinian. Theorem A.33. Let (R, m, k) be a local ring and M an R-module. The following conditions are equivalent: (1) M is m-torsion and rank k soc(m) is finite of rank t; (2) M is an essential extension of a k-vector space of finite rank t; (3) M can be embedded in a direct sum of t copies of E R (k); (4) M is artinian. Proof. (1) = (2): Since M is m-torsion, soc(m) M is essential by example A.9. (2) = (3): Suppose that M is essential over a k-vector space V of finite rank t. Let e 1,..., e t be a k-basis for V. Then V = t i=1 ke i, and so E R (V ) = E R ( t i=1 ke i ) = t E R (k). Since M is an essential extension of V, we know that M E(V ), so we re done. (3) = (4): By Corollary A.32, we know E R (k) is artinian, therefore any finite direct sum of copies of E R (k) is artinian. Therefore any submodule of a finite direct sum of copies of E R (k) is artinian. (4) = (1): If m M is any element, Rm mm m 2 m is a descending chain of modules, hence it stabilizes. Therefore there exists j 0 such that m j+1 m = m j m, and therefore m(m j m) = m j m. By Nakayama s lemma, m j m = 0, so M is m-torsion. Since M is artinian and soc(m) M, soc(m) is artinian. But then rank k soc(m) is equal to the length of soc(m) as a k-module. Since ever element of soc(m) is killed by m, soc(m) R k = soc(m) R R/m = soc(m) R R = = soc(m), and so the length of soc(m) as a k-module is equal to the length of soc(m) as an R-module. Example A.34. Let (R, m, k) be a DVR with m = xr (e.g., R = k[[x]]). Then E R (k) = R x /R: R x /R is divisible since if 0 f R is not a unit, then f m j \ m j+1 for some j, and so f x is a unit, hence j ( ) f f (R x /R) = (f R x )/R = x j R x /R = R x /R. Therefore R x /R is an injective R-module since R is a P.I.D. Furthermore, it is f (x)-torsion: if + R R x k x /R, then (x) k annihilates it. The socle is { } f xf { a } soc(r x /R) = + R f R, xk x k R = x + R a R \ (x) and so it is generated by 1 x as a R/(x)-module. Therefore R x/r is (x)-torsion and rank k soc(r x /R) = 1. Therefore R x /R is an essential extension of k and R x /R E R (k) by Theorem A.33. Since R x /R is injective, R x /R = E R (k). i=1

11 A.4. MATLIS DUALITY 11 Theorem A.35 (Matlis Duality). Let (R, m, k) be a complete local ring, let ( ) = Hom R (, E R (k)), and let M be an R-module. Then (1) If M is noetherian (resp. artinian) then M is noetherian (resp. artinian); (2) If M is artinian or noetherian, then the map M M is an isomorphism. Remark A.36. Let M be a finitely generated module over a complete local ring (R, m, k). One has isomorphisms Hom R (k, M ) = Hom R (k R M, E R (k)) = Hom R (M/mM, E R (k)) = Hom k (M/mM, k). Thus the number of generators of M as an R-module is rank k soc(m ); this number is the type of M.

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