LOCAL COHOMOLOGY AND D-MODULES

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1 LOCAL COHOMOLOGY AND D-MODULES talk gven at the Unversty of Mchgan Student Commutatve Algebra Semnar 4 & 11 October, 2011 by Ashley K. Wheeler Ths talk follows the frst half of Mel Hochster s notes D-modules and Lyubeznk s Fnteness Theorems for Local Cohomology, whch you can fnd on hs webste The materal s hard, so much so that I do not cover any of Lyubeznk s results. In fact, my best judgment found the seemngly excessve number of exercses and footnotes whch follow the most effectve way to present such a breadth of materal, wthout too many tangents. Basc Local Cohomology In ths secton the symbol I wll denote an deal of a Noetheran 1 rng R and M wll denote an R-module, whch may or may not be Noetheran 2. Recall 3 one way to defne local cohomology s to take a drect lmt 4 of Ext modules 5 : H IM = lm t Ext RR/I t, M. The 0th local cohomology module has the dentfcaton HI 0 M = lmthom R R/I t, M = Ann M I t t 1 A rng R s Noetheran means all of ts deals are fntely generated. Equvalently, the deals n R satsfy the ascendng chan condton ACC, that every strctly ascendng chan of deals must termnate, and ths s equvalent to the condton that every nonempty collecton of deals n R has a maxmal element. For a proof of the equvalence of these condtons, see Chapter 1 of [Es], who also shows these statements are equvalent to Hlbert s orgnal defnton. A polynomal rng over a feld s an example of a Noetheran rng. 2 An R-module M s Noetheran means all of ts submodules are fntely generated. For reasons analogous to the case of a rng, ths s equvalent to M satsfyng ACC on ts submodules, whch s equvalent to every nonempty collecton of submodules havng a maxmal element. Proposton. If R s a Noetheran rng and M s a fntely generated R-module, then M s Noetheran. Proof. See [Es] p Lus Nuñez-Betancourt and Lnquan Ma each defned local cohomology for M wth support n I n semnar talks pror to ths one. A good source s [Iyen]; local cohomology s fnally defned n Chapter 7. However, the chapters leadng up to t provde an excellent explanaton of the dffcult concepts needed to even defne t. 4 For a constructon of drect lmts and nverse lmts, see [Hoch] p Ext s called the extenson functor. Appendx 3 of [Es] gves the constructon of ths homologcal algebra tool, as well as the constructon of the Tor functor. For a much more general ntroducton to δ-functors see Chapter 2 of [Web]. 1

2 2 where Ann M I t s the annhlator of I t n M. So another way we can defne local cohomology s to use the rght derved functors of Γ I? = t Ann?I t. Explctly, create an exact sequence of R-modules 0 M N 1 N 2 where the N satsfy the followng property: for every deal J R, every homomorphsm J N s actually the restrcton of a homomorphsm defned on all of R. 6 Omt M and apply the functor Γ I? to each term of the sequence. The cohomology of the resultng sequence wll be the same as the local cohomology defned above. Both of these defntons nvolve drected systems wth deals I t, and ths observaton makes t clear why local cohomology s the same up to radcals of I. Suppose f 1,..., f n R and rad I = rad f 1,..., f n R. A thrd way to compute local cohomology s to use the drect lmt of the Kozsul complexes K f t 1,..., f t n; M. Secton 7 of [HW11] descrbes the constructon explctly whle Chapter 17 of [Es] defnes more general Koszul complexes and ther propertes beyond the scope of our context 7. We frst need the noton of the tensor product of complexes. For K and L complexes of R-modules wth respectve dfferentals d K and d L, let M = K R L denote the complex 1 M h = +j=h K R L j wth 2 dfferental d h a b = d K a b + 1 a d L b, where a K, b L j, and + j = h. Exercse. M s a complex of R-modules. For N complexes K 1,..., K N defne the tensor product recursvely as K 1 R R K N K = 1 R R K N 1 R K N. Now for f R, defne the Koszul complex K f; R = 0 R R 0, where the mddle map s multplcaton by f. Then for f 1,..., f n R, defne the Koszul complex n K f 1,..., f n ; M = K f ; R R M, meanng, tensor M wth each term of the complex K f 1,..., f n ; R = n K f ; R. 6 Such a module s called njectve. The exact sequence 0 M N 1 N 2 s called an njectve resoluton of M. See Appendx 3.4 of [Es] for a more detaled explanaton of njectve resolutons. 7 The Koszul complex we care about wll be a Cech complex.

3 Notaton. When the context s clear, wrtng f n place of f 1,..., f n s easer and qucker, as n K f; M. Let Let H f ; M denote ts cohomology. K f ; M = lm t K f t ; M = lm t K f t 1,..., f t n; M. Theorem. Suppose rad I = rad f 1,..., f n R. Then canoncally as functors of M. Proof. See [HW11]. H j I M = H f ; M An mmedate consequence s that local cohomology HI M vanshes when exceeds the number of generators for any deal wth the same radcal as I. There are other, slghtly less obvous consequences whch follow from the varous defntons of local cohomology: Exercse. A short exact sequence of R-modules nduces a long exact sequence 0 M M M 0 0 H 0 I M H 0 I M H 0 I M H 1 I M H I M H I M H I M H +1 I M. 3 Exercse. If R S s a rng homomorphsm, IS denotes the deal generated by the mage of I n S, and M s an S-module hence an R-module, then H I M = H IS M. Exercse. Localzaton at a maxmal deal does not change local cohomology. The Rng of Dfferental Operators Hochster s treatment of D-modules n [HD] s a lght survey of the results n the theory and omts many techncal detals. Hs man source s Jan-Erk Björk s book Rngs of Dfferental Operators. The ntent of ths talk s to keep the materal approachable to a Mchgan graduate student who has passed the Qualfyng Revew.e., a second or thrd year student. The rng we care about wll be D = DR, K, constructed as follows. Let K be a feld of characterstc 0 and let R denote the formal power seres rng K[[x 1,..., x n ]] n n varables over K. The rng of dfferental operators DR, K conssts of the K-vector space endomorphsms of R generated by multplcaton of elements n R and the usual

4 4 dfferental operators δ 1,,..., δ n defned formally 8. For the rest of ths talk, D, R, and K are defned as above. 9 Exercse. The δ commute wth eachother. Exercse. Thnkng of the x j as operators on R, meanng for f R, x j f s just the product x j f, f j then δ x j x j δ = 0. Exercse. Agan, thnkng of the x as operators on R, δ x x δ = 1. Exercse. More generally, for any f R, thought of as an operator on R, δ f fδ = f. Exercse. D s R-free on the monomals n the δ, both as a left and as a rght R-module. Holonomc D-Modules Say A s an assocatve not necessarly commutatve rng wth 1, and has an ascendng fltraton Σ = Σ 0 Σ 1 Σ 2, meanng the followng condtons hold: 1 each Σ s an addtve subgroup, 2 1 Σ 0, 3 Σ = A, and 4 Σ Σ j Σ +j for all, j. Furthermore, we nsst the fltraton s such that the assocated graded rng gra = Σ 0 Σ 1 /Σ 0 Σ /Σ 1 s commutatve and Noetheran. Such a rng A s called a fltered rng. Exercse. The fltraton on A actually does gve gr A a rng structure. D as defned above has a fltraton Σ n whch Σ conssts of all R-lnear combnatons of monomals n degree at most n the δ j. 8 When dfferental operators δ are defned formally t means for f R, δ f = f where the partals are not really lmts as n the usual defnton of a dervatve. Rather, they are smply defned by the power rule for dfferentaton. Ths s why we need to be n characterstc 0. A rng of dfferental operators does exst when K s characterstc p > 0, but ts constructon s more complcated. 9 D s sometmes called the Weyl algebra, though the usual defnton uses a polynomal rng R nstead of a formal power seres rng.

5 Exercse. The assocated graded rng gr D s somorphc to a polynomal rng R[ζ 1,..., ζ n ] n n varables over R, where ζ s the mage of δ n Σ 1 /Σ 0. In partcular, gr D s commutatve, Noetheran, and regular 10. Its Krull dmenson s 2n. A fntely generated left A-module M has a good fltraton Γ = Γ 0 Γ 1 Γ 2 means t satsfes the followng: 1 {Γ } are abelan subgroups, 2 Γ = M, 3 Σ Γ j Γ +j for all, j N so far these are just the condtons that make t a fltraton, and the assocated graded module gr Γ M = Γ /Γ 1 s fntely generated over gr A ths last condton s the condton for the fltraton to be good. Björk proved the followng: Proposton. A as descrbed above s both left and rght Noetheran 11. Moreover, an A- module M has a good fltraton f and only f t s fntely generated as an A-module and for {Γ } and {Γ } fltratons on M there exsts an nteger c wth 5 for all. Γ Γ +c and Γ Γ +c Corollary. By the above proposton, the Krull dmenson of gr Γ M s ndependent of the choce of good fltraton Γ. Proof. See p. 5 of [HD], where Hochster ctes Björk. From the above dscusson, A = D = DR, K s a fltered rng. The fntely generated D-modules whose assocated graded modules have Krull dmenson n, together wth the 0 module, consttute the left Bernsten class 12. There s a smlar noton for the rght Bernsten class and n fact, a dualty between these two categores holds. 10 A local rng wth maxmal deal m s regular means m can be generated by exactly d elements, where d s the Krull dmenson of the rng. The Krull dmenson of a rng R s the supremum of the lengths of chans of prme deals n R; the Krull dmenson of an R-module M s the Krull dmenson of the rng R/Ann M. More generally, a rng R s regular means all of ts localzatons at maxmal deals are regular. 11 When A s not necessarly commutatve the Noetheran condton may hold for left or rght deals. 12 To defne the more general noton for a fltered rng A requres understandng of weak global dmenson.

6 6 Theorem 13. The modules n the Bernsten class have fnte length as D-modules,.e., each nonzero module n the Bernsten class has a fnte fltraton by smple D-modules each of whch s agan n the Bernsten class. Proof. See p. 7 of [HD]. Notaton. The D-modules n the Bernsten class are called holonomc. Proposton. 1 Submodules and quotents of holonomc D-modules are also holonomc. 2 For a short exact sequence of D-modules 0 M 1 M 2 M 3 0 M 2 s a holonomc D-module f and only f both M 1 and M 3 are. Proof. See p. 6 of [HD]. Exercse. R s a holonomc D-module. Exercse. If W s any multplcatve system n D and M s a D-module, then W 1 M has the structure of a D-module n such a way that the map M W 1 M s a homomorphsm of D-modules. Applcatons to Local Cohomology Theorem Björk. Wth D = DR, K as defned above, f M s a holonomc D-module and f R, then the localzaton M f s a holonomc D-module. Ths dffcult result leads to many applcatons to local cohomology theory. Corollary. The local cohomology modules HI M all have the structure of D-modules n such a way that f M s holonomc, then so are HI M. Proof. Wrte I = f 1,..., f s R. We can use the Kozsul complex K f ; M to compute HI M. Ths s 0 M M f <j M f f j M f1 f s 0 where the maps are alternatng sums of the natural localzaton maps. Thus by an above exercse the Koszul complex s a complex of D-modules. In partcular, the local cohomology modules are D-modules. If M s holonomc then the theorem says so are the M f. To see drect sums of holonomc D-modules are holonomc, consder the splt exact sequence 0 M 1 M 1 M 2 M A more general statement s true for any fltered rng A.

7 where M 1 and M 2 are holonomc. Another exercse from above shows the mddle term M 1 M 2 s holonomc as well. Then for more than two summands use nducton to show holonomcty. Fnally, one of the above exercses shows the kernels, whch are submodules, and the cohomology modules, whch are quotents, must be holonomc as well. In partcular, HI M must be holonomc f M s. For any rng S and S-module N, a prme deal P S s assocated to N means P annhlates klls an element of N. The set of all prmes assocated to N s denoted Ass S N and s sometmes called the assassnator of N. Exercse. If S s Noetheran then Ass S N s nonempty. Corollary. If M s a smple 14 D-module then the assassnator of M as an R-module contans a unque element P note by constructon any D-module s also an R-module. Hence, f M s a holonomc D-module then Ass R M s fnte. Proof. Suppose M s smple. Frst of all, R s Noetheran mples M has an assocated prme, P. Then HP 0 M = Ann M P t t s a nonzero submodule of M, hence s equal to M. If Q s another assocated prme then H 0 QM = M = H 0 P M. Local cohomology modules wth support n dfferent deals are the same as long as those deals have the same radcal. But P and Q are prme and so they are already radcal. Therefore, P and Q are the same deal. When M s holonomc t has a fnte fltraton by smple modules. Therefore f P annhlates an element of M, t annhlates an element n one of those smple modules and P s the unque assocated prme for that smple module. So M can only have fntely many assocated prmes. By nducton, f I 1,..., I s s a sequence of deals of R, 1,..., s s a sequence of ntegers, and M s a holonomc D-module, then the terated local cohomology module H s I s H s 1 I s 1 H 1 I 1 M s a holonomc D-module. In partcular, H s has a fnte assassnator. I s H s 1 I s 1 H 1 I 1 R s holonomc and so Proposton. Let m = x 1,..., x n R denote the homogeneous maxmal deal 15 n R. Then D/Dm = H n mr as a D-module A module s smple means t has no nontrval submodule. 15 Homogeneous means every element n m has the same degree n the ndetermnates x. 16 Hm n R s also the unque smallest njectve R-module contanng R/m, called the njectve hull of R/m n R. 7

8 8 Proof. Let S denote the polynomal rng K[x 1,..., x n ] and let Q = x 1,..., x n S. Exercses from earler gve H n mr = H n QS Q S Q = H n QS. So we can nstead show D/Dm = HQ ns. The Koszul complex K x ; S gves n HQS n = Coker S x1 ˆx x n S x1 x n where the hat means that ndetermnant s omtted. It turns out HQ n S s the K-span of the monomals x j 1 1 x j n n where j are strctly negatve. From an earler exercse D s R-free as a rght R-module on the monomals n δ 1,..., δ n ; therefore D/Dm s K-free on the span of the mages of the monomals n the δ. There s a D-lnear map D HQ n S that sends 1 x 1 1 x 1 n. Ths map must kll annhlate m and hence also klls the left deal Dm, nducng a D-lnear map D/Dm HQ ns. For ths map the mage n Hn Q S of the element represented by δ 1 1 δ n n s represented by 1 n n 1 1 x 1 1 x 1 n = n j!x j+1 j. n Ths means the nduced D-lnear map carres a K-vector space bass to a K-vector space bass, so s a D-somorphsm. The fnal results frst requre some structure theory for njectve modules 17. Let S be a commutatve, assocatve rng wth 1. A homomorphsm of S-modules h : N M s an essental extenson means t s njectve and the followng equvalent condtons hold: Every nonzero submodule of M has a nonzero ntersecton wth hn. Every nonzero element of M has a nonzero multple n hn. If ϕ : M Q s a homomorphsm such that ϕ h s njectve then ϕ s njectve. Suppose S s also Noetheran and local wth maxmal deal m, and suppose M s an S- module such that every element of M s klled by a power of m. The socle n M s defned as SocM = Ann M m. In ths specal case where every element of M s klled by a power of m, Soc M s the largest submodule of M whch may be vewed as a vector space over S/m; any larger submodule would contan an element not klled by m, so could not be a vector space over S/m. Furthermore, the ncluson Soc M M s an essental extenson. To see ths, choose x M nonzero and let t denote the largest nteger such that m t x 0. Then we can choose y m t such that yx 0. The ncluson 17 See [HW11]. my mm t x = 0 j=1

9 9 mples y Soc M and hence the nonzero multple yx Soc M. Exercse. In general, f N M s an essental extenson then Soc M N. Here s the result on D-modules: Proposton. Wth m as before, the homogeneous maxmal deal of R = K[[x 1,..., x n ]], f M s any D-module no fnteness condtons on M ths tme such that every element s klled by a power of m, then M s somorphc wth a drect sum of copes of D/Dm. When M s holonomc the drect sum s fnte. Proof. By the dscusson above M s an essental extenson of a K-vector space V M, and we may choose a K-vector space bass {v λ } λ Λ for V. The use of Λ to denote the ndexng set ndcates the bass may be nfnte. Consder a free D-module G wth free generators {u λ }, also ndexed by Λ. Each copy of D n G s both a left and rght module over D, so G s both a left and rght module over D. Therefore G s both a left an a rght module over R. Ths wll matter later n the proof when we are consderng G/Gm as an R-module. Defne the D-lnear map G M by u λ v λ for each λ Λ. Ths nduces a map G/Gm M whch sends the mages of the u λ to the correspondng elements n the bass v λ for V. The goal s to show ths nduced map s the somorphsm we want. Because G was chosen as a drect sum of copes of D ndexed by Λ, G/Gm may be dentfed wth a drect sum of copes of D/Dm ndexed by Λ. Thus, G/Gm s an essental extenson of a K-vector space, SocG/Gm. The bass on SocG/Gm s nduced by the mages of the u λ n G/Gm, so s ndexed by Λ. Each of these bass elements maps to ts correspondngly ndexed bass element v λ n V. That bjecton between vector space bases shows SocG/Gm s mapped somorphcally onto V. As an somorphsm, n partcular, the map G/Gm M s njectve on SocG/Gm. Suppose a nonzero element n G/Gm maps to zero. Then because SocG/Gm G/Gm s an essental extenson, a nonzero multple of that element n SocG/Gm s mapped to zero, contradctng njectvty. So G/Gm M must also be njectve. In fact, the map splts as a map of R-modules: If I s an deal n R and ϕ : I G/Gm an R-lnear map, then φ must kll m. So we can actually thnk of ϕ as a map I/Rm G/Gm. Snce m s a maxmal deal, I/Rm must be ether zero or the unt deal. In ether case though, φ wll extend to a map from R. From ths we can conclude G/Gm s an njectve R-module, therefore ts njecton nto M splts. 18 Wrte M = G/Gm M 0 as the R-module splttng. As a submodule of M, every element of M 0 s klled by a power of m that was a hypothess on M. Therefore f M 0 0, ts socle s nonzero. The socle of M 0 s contaned n the socle of M, whch s contaned n V. But V s contaned n the mage of G/Gm, whch only meets M 0 at zero, by constructon. Therefore M 0 = 0 and we have M = G/Gm 0 = G/Gm, 18 The splttng of an njectve map s a general property of njectve modules.

10 10 as requred. Notaton. To save wrtng space, let the symbol T denote an teraton of local cohomology functors. Corollary. Let T be a composton of local cohomology functors as descrbed just above. If T R has the property that every element s klled by a power of m, then T R s somorphc as a D-module to a fnte drect sum of copes of D/Dm, and so s njectve. Proof. T R s a holonomc D-module by a result above. Then the desred result follows from the last two propostons about D-modules. For more results, agan, see Mel Hochster s notes. APPENDIX: SOLUTIONS TO EXERCISES Exercse. M s a complex of R-modules. Soluton. We constructed M so that ts terms are R-modules and ts dfferentals are R-lnear. We just need to check that the compostons M h 1 M h M h+1 are all zero. Say a K and b j L j, for all, j. Apply d h 1 to a typcal element n M h 1 : d h 1 +j=h 1 a b j = +j=h 1 = +j=h 1 The resultng terms are n M h. Now apply d h : d h +j=h 1 = +j=h 1 = +j=h 1 = +j=h 1 = = 0. +j=h 1 d h 1 a b j dk a b j + 1 a d L b j d h d k a b j + 1 a d L b j dk a b j + 1 a d L b j d h d K a b j + d h 1 a d L b j dk d K a b j d K a d L b j + dk 1 a d L b j + 1 a d L d L b j d K a d L b j + 1 d K a d L b j + 0

11 We appled the composton to an arbtrary element and got zero, so the composton map tself must be zero. Exercse. A short exact sequence of R-modules nduces a long exact sequence 0 M M M 0 0 H 0 I M H 0 I M H 0 I M H 1 I M H I M H I M H I M H +1 I M. 11 Soluton. Ext s a rght derved functor, hence a unversal cohomologcal δ-functor. Thus, by defnton 19, nduces the long exact sequence 0 Hom R R/I t, M Hom R R/I t, M Hom R R/I t, M Ext 1 RR/I t, M Ext 1 RR/I t, M Ext 1 RR/I t, M Ext RR/I t, M Ext RR/I t, M Ext RR/I t, M. To actually get local cohomology though, we have to take the drect lmt. But the drect lmt s a fltered colmt 20, so s exact. In other words, applyng lmt to each term of the above long exact sequence produces precsely the desred long exact sequence on local cohomology. Exercse. If R S s a rng homomorphsm, IS denotes the deal generated by the mage of I n S, and M s an S-module hence an R-module, then H I M = H IS M. Soluton. M s an R-module by restrcton of scalars 21,.e., multplcaton of M by f R s dentcal to multplcaton by the mage of f n S. Let f 1,..., f n denote the generators of I and g 1,..., g n the respectve generators of IS. Both Koszul complexes K f ; M and K g ; M consst of localzaton maps and the remark above gves an dentfcaton of M f wth M g. Therefore the Koszul complexes and hence, the cohomology, are the same. Exercse. Localzaton at a maxmal deal does not change local cohomology. Soluton. Localzaton at a maxmal deal s a rng homomorphsm, so ths s a specal case of the prevous exercse. 19 Though very formal, [Web] has the best justfcaton for these statements. 20 See [Web]. Ths exercse also appears as a Dscusson n [HW11]. 21 See Chapter 2 of [AtyM].

12 12 Exercse. The δ commute wth eachother. Soluton. Choose an arbtrary element f R and do the computaton: as desred. δ δ j δ j δ f = δ δ j f δ j δ f = δ f δ j f x j = f x j = 2 f x j = 0, x j f 2 x j f Exercse. Thnkng of the x j as operators on R, meanng for f R, x j f s just the product x j f, f j then δ x j x j δ = 0. Soluton. As n the prevous exercse, choose an arbtrary element f R and do the computaton: as desred. δ x j x j δ f = δ x j f x j δ f = x j f x j f = 0 f + x j f x j f = 0, Exercse. Agan, thnkng of the x as operators on R, δ x x δ = 1. Soluton. Applyng the left hand sde to arbtrary f R, we want the result to be f. δ x x δ f = δ x f x δ f = x f x f = 1 f + x f x f = f, so δ x x δ s n fact the dentty operator. Exercse. More generally, for any f R, thought of as an operator on R, δ f fδ = f.

13 13 Soluton. Ths tme f s thought of as an operator, so take g R: the desred result. δ f fδ g = δ fg fδ g = fg f g = f g + f g f = f g f = g, g Exercse. D s R-free on the monomals n the δ, both as a left and as a rght R-module. Soluton. The prevous few exercses gve a way to wrte an element of D that has an expresson that s not a sum of two or more elements n D as an element of R multpled on the rght by a monomal n the δ : as shown above, the δ commute wth eachother; R s already commutatve so we only need a way to wrte δ j f, for f R, as an expresson consstng of an element of R multpled on the rght by a monomal n the δ. The last exercse gves δ f = fδ + f. The commutatvty of elements n R wth eachother and of the δ wth eachother mples such an expresson s unque. So D s R-free on the monomals n the δ, as a left R-module. Lkewse, for f R we have fδ = δ f f by the prevous exercse. Ths fact, along wth the commutatvty of elements n R wth eachother and of the δ wth eachother gves every element of D whch has an expresson not a sum of two or more elements n D a unque expresson as an element of R multpled on the left by a monomal n the δ. Thus D s also R-free on the monomals n the δ, as a rght R-module. Exercse. The fltraton on A actually does gve gr A a rng structure. Soluton. We just have to check the rng axoms for gr A. Defne addton and multplcaton n gr A as the respectve nduced operatons from A. Any two elements n A le n Σ, for some, and so upon takng quotents, gr A s stll a commutatve group under addton. Smlarly, the property Σ Σ j Σ +j ensures any two elements can be multpled wthn one of the subgroups n the fltraton Σ, and so assocatvty s also preserved upon takng quotents. Snce 1 Σ 0, t remans the unt element n gr A. Fnally, snce both rng

14 14 operatons can be done n one common subgroup Σ, the dstrbutve laws reman ntact upon takng quotents. Exercse. The assocated graded rng gr D s somorphc to a polynomal rng R[ζ 1,..., ζ n ] n n varables over R, where ζ s the mage of δ n Σ 1 /Σ 0. In partcular, gr D s commutatve, Noetheran, and regular. Its Krull dmenson s 2n. Soluton. For each 1, elements of Σ /Σ 1 are n bjecton wth the R-lnear combnatons of monomals n degree exactly n the δ j. In partcular, Σ 1 /Σ 0 s generated over R by the mages of the δ. The suggested map δ + Σ 0 ζ s automatcally R-lnear, so addton and multplcaton agree wth the respectve operatons n R. Commutatvty between elements of R and the varables wll follow by nducton. The base case s straghtforward: for f R, δ f fδ = f Σ 0, so vanshes n the quotent. To smplfy notaton n the nductve case, for any d D, let d j denote the mage of d n Σ j /Σ j 1. So n the base case, for example, δ f fδ = f = 0. Now say d D has the expresson d = f n where f R and j j n = j. Assume commutatvty holds for elements n Σ j /Σ j 1. Then j j 1 j j 1 1 n n 1 δ l d d δl = δl f f δ l 1 0 n = δ l f = f 0 1 n δ l = f 0 n = 0. δ l j j j+1 f 0 n f 0 n f 0 n j j δ l δ l 1 δ l 1 j+1 Verfyng the commutatvty of the rng gr D ensures the map to R[ζ 1,..., ζ n ] preserves rng operatons. Furthermore, the bjecton between δ 1 and ζ ensures the map s a rng somorphsm.

15 The addtonal statements commutatve, Noetheran, regular, Krull dmenson 2n are already true for R[ζ 1,..., ζ n ] and do not change under rng somorphsm, hence are true for gr D. 15 Exercse. R s a holonomc D-module. Soluton. Frst of all, R s fntely generated as a D-module: defne the D-map D R d d1 for each d D. Then R = D/Ann D 1, so only needs one generator. We have to check the Krull dmenson of gr Γ R, assumng a good fltraton Γ exsts. Defne Γ = {Γ } by Γ = Σ R, where Σ R s agan the subgroup of all R-lnear combnatons of monomals n degree at most n the δ. Then Σ R = R for all and hence Γ satsfes all the condtons to be a fltraton. The assocated graded module s then gr Γ R = Γ /Γ 1 = R/0 R/R R/R = R where by conventon Γ = 0 for all < 0. Recall the assocated graded rng for D s a polynomal rng n n varables over R, so R s agan generated by one element over D. The Krull dmenson of R s n because R = K[[x 1,..., x n ]]. Therefore R s a holonomc D-module. Exercse. If W s any multplcatve system n R and M s a D-module, then W 1 M has the structure of a D-module n such a way that the map M W 1 M s a homomorphsm of D-modules. Proof. Frst extend the acton of D to an element m w W 1 M, where m M and w W. Elements n R already act as usual, by multplcaton, snce w R. For = 1,..., n, δ m w = w δ m m w w 2 W 1 M. The D-lnearty s carred over from D-lnearty over M and the lnearty of multplcaton and partal dervatves. The natural map M W 1 M gven by m m/1 s already R-lnear. To get D-lnearty, take = 1,..., n and δ m δ m 1 = δ m 1 = 1 δ m m = δ m 1,

16 16 as t should. Exercse. If S s Noetheran then Ass S N s nonempty. Proof. Choose a nonzero element u N. The followng set A = {Ann ru r S, ru 0} s a famly of deals n S. A s nonempty because t contans the zero deal. All the deals n A are proper, because f S klls u then so must 1, meanng u = 0, a contradcton to the choce of u. S s Noetheran, so A must have a maxmal element, I = Ann ru, for some r S. By constructon ru 0 so we can just rechoose u so that maxmal deal n A wll be I = Ann u. Showng I s prme wll show Ass S N s nonempty. Suppose a, b / I and ab I. Snce b s not n I, bu 0. On the other hand, I bu = 0, because abu = 0. Therefore I + Sa klls u, so s also n A. But I I + Sa S was chosen to be maxmal, so a must also be n I. Exercse. In general, f N M s an essental extenson then Soc M N. Proof. It s enough to show any nonzero element n Soc M les n N, because zero already does. So choose nonzero u Soc M. By defnton, the maxmal deal m S klls u, but S does not by the selecton of u. So Su = S/Ann u = S/m, a feld. On the other hand, N M s an essental extenson, so n partcular, u les n a nonzero submodule contaned n N. But u generates S/m, the smallest nonzero S-module, and so S/m N. In partcular, u N. Bblography [AtyM] M.F. Atyah and I.G. MacDonald, Introducton to Commutatve Algebra, Advanced Book Program, Westvew Press, A Member of the Perseus Books Group, Boulder, Colorado, [Es] Davd Esenbud, Commutatve Algebra wth a Vew Toward Algebrac Geometry, Graduate Texts n Mathematcs 150, Sprnger Scence+Busness Meda, Inc., New York, NY, [HD] Mel Hochster, D-modules and Lyubeznk s Fnteness Theorems for Local Cohomology, Notes, see Unpublshed, [Hoch] Mel Hochster, Math 614 Lecture Notes, Fall, 2010, Course Notes, see umch.edu/~hochster/614f10/614.html, Unpublshed, [HW11] Mel Hochster, Local Cohomology, Course Notes, see 615W11/loc.pdf, Unpublshed, [Iyen] Srkanth B. Iyengar, Graham J. Leuschke, Anton Leykn, Clauda Mller, Ezra Mller, Anurag K. Sngh, Ul Walther, Twenty-Four Hours of Local Cohomology, Graduate Studes n Mathamatcs, Volume 87, Amercan Mathematcal Socety, Provdence, RI, [Web] Charles A. Webel, An ntroducton to homologcal algebra, Cambrdge studes n advanced mathematcs, 38, Cambrdge Unversty Press, Cambrdge, Unted Kngdom, 1994.