# APPLICATIONS OF LOCAL COHOMOLOGY

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2 2 TAKUMI MURAYAMA () H i J (R) = Hi J (R), and (2) if J is generated by k elements, then HJ i (R) = 0 for all i > k. These modules are what we will call local cohomology modules. 2. Local cohomology as a derived functor Just like sheaf cohomology, there are multiple ways to define local cohomology. We will describe the derived functor construction first. Let R be a noetherian ring, I R an ideal, and M an R-module. We define the I-power torsion module of M to be HI 0 (M) := {m M I d m = 0 for some d N} = lim Hom(R/I d, M). Note that Hom(R/I d, ) is left exact and lim is exact, and so H 0 I is a left exact functor. Like any left exact functor, we can define its right derived functors, by choosing an injective resolution 0 M E 0 E, applying the functor HI 0( ) term by term to get a new complex H0 I (E ), and then computing cohomology to get modules HI i (M). We automatically get some really nice properties! Properties 4. () If I and J have the same radical, they define the same functor HI i ( ) (Property () from before). (2) If 0 N M L 0 is a short exact sequence of R-modules, there is a long exact sequence 0 H 0 I (N) H 0 I (M) H 0 I (L) H I (N) H I (M). (3) Every element of HI i (M) is killed by some power of I. (4) Since lim is exact, we can compute local cohomology as HI(M) i = lim Ext i (R/I d, M). Before we get too carried away with technical details, let s compute some examples. Example 5. Let p Z be a prime number. We will compute HI i (Z), where I = (p). Since Z is a PID, the injective modules are exactly the divisible modules. We have the following injective resolution of Z: 0 Z Q Q/Z 0. Now since HI 0 simply computes pd -torsion for all d, the only non-vanishing term after applying HI 0 ( ) is HI 0(Q/Z), and so we have HI (Z) = HI 0 (Q/Z) = Z[p ]/Z, where the last isomorphism is by unique factorization. Similarly, we can compute the local cohomology of R = k[x], where k is a field: Exercise 6 ([Hun07, Ex. 2.2]). Let R = k[x] be a polynomial ring in one variable over a field k. Our goal is to completely describe HI i (M) for I = (x) and M an arbitrary finitely generated module. () Using the structure theorem for finitely generated modules over a PID, reduce to the case of computing HI i (R/(g)) for g R. Proof. The structure theorem says M = s R/(g j ) j= for some elements g j R, possibly equal to zero. We conclude by the fact that computing local cohomology commutes with direct sums. (2) First compute the case when g = 0 by considering the short exact sequence where K = k(x) is the fraction field of k[x]. 0 R K K/R 0,

3 APPLICATIONS OF LOCAL COHOMOLOGY 3 Proof. As was the case for HI i(z), the only nonzero term after applying H0 I ( ) is H0 I (K/R), and so H I (R) = H 0 I (K/R) = k[x, x ]/k[x], using unique factorization as before. Note this has a k-basis x n n such that x x = 0. (3) Now compute the case when g 0 by considering the short exact sequence 0 R g R R/(g) 0. Hint: Writing g = x n h for (h, x) =, note that h acts as a unit on HI (R) since there exist a, b R such that ah = bx, and bx acts as a unit on this module. Proof. Using (b), we have that RI i (R) is nonzero if and only if i =, and so the long exact sequence on cohomology is 0 H 0 I (R/(g)) H I (R) g H I (R) H I (R/(g)) 0. Now consider the hint; bx indeed acts as a unit on HI (R) since it does on each basis element: x n ( bx)( + bx + b2 x b n x n ) = x n ( bn x n ) = x n. Now, HI 0(R/(g)) is the kernel by multiplication by xn, and HI (R/(g)) is the cokernel by multiplication by x n. The set of elements in H I (R) annihilated by xn is generated by x n, hence H I (R/(g)) = R/(x n ). Finally, HI (R) = k[x, x ]/k[x] is divisible by R, hence multiplication by x n is surjective, and so H (R/(g)) = Computing local cohomology One disadvantage of working with derived functors is that computing injective resolutions in practice is rather difficult. We already saw how the long exact sequence associated to a short exact sequence can be very useful, since we can break apart local cohomology modules into pieces we can understand. We first give another example where a similar breaking apart procedure is useful. 3.. The Mayer Vietoris sequence. In algebraic topology, we learn about the Mayer Vietoris sequence, which allows us to break apart a topological space into smaller pieces whose (co)homology we hopefully understand. We have a similar result for local cohomology: Theorem 7. Let I and J be ideals in a noetherian ring R, and let M be a finitely generated module M. Then, there is a long exact sequence in local cohomology 0 H 0 I+J(M) H 0 I (M) H 0 J(M) H 0 I J(M) H I+J(M) H J(M) H J(M). Proof. We will use the identification H i I (M) = lim Ext i (R/I d, M) from before. Take the short exact sequences 0 R/(I n J n ) R/I n R/J n R/(I n + J n ) 0 for each n, and consider their corresponding long exact sequences on Ext modules. Then, since the system {I n J n } is cofinal with {(I + J) n }, and since the system {I n + J n } is cofinal with {(I + J) n }, we can take direct limits over these long exact sequences to get the desired long exact sequence Čech cohomology. Just like with sheaf cohomology in algebraic geometry, in nice cases we have a very concrete way of computing local cohomology. This is called Čech cohomology. Let R be a noetherian ring, and let I be an ideal in R generated by elements (x,..., x t ). Write [t] = {,..., t} for the set of integers from to t, and for any subset J [t] let x J = j J x j. Denote M[x J ] to be the localization of M by inverting x J. Theorem 8. For any R-module M, the local cohomology module HI i (M) is the i-th cohomology of the complex t C(x,..., x t ; M) := 0 M d M[x d d i ] M[x J ] d d M[x {,...,t} ] 0 i= J =s

4 4 TAKUMI MURAYAMA where the differential takes an element to the element m J M[x J ] M[x J ] J =s d(m J ) = k / J( ) o J (k) m J {k}, where o J (k) denotes the number of elements of J less than k, and m J {k} denotes the image of m J in the further localization M[x J {k} ] = M[x J ][x k ]. We omit the proof, and instead point the reader to [Eis05, Thm. A.3] or [Hun07, Prop. 2.3]. We point out that this theorem has the following nice corollary, which is Property (2) from before: Corollary 9. If I = (x,..., x t ), then HI i (M) = 0 for all i > t. Proof. The length of the Čech complex C(x,..., x t ; M) is t. We have therefore found a nice functor that satisfies the properties we wanted! We now turn back to Example Hartshorne s example Exercise 0 (Hartshorne s Example [Har70, Exc. III.5.2; Har77, Exc. III.4.9]). I(Y ) = (x, y) (u, v) = (xu, xv, yu, yv) k[x, y, u, v]. We want to show that I(Y ) cannot be generated, up to radical, by two elements. () Using our two main properties, show that it suffices to show HI(Y 3 )(R) 0. Consider the ideal Proof. Suppose J is an ideal generated by two elements such that J = I(Y ). Then, by Corollary 9 of the identification of local cohomology with Čech cohomology, we have H3 J = H3 I(Y ) = 0, using our other main property that local cohomology does not change under taking the radical of J. (2) Use the Mayer Vietoris sequence on I(Y ) and the vanishing property to show H 3 I(Y ) (R) = H 4 m(r), where m = (x, y) + (u, v). Proof. The Mayer Vietoris sequence gives H 3 (x,y) (R) H3 (u,v) (R) H3 I(Y ) (R) H4 m(r) H 4 (x,y) (R) H4 (u,v) (R). The two edge modules vanish since (x, y) and (u, v) are generated by two elements each, and by applying Corollary 9. (3) Compute Hm(R) 4 using Čech cohomology. Hint: The relevant part of the Čech complex is R[y, u, v ] R[x, u, v ] R[x, y, v ] R[x, y, u ] R[x, y, u, v ] 0 Proof. The hard part is writing down the complex. Using the above representation of the complex, we obtain HI(Y 3 ) (R) = Hm(R) 4 = k x a x b 2x c 3x d 4 a, b, c, d < 0 0.

5 REFERENCES 5 5. Back to Curves In Example, we saw that it was fairly easy to construct examples of non-complete intersections by considering monomial curves in P 3. The twisted cubic ended up being a set-theoretic complete intersection. However, the generalization of this fact is not known: Open Question ([Har77, Exc. I.2.7(d)]). Can all irreducible curves in P 3 be defined set-theoretically by two equations? This question is surprisingly difficult: while local cohomology gives us an obstruction for a variety to be a set-theoretic complete intersection, if the relevant local cohomology module vanishes, we get no information. Even for the following simple example, we have no idea: Example. Consider the smooth rational quartic curve C in P 3 defined as the image of the map P P 3 [s : t] [s 4 : s 3 t : st 3 : t 4 ] It is known that HI(C) i (M) = 0 for all i > 2 and all modules M [Har70, Ch. III], so local cohomology does not provide an obstruction to C being a set-theoretic complete intersection. On the other hand, it is known that if we work over a field of characteristic p > 0, we do get a set-theoretic complete intersection [Har79]. References [Eis05] D. Eisenbud. The geometry of syzygies: A second course in commutative algebra and algebraic geometry. Graduate Texts in Mathematics 229. New York: Springer-Verlag, [Har67] R. Hartshorne. Local cohomology. A seminar given by A. Grothendieck, Harvard University, Fall Berlin: Springer-Verlag, 967. [Har70] R. Hartshorne. Ample subvarieties of algebraic varieties. Lecture Notes in Mathematics 56. Berlin: Springer-Verlag, 970. [Har77] R. Hartshorne. Algebraic geometry. Graduate Texts in Mathematics 52. New York: Springer-Verlag, 977. [Har79] R. Hartshorne. Complete intersections in characteristic p > 0. Amer. J. Math. 0.2 (979), pp [Hun07] C. Huneke. Lectures on local cohomology. Appendix by A. Taylor. Interactions between homotopy theory and algebra. Contemp. Math Amer. Math. Soc., Providence, RI, 2007, pp [Wei94] C. A. Weibel. An introduction to homological algebra. Cambridge Studies in Advanced Mathematics 38. Cambridge: Cambridge University Press, 994.

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