IDEALS AND THEIR INTEGRAL CLOSURE

Transcription

1 IDEALS AND THEIR INTEGRAL CLOSURE ALBERTO CORSO (joint work with C. Huneke and W.V. Vasconcelos) Department of Mathematics Purdue University West Lafayette, San Diego January 8,

2 SETTING Let R be a Noetherian ring a and I one of its ideals. The integral closure of I is the ideal I of all elements of R that satisfy an equation of the form X n + a 1 X n a n 1 X + a n = 0, a i I i. The radical I consists of all the solutions in R of equations of the form X m b = 0, b I. In particular: I I I. I is integrally closed (resp. normal) if I = I (resp. I n = I n for all n). a Later we may even want to assume the ring to have other additional properties (such as: Cohen Macaulay, with infinite residue field, etc. 2

3 PROBLEMS For an ideal I of R we would like to give an answer to questions of the following kind: Find efficient and global criteria to test whether or not I is integrally closed. Comment: The criteria should involve natural objects associated with I: e.g. the powers of I, the radical of I, modules of syzygies, etc. If I fails the above tests, find a procedure to compute (part of) the integral closure of I. Comment: It is not an easy task. For example, if R is a polynomial ring over a field and I is a monomial ideal, I is then the monomial ideal defined by the integral convex hull of the exponent vectors of I. However, if I is binomial then I need not be binomial. When I integrally closed implies I normal. 3

4 POSSIBLE APPROACHES A consequence of the determinant trick is that for every finitely generated faithful R-module M then I IM: M I. In particular if I is integrally closed, for any such M, IM: M = I. We then need appropriate test modules for a given ideal I. An alternative approach is through the Rees algebra R[It] = R + It + I 2 t I n t n + of I; one then looks for its integral closure inside R[t]: R + It + I 2 t 2 + R[t]. This is obviously wasteful of resources since the integral closure of all powers of I will be computed. Of course this could be profitably taken if it turns out that I is normal. 4

5 CRITERION Let I be a height unmixed ideal in a Cohen Macaulay ring R. Suppose that I is generically a complete intersection. Then the following conditions are equivalent: I is integrally closed; I = IL: L, where L = I: I. Remarks: An earlier version of this criterion had the condition I = IL: L replaced by I = IL: L 2. The proof in both cases is essentially the same, and it is based on the following two local criteria. 5

6 Theorem a : A LOCAL RESULT OF GOTO Let I be an ideal in a Noetherian ring R and assume that µ R (I) = height R (I) = g. Then the following conditions are equivalent: I = I, i.e. I is integrally closed; I n = I n, i.e. I is normal; for each p Ass R (R/I), the local ring R p is regular and ( IRp + p 2 ) R p λ Rp p 2 g 1. R p When this is the case, Ass R (R/I) = Min R (R/I) and I is generated by an R-regular sequence. a S. Goto: Integral closedness of complete-intersection ideals, J. Algebra 108 (1987),

7 Theorem a+b : LINKAGE AND REDUCTION NUMBER Let (R, m) be a Cohen Macaulay local ring and let I = (z 1,..., z g ) be an ideal generated by a regular sequence inside a prime ideal p of height g. If we set L = I: p then L 2 = IL if one of the following two conditions holds: (l 1 ) R p is not a regular local ring; (l 2 ) R p is a regular local ring with dimension at least 2 and two of the z i s in p (2). a A. Corso, C. Polini and W. V. Vasconcelos: Links of prime ideals, Math. Proc. Camb. Phil. Soc. 115 (1994), b A. Corso and C. Polini: Links of prime ideals and their Rees algebras, J. Algebra 178 (1995),

8 ANOTHER USEFUL CRITERION Let R be a Gorenstein ring and let I be a Gorenstein ideal of codimension 3. Then the following conditions are equivalent: I is generically a complete intersection; I 2 : I = I. Remarks: In particular, if I is integrally closed then it is generically a complete intersection. Furthermore, one could combine this and the previous criterion to get a condition for a perfect Gorenstein ideal of codimension 3 to be integrally closed. However, this is not necessary as the next result shows. 8

9 GORENSTEIN IDEALS Let I be a Gorenstein ideal of codimension g 2 of a regular local ring R. Then the following conditions are equivalent: I is an integrally closed ideal; I = IL: L, where L = I: I. Comments: As a consequence of the proof, I is generically a complete intersection of the kind described in Goto s paper. Is it possible to find an easy, global criterion to characterize integrally closed ideals with type 2 (3, etc...)? 9

10 INTEGRAL CLOSEDNESS AND NORMALITY: ideals with linear presentation Theorem: Let k be a field of characteristic zero and let I R = k[x 1,..., x d ] be a Gorenstein ideal defined by the Pfaffians of a n n skew symmetric matrix ϕ with linear forms as entries. Suppose n = d + 1 (hence d is even) and that I is a complete intersection on the punctured spectrum. If I is integrally closed it is also normal. Remark: The proof is based on the Jacobian criterion. Corollary: Let k be a field of characteristic zero and let I R = k[x 1, x 2, x 3, x 4 ] be a Gorenstein ideal defined by the Pfaffians of a five by five skew symmetric matrix ϕ with linear forms as entries. If I is integrally closed it is also normal. 10

11 (COUNTER)EXAMPLE Let R = k[a, b, c, d] with a, b, c, d be variables and char(k) = 0. The Pfaffians of the 5 5 matrix ϕ 0 a 2 b 2 c 2 d 2 a 2 0 d 2 ab c 2 ϕ = b 2 d 2 0 a 2 ab, c 2 ab a 2 0 b 2 d 2 c 2 ab b 2 0 define an height 3 Gorenstein ideal I such that I 2 : I = I and I = IL: L, where L = I: I. Hence I is integrally closed BUT it is not normal. Note that I 2 is integrally closed as well. Is I 3 BAD? 11

12 INTEGRAL CLOSEDNESS AND NORMALITY: complete intersections of codimension 2 Theorem: Let R be a regular local ring and I = (a, b) a complete intersection of codimension 2. Then I is normal, i.e., for all n 1. I n = (I) n Remark: For dimension 2 this is a very well known result of Zariski. 12

13 A METHOD TO COMPUTE THE INTEGRAL CLOSURE If I = (a 1,..., a n ) then one can represent its Rees algebra as R[It] = R[T 1,..., T n ]/P, where P is the kernel of the map ϕ : R[T 1,..., T n ] R[It] that sends T i to a i t. If R[It] is an affine domain over a field of characteristic zero and Jac denotes its Jacobian ideal, then the ring Hom R[It] (Jac 1, Jac 1 ) = = (Jac Jac 1 ) 1, is guaranteed to be larger than R[It] if the ring is not already normal. Naturally, this process can be repeated several times until the integral closure of R[It] has been reached. The degree one component of the final output gives the desired integral closure of I. 13

14 EXAMPLE Let k be a field of characteristic zero and let I R = k[x, y] be the codimension 2 complete intersection I = (x 3 + y 6, xy 3 y 5 ). Iterating three times the method outlined before we can compute I. To be precise, the three outputs are: J 1 = (x 3 + y 6, xy 3 y 5, y 8 ), J 2 = (x 3 + y 6, xy 3 y 5, x 2 y 2 y 6, y 7 ), J 3 = I = (xy 3 y 5, y 6, x 3, x 2 y 2 ). Note that I is also a normal ideal. For the records, despite the fact that the original setting for the problem is a polynomial ring in 2 variables over a field of characteristic zero, overall we had to make use of 18 additional variables: quite a waste! 14

15 FINAL COMMENT If I is a complete intersection, several initial iterations of the process may be avoided (see [CP2] and [PU]). If I is an m-primary complete intersection of a Gorenstein local ring (R, m) and I m s but I m s+1 then one has an increasing sequence of ideals I k = I: m k satisfying I 2 k = II k for k = 1,..., s if dim(r) 3 or for k = 1,... s 1 if R is a regular local ring and dim(r) = 2. This says that the I k s are contained in the integral closure of I. Hence, instead of computing the integral closure of R[It] one may start directly from R[I s t] (or R[I s 1 t] if R is a regular local ring and dim(r) = 2). 15

16 If I is not primary to the maximal ideal one may use instead the sequence of ideals I k = I: ( I) (k) = I: ( I) k, provided that at each localization at the associated primes of I the conditions in [CP2] and [PU] are satisfied so that I 2 k = II k for all the k s in the appropriate range. In the case of the example: R = k[x, y] and I = (x 3 + y 6, xy 3 y 5 ) one has that I (x, y) 3. Since dim(r) = 2 one can consider only the cases k = 1, 2 = s 1. We can check that J 1 = I: m J 2 = I: m 2. 16