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Math 711: Lecture of November 7, 2007
Our current theory of test elements permits the extension of many results proved under other hypotheses, such as the condition that the ring under consideration be a homomor- phic image of a Cohen-Macaulay ring, to the case of excellent local rings or, more generally, rings for which we have completely stable test elements (or completely stable big test ele- ments, depending on whether the result being proved is for finitely generated modules or for arbitrary modules).
Here is one example:
Theorem (colon-capturing). Let (R, m, K) be an excellent reduced equidimensional local ring of prime characteristic p > 0 , and let x 1 ,... , xk+1 be part of a system of parameters. Let Ik = (x 1 ,... , xk)R. Then I k∗ :R xk+1 = I∗ k. In particular, Ik :R xk+1 ⊆ I k∗.
Proof. Note that R has a completely stable test element c. Suppose that uxk+1 ∈ Ik but u ∈ R − Ik. This is also true when we pass to R̂ , which is a homomorphic image of a regular ring and, hence, of a Cohen-Macaulay ring. Therefore, from the result on colon- capturing from p. 9 of the Lecture Notes from October 5, we have that u ∈ (Ik R̂ )∗, whence
cuq^ ∈ (I R̂ )[q]^ = I[q]^ R̂ for all q, and it follows that cuq^ ∈ I[q]^ R̂ ∩ R = I[q]^ for all q. Thus, u ∈ I∗.
Now suppose that uxk+1 ∈ I k∗. Then uq^ xqk+1 ∈ (I k∗ )[q]^ ⊆ (I k[q ])∗^ for all q, and so
cuq^ xqk+1 ∈ I k[q ]for all q, and cuq^ ∈ I k[q ]:R xqk+1 ⊆ (I k[q ])∗^ by the result of the first paragraph
applied to xq 1 ,... , xqk+1. Hence, c^2 uq^ ∈ (Ik)[q]^ for all q, so u ∈ I k∗. The opposite conclusion is obvious.
A Noetherian ring is called locally excellent if its localization at every maximal ideal (equivalently, at every prime ideal) is excellent.
Corollary. If R is weakly F-regular and locally excellent, then R is Cohen-Macaulay.
Proof. Both weak F-regularity and the Cohen-Macaulay property are local on the maximal ideals of R. Hence, we may assume that R is local. Since weakly F-regular rings are normal, R is certainly equidimensional. Since colon-capturing holds for systems of parameters in R, the result is immediate.
We can also prove a global version of this Theorem above that is valid even in case the ring is not equidimensional. We need one additional fact. 1
Lemma. Let (R, m, K) be an excellent local ring and let I be an ideal of R that has height
at least k modulo every minimal prime of R. Then I R̂ has height at least k modulo every minimal prime of R̂.
Proof. If pi is a minimal prime of R, pi R̂ is a radical ideal and is the intersection of certain minimal primes qij. The intersection of all qij is the same as the intersection of the pi R̂. Since finite intersection commutes with flat base change, this is 0. Thus, it will suffice to show that the height of I is at least k modulo every qij. To this end, we can replace R by R/pi. Thus, it is enough to show the result when R is an excellent local domain. In
this case, R̂ is reduced and equidimensional. Any prime Q of R̂ containing I R̂ lies over a prime P of R containing I. The height of Q is at least the height of Q 0 where Q 0 is a
minimal prime of P R̂. Hence, it suffices to show that if Q is a minimal prime of P R̂ then height (Q) = height (P ). Since R and R̂ are equidimensional and catenary,
height P = dim (R) − dim (R/P ) = dim ( R̂ ) − dim ( R/P̂ R̂ ).
Since the completion of R/P is equidimensional,
dim ( R/P̂ R̂ ) = dim
R/Q
Hence, height (P ) = dim ( R̂ ) − dim
R/Q
= height (Q),
as required.
Theorem (colon-capturing). Let R be a reduced Noetherian ring of prime characteristic p > 0 that is locally excellent and has a completely stable test element c. This holds, for example, if R is reduced and essentially of finite type over an excellent semilocal ring. Let x 1 ,... , xk+1 be elements of R. Let It denote the ideal (x 1 ,... , xt)R, 0 ≤ t ≤ k + 1. Suppose that the image of the ideal Ik has height k modulo every minimal prime of R, and that the image of the ideal Ik+1R has height k + 1 modulo every minimal prime of R. Then I k∗ :R xk+1 = I k∗.
Proof. We first prove that Ik :R xk+1 ⊆ I k∗. The stronger conclusion then follows exactly as in the Theorem above because c is a test element.
If xk+1u ∈ Ik but u /∈ I k∗ , we can choose q so that cuq^ ∈/ I k[q ]. This is preserved when we
localize at a maximal ideal in the support of (I k[q ]+ cuq^ )/I k[q ]. We have therefore reduced to the case of an excellent local ring Rm. By the Lemma above, the hypotheses are preserved
after completion, and we still have cuq^ ∈/ I k[q ]S = (IkS)[q], where S is the completion of Rm. Since c is a test element in S, we have that u /∈ (IkS)∗. Since S is a homomorphic image of a Cohen-Macaulay ring, this contradicts the Theorem on colon-capturing from p. 9 of the Lecture Notes from October 5.
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inclusion of domains. If R′^ is regular, then the element is in the image of the submodule, and this is preserved when we pass to R′/Q. If not, because R′^ is normal, the defining ideal of the singular locus has depth at least two, and we can find a regular sequence b, c in R′^ such that R′ b and R′ c are both regular. Then b and c are not both in Q, and hence at least one of them has nonzero image in R/Q: say that c has nonzero image. For some s, cs^ is a test element (respectively, big test element) in R, and it has nonzero image in R′/Q. It follows that the base change R′^ → R′/Q preserves tight closure, and, hence, so does the composite base change R → R′/Q.
Now suppose that the base change R → R/P fails to preserve tight closure. By the argument above, the further base change R/P ↪→ R′/Q restores the image of the element to the tight closure. This contradicts the fourth problem in Problem Set #4. Hence, R → R/P preserves tight closure.
Corolllary. Let R → S be a homomorphism of Noetherian rings such that S has a com- pletely stable (respectively, completely stable big) test element c and suppose that N ⊆ M are finitely generated (respectively, arbitrary) R-modules. Let u ∈ N (^) M∗. Then 1 ⊗ u ∈ 〈S ⊗R N 〉∗ S⊗RM.
Proof. Suppose that we have a counterexample. Then for some q, c(1 ⊗ u)q^ ∈ 〈/ S ⊗R N 〉[q]. This continues to be the case after localization at a suitable maximal Q ideal of S, and after completing the local ring SQ. Hence, we obtain a counterexample such that (S, n, L) is a complete local ring. Let m be the contraction of n to R and let let R 1 be the completion of Rm. Then the initial instance of tight closure is preserved by the base change from R to R 1 but not by the base change R 1 → S. This is a contradiction, since R 1 is in Rbig.
Although the theory of test elements that we have developed thus far is reasonably satisfactory for theoretical purposes, it is useful to have theorems that assert that specific elements of the ring are test elements: not only that some unknown power is a test element.
For example, the following result is very useful.
Theorem. Let R be a geometrically reduced, equidmensional algebra finitely generated over a field K of prime characteristic p > 0. Then the elements of the Jacobian ideal J (R/K) that are in R◦^ are completely stable big test elements for R.
It will be a while before we can prove this. We shall discuss the definition and properties of the Jacobian ideal in detail later. For the moment, we make only two comments. First, the Jacobian ideal defines the geometrically regular locus in R. Second, if R = K[x 1 ,... , xn]/(f ) is a hypersurface, then J (R/K) is simply the ideal generated by the images of the partial derivatives ∂f /∂xi in R.
For example, suppose that K has characteristic p > 0 with p 6 = 3. The Theorem above tells us that if R = K[x, y, z]/(x^3 +y^3 +z^3 ) then the elements x^2 , y^2 , and z^2 are completely stable big test elements (3 is iinvertible in K). This is not the best possible result: the test ideal turns out to be all of m = (x, y, z). But it gives a good starting point for computing τ (R).
The situation is essentially the same in the local case, where we study Rm instead. In this case, once we know that x^2 , y^2 are test elements, we can calculate the test ideal as (x^2 , y^2 ) :R (x^2 , y^2 )∗, since this ring is Gorenstein and we may apply problem 4 of Problem Set #3. The socle generator modulo I = (x^2 , y^2 ) turns out to be xyz^2 , and the problem of showing that the test ideal is mRm reduces to showing that the ideal (x^2 , y^2 , xyz^2 ) is tightly closed in Rm. The main point here is that the Theorem above can sometimes be used to make the calculation of the test ideal completely down-to-earth. The proof of the Theorem above involves several ingredients. One is the Lipman-Sathaye Jacobian Theorem, which we will state but not prove. The Theorem is proved in [J. Lipman and A. Sathaye, Jacobian ideals and a theorem of Brian¸con-Skoda, Michigan Math. J. 28 (1981) 199–222]. Moreover, a complete treatment of the Lipman-Sathaye argument is given in the Lecture Notes from Math 711, Fall 2006. See specifically, the Lectures of September 25, 27, and 29, as well as the Lectures of October 2, 4, 6, 9, 11, and 13. Another ingredient is the Theorem stated just below. We shall say that an extension of a domain S of a field K is ´etale if S is a finite product of finite separable algebraic extension fields of K. We shall say that an extension A → R of a domain A is generically ´etale if the generic fiber is ´etale, i.e., frac (A) ⊗A R is a finite product of finite separable algebraic extension fields of frac (A).
Theorem. Let R be a module-finite and generically ´etale extension of a regular ring A of prime characteristic p > 0. Let r ∈ R◦^ be such that cR∞^ ⊆ R[A∞]. Then c is a completely stable big test element for R.
We shall see that elements c as above exist, and that the Lipman-Sathaye Theorem can be used to find specific elements like this. The reason that it is very helpful that cR∞^ ⊆ R[A∞] is that it turns out that R[A∞] ∼= R ⊗A A∞^ is faithfully flat over R, since A∞^ is faithfully flat over A. This makes it far easier to work with R[A∞] than it is to work with R∞, and multiplication by c can be used to “correct” the error in replacing R∞^ by R[A∞]. We begin by proving the following preliminary result.
Lemma. Let (A, mA, K) be a normal local domain and let R be a module-finite extension domain of A that is generically ´etale over A. Let ord be a Z-valued valuation on A that is nonnegative on A and positive on mA. Then ord extends uniquely to A∞^ by letting ord (a^1 /q^ ) = (1/q)ord (a) for all a ∈ A − { 0 }. The extended valuation takes values in Z[1/p]. Let m be a proper ideal of R. Let d be the torsion-free rank of R as an A-module. Let u ∈ mR[A∞] ∩ A∞. Then ord (u) ≥ 1 /d!.
Proof. We know that frac (R) is separable of degree d over frac (A) = K. Let θ be a primitive element. The splitting field L of the minimal polynomial f of θ is generated by the roots of f , and is Galois over K, with a Galois group that is a subgroup of the permutations on d, the roots of f. We may replace R by the possibly larger ring which is
