Tight Closure/Localization problem/Counterexample/Section

From Wikiversity
Jump to navigation Jump to search

Let be a multiplicative system and an ideal in . Then the localization problem of tight closure is the question whether the identity


Here the inclusion is always true and is the problem. The problem means explicitly:

If , can we find an such that holds in ?

Let be a one-dimensional domain, of finite type and an ideal in . Suppose that localization holds and that

( is the multiplicative system). Then holds in for almost all in Spec .

By localization, there exists , , such that . By persistence of tight closure (under a ring homomorphism), we get

The element does not belong to for almost all , so is a unit in and hence

for almost all .

In order to get a counterexample to the localization property we will look now at geometric deformations:

where has degree and have degree and is homogeneous. Then (for every homomorphism to a field)

is a two-dimensional standard-graded ring over . For the residue class fields of points of we have basically two possibilities.

    • ,
    the function field. This is the generic or transcendental case.
    • ,
    the special or algebraic or finite case.

How does vary with ? To analyze the behavior of tight closure in such a family we can use what we know in the two-dimensional standard-graded situation.

In order to establish an example where tight closure does not behave uniformly under a geometric deformation, we first need a situation where strong semistability does not behave uniformly. Such an example was given, in terms of Hilbert-Kunz theory, by Paul Monsky in 1998.



Then Monsky proved the following results on the Hilbert-Kunz multiplicity of the maximal ideal in , a field:

We consider as an -algebra, the corresponding morphism and the corresponding smooth projective relative curve . The fibers are and respectively.

By the geometric interpretation of Hilbert-Kunz theory, the computations mentioned in example mean that the restricted cotangent bundle

is strongly semistable in the transcendental case, but not strongly semistable in the algebraic case. In fact, for , , where , the -th Frobenius pull-back destabilizes (meaning that it is not semistable anymore).

The maximal ideal can not be used directly, as it is tightly closed. However, we look at the second Frobenius pull-back which is (characteristic two) just

By the degree formula, we have to look for an element of degree . Let's take . This is our example ( does not work). First, by strong semistability in the transcendental case, we have

by the degree formula. If localization would hold, then by fact, would also belong to the tight closure of for almost all algebraic instances , . Contrary to that we show that for all algebraic instances, the element never belongs to the tight closure of .

Let , , . Set . Then

This is an elementary but tedious computation.

Tight closure does not commute with localization.

One knows in our situation that is a so-called test element. Hence fact shows that

In terms of affineness of quasiaffine schemes (or local cohomology), this example has the following properties: the open subset given by the ideal

has cohomological dimension if is transcendental and has cohomological dimension (equivalently, is an affine scheme) if is algebraic.

Tight closure is not plus closure in graded dimension two for fields with transcendental elements.


In this ring , but it can not belong to the plus closure. Else there would be a curve morphism

which annihilates the cohomology class and this would extend to a morphism of relative curves over almost everywhere.