# Tight Closure/Localization problem/Counterexample/Section

Let ${\displaystyle {}S\subseteq R}$ be a multiplicative system and ${\displaystyle {}I}$ an ideal in ${\displaystyle {}R}$. Then the localization problem of tight closure is the question whether the identity

${\displaystyle {}(I^{*})_{S}=(IR_{S})^{*}\,}$

holds.

Here the inclusion ${\displaystyle {}\subseteq }$ is always true and ${\displaystyle {}\supseteq }$ is the problem. The problem means explicitly:

If ${\displaystyle {}f\in (IR_{S})^{*}}$, can we find an ${\displaystyle {}h\in S}$ such that ${\displaystyle {}hf\in I^{*}}$ holds in ${\displaystyle {}R}$?

## Proposition

Let ${\displaystyle {}\mathbb {Z} /(p)\subset D}$ be a one-dimensional domain, ${\displaystyle {}D\subseteq R}$ of finite type and ${\displaystyle {}I}$ an ideal in ${\displaystyle {}R}$. Suppose that localization holds and that

${\displaystyle f\in I^{*}{\text{ holds in }}R\otimes _{D}Q(D)=R_{D^{*}}=R_{Q(D)}}$

(${\displaystyle {}S=D^{*}=D\setminus \{0\}}$ is the multiplicative system). Then ${\displaystyle {}f\in I^{*}}$ holds in ${\displaystyle {}R\otimes _{D}\kappa ({\mathfrak {p}})}$ for almost all ${\displaystyle {}{\mathfrak {p}}}$ in Spec ${\displaystyle {}D}$.

### Proof

By localization, there exists ${\displaystyle {}h\in D}$, ${\displaystyle {}h\neq 0}$, such that ${\displaystyle {}hf\in I^{*}{\text{ in }}R}$. By persistence of tight closure (under a ring homomorphism), we get

${\displaystyle hf\in I^{*}{\text{ in }}R_{\kappa ({\mathfrak {p}})}.}$

The element ${\displaystyle {}h}$ does not belong to ${\displaystyle {}{\mathfrak {p}}}$ for almost all ${\displaystyle {}{\mathfrak {p}}}$, so ${\displaystyle {}h}$ is a unit in ${\displaystyle {}R_{\kappa ({\mathfrak {p}})}}$ and hence

${\displaystyle f\in I^{*}{\text{ in }}R_{\kappa ({\mathfrak {p}})}}$

for almost all ${\displaystyle {}{\mathfrak {p}}}$.

${\displaystyle \Box }$

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

${\displaystyle {}D={\mathbb {F} }_{p}[t]\subset {\mathbb {F} }_{p}[t][x,y,z]/(g)=S\,,}$

where ${\displaystyle {}t}$ has degree ${\displaystyle {}0}$ and ${\displaystyle {}x,y,z}$ have degree ${\displaystyle {}1}$ and ${\displaystyle {}g}$ is homogeneous. Then (for every homomorphism ${\displaystyle {}{\mathbb {F} }_{p}[t]\rightarrow K}$ to a field)

${\displaystyle S\otimes _{{\mathbb {F} }_{p}[t]}K}$

is a two-dimensional standard-graded ring over ${\displaystyle {}K}$. For the residue class fields of points of ${\displaystyle {}{\mathbb {A} }_{{\mathbb {F} }_{p}}^{1}=\operatorname {Spec} {\left({\mathbb {F} }_{p}[t]\right)}}$ we have basically two possibilities.

• ${\displaystyle {}K={\mathbb {F} }_{p}(t)}$,
the function field. This is the generic or transcendental case.
• ${\displaystyle {}K={\mathbb {F} }_{q}}$,
the special or algebraic or finite case.

How does ${\displaystyle {}f\in I^{*}}$ vary with ${\displaystyle {}K}$? 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.

## Example

Let

${\displaystyle {}g=z^{4}+z^{2}xy+z{\left(x^{3}+y^{3}\right)}+{\left(t+t^{2}\right)}x^{2}y^{2}\,.}$

Consider

${\displaystyle {}S={\mathbb {F} }_{2}[t,x,y,z]/(g)\,.}$

Then Monsky proved the following results on the Hilbert-Kunz multiplicity of the maximal ideal ${\displaystyle {}(x,y,z)}$ in ${\displaystyle {}S\otimes _{{\mathbb {F} }_{2}[t]}L}$, ${\displaystyle {}L}$ a field:

${\displaystyle e_{HK}(S\otimes _{\mathbb {F} _{2}[t]}L)={\begin{cases}3{\text{ for }}L={\mathbb {F} }_{2}(t)\\3+{\frac {1}{4^{d}}}{\text{ for }}L={\mathbb {F} }_{q}={\mathbb {F} }_{2}(\alpha ),\,(t\mapsto \alpha ,\,d=\deg(\alpha ))\,.\end{cases}}\,}$

We consider ${\displaystyle {}S}$ as an ${\displaystyle {}{\mathbb {F} }_{2}[t]}$-algebra, the corresponding morphism ${\displaystyle {}\operatorname {Spec} {\left(S\right)}\rightarrow {\mathbb {A} }_{{\mathbb {F} }_{2}}^{1}}$ and the corresponding smooth projective relative curve ${\displaystyle {}C=\operatorname {Proj} {\left(S\right)}\rightarrow {\mathbb {A} }_{{\mathbb {F} }_{2}}^{1}}$. The fibers are ${\displaystyle {}\operatorname {Spec} {\left(S_{\kappa ({\mathfrak {p}})}\right)}}$ and ${\displaystyle {}C_{\kappa ({\mathfrak {p}})}}$ respectively.

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

${\displaystyle {}\operatorname {Syz} {\left(x,y,z\right)}=(\Omega _{{\mathbb {P} }_{}^{2}}){|}_{C}\,}$

is strongly semistable in the transcendental case, but not strongly semistable in the algebraic case. In fact, for ${\displaystyle {}d=\deg(\alpha )}$, ${\displaystyle {}t\mapsto \alpha }$, where ${\displaystyle {}L=\mathbb {F} _{2}(\alpha )}$, the ${\displaystyle {}d}$-th Frobenius pull-back destabilizes (meaning that it is not semistable anymore).

The maximal ideal ${\displaystyle {}(x,y,z)}$ can not be used directly, as it is tightly closed. However, we look at the second Frobenius pull-back which is (characteristic two) just

${\displaystyle {}I={\left(x^{4},y^{4},z^{4}\right)}\,.}$

By the degree formula, we have to look for an element of degree ${\displaystyle {}6}$. Let's take ${\displaystyle {}f=y^{3}z^{3}}$. This is our example (${\displaystyle x^{3}y^{3}}$ does not work). First, by strong semistability in the transcendental case, we have

${\displaystyle f\in I^{*}{\text{ in }}S\otimes {\mathbb {F} }_{2}(t)}$

by the degree formula. If localization would hold, then by fact, ${\displaystyle {}f}$ would also belong to the tight closure of ${\displaystyle {}I}$ for almost all algebraic instances ${\displaystyle {}{\mathbb {F} }_{q}={\mathbb {F} }_{2}(\alpha )}$, ${\displaystyle {}t\mapsto \alpha }$. Contrary to that we show that for all algebraic instances, the element ${\displaystyle {}f}$ never belongs to the tight closure of ${\displaystyle {}I}$.

## Lemma

Let ${\displaystyle {}{\mathbb {F} }_{q}={\mathbb {F} }_{p}(\alpha )}$, ${\displaystyle {}t\mapsto \alpha }$, ${\displaystyle {}\deg(\alpha )=d}$. Set ${\displaystyle {}Q=2^{d-1}}$. Then

${\displaystyle {}xyf^{Q}\notin I^{[Q]}\,.}$

### Proof

This is an elementary but tedious computation.

${\displaystyle \Box }$

## Theorem

Tight closure does not commute with localization.

### Proof

One knows in our situation that ${\displaystyle {}xy}$ is a so-called test element. Hence fact shows that ${\displaystyle {}f\notin I^{*}}$

${\displaystyle \Box }$

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

${\displaystyle (x,y,z)\subseteq {\mathbb {F} }_{2}(t)[x,y,z,s_{1},s_{2},s_{3}]/{\left(g,s_{1}x^{4}+s_{2}y^{4}+s_{3}z^{4}+y^{3}z^{3}\right)}\,}$

has cohomological dimension ${\displaystyle {}1}$ if ${\displaystyle {}t}$ is transcendental and has cohomological dimension ${\displaystyle {}0}$ (equivalently, ${\displaystyle D(x,y,z)}$ is an affine scheme) if ${\displaystyle {}t}$ is algebraic.

## Corollary

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

### Proof

Consider

${\displaystyle {}R={\mathbb {F} }_{2}(t)[x,y,z]/(g)\,.}$

In this ring ${\displaystyle {}y^{3}z^{3}\in I^{*}}$, but it can not belong to the plus closure. Else there would be a curve morphism

${\displaystyle Y\longrightarrow C_{{\mathbb {F} }_{2}(t)}}$

which annihilates the cohomology class ${\displaystyle {}c}$ and this would extend to a morphism of relative curves over ${\displaystyle {}{\mathbb {A} }_{{\mathbb {F} }_{2}}^{1}}$ almost everywhere.

${\displaystyle \Box }$