Tight Closure/Localization problem/Counterexample/Section

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

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

holds.

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

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

Proposition

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

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

(${}S=D^{*}=D\setminus \{0\}$ is the multiplicative system). Then ${}f\in I^{*}$

holds in ${}R\otimes _{D}\kappa ({\mathfrak {p}})$ for almost all ${}{\mathfrak {p}}$ in Spec ${}D$.

Proof

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

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

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

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

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

$\Box$

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

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

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

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

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

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

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

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

Consider

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

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

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

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

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

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

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

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

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

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

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

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

Lemma

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

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

Proof

This is an elementary but tedious computation.

$\Box$

Theorem

Tight closure does not commute with localization.

Proof

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

$\Box$

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

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

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

Corollary

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

Proof

Consider

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

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

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

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

$\Box$