Benutzer:Bocardodarapti/Talk in Yokohama March 2008

TIGHT CLOSURE IS DEAD - LONG LIVE TIGHT CLOSURE

Let ${}R$ be a noetherian domain of positive characteristic, let

$F\colon R\longrightarrow R,\,f\longmapsto f^{p},$

be the Frobenius homomorphism and

$F^{e}\colon R\longrightarrow R,\,f\longmapsto f^{q}$

(mit ${}q=p^{e}$) its ${}e$th iteration. Let ${}I$ be an ideal and set

${}I^{[q]}={\text{ extended ideal of }}I{\text{ under }}F^{e}\,.$

Then define the tight closure of ${}I$ to be the ideal

${}I^{*}={\left\{f\in R\mid {\text{ there exists }}z\neq 0{\text{ such that }}zf^{q}\in I^{[q]}{\text{ for all }}q=p^{e}\right\}}\,.$

LOCALIZATION PROBLEM

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 ${}R$?

Why does localization often holds?

Because an element belongs to the tight closure because of a certain reason, and this reason often localizes/globalizes.

A typical reason for belonging to the tight closure is through plus closure.

For an ideal ${}I\subseteq R$ in a domain ${}R$ define its plus closure by

${}I^{+}={\left\{f\in R\mid {\text{there exists a finite domain extension }}R\subseteq T{\text{ such that }}f\in IT\right\}}\,.$

Equivalent: Let ${}R^{+}$ be the absolute integral closure of ${}R$. This is the integral closure of ${}R$ in an algebraic closure of the quotient field ${}Q(R)$ (first considered by Artin). Then

$f\in I^{+}{\text{ if and only if }}f\in IR^{+}.$

The plus closure commutes with localization.

We also have the inclusion ${}I^{+}\subseteq I^{*}$. Here the question arises:

Question: Is ${}I^{+}=I^{*}$?

This question is known as the tantalizing question in tight closure theory.

One should think of the left hand side as a geometric condition and of the right hand side as a cohomological conditon (later). The two main results on this question are the following:

Theorem ((K. Smith))

Let ${}R$ be local and excellent. If ${}I$ is a parameter ideal (generated by a (sub-)system of parameters), then

$I^{*}=I^{+}\,$

and the tight closure of $I$ commutes with localization.

Theorem ((Brenner))

Let ${}R$ be a standard-graded, two-dimensional normal domain over (the algebraic closure of) a finite field. Let ${}I$ be an ${}R_{+}$-primary graded ideal.

Then

${}I^{*}=I^{+}\,.$

In dimension two tight closure always localizes. However, a positive answer to the tantalizing question in dimenson two in general would imply the localization property in dimension three.

The result just mentioned rests on a detailed study of tight closure in dimension two with the help of vector bundles on the corresponding curve.

The first chance for a counterexample is thus in dimension three, no parameter ideal. We look at a special case of the localization problem:

GEOMETRIC DEFORMATIONS OF 2-DIMENSIONAL TIGHT CLOSURE PROBLEMS

Let ${}D\subseteq R$ flat, ${}f$ and ${}I$ in ${}R$. For every ${}D\rightarrow K$, ${}K$ a field, we can consider ${}f$ and ${}I$ in ${}R\otimes _{D}K$. In particular for ${}K=\kappa ({\mathfrak {p}})$, ${}{\mathfrak {p}}\in \operatorname {Spec} \,D$.

How does the property

$f\in I^{*}{\text{ in }}R\otimes _{D}\kappa ({\mathfrak {p}})\,$

vary with ${}{\mathfrak {p}}$?

There are two cases:

• ${}D$ contains ${}\mathbb {Z} /(p)$. This is a geometric or equicharacteristic deformation. Example ${}D=\mathbb {Z} /(p)[t]$.

• ${}D$ contains ${}\mathbb {Z} \subseteq D$. This is an arithmetic or mixed characteristic deformation. Example ${}D=\mathbb {Z}$.

Proposition

Let

${}\mathbb {Z} /(p)\subset D$ be a one-dimensional domain and ${}D\subseteq R$ 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 ${}D$.

Beweis

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 ${}{\mathfrak {p}}$.

$\Box$

We will look for the easiest possibilty of such a deformation:

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

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

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

is a two-dimensional standard-graded ring over $K$. For residue class fields of points of ${\mathbb {A} }_{{\mathbb {F} }_{p}}^{1}=\operatorname {Spec} {{\mathbb {F} }_{p}[t]}$ 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.

Note that in the second case tight closure is plus closure. To analyze the behavior of tight closure in such a family we can use what we know in the two-dimensional standard-graded situation.

GEOMETRIC INTERPRETATION IN DIMENSION TWO

Let ${}R$ be a two-dimensional standard-graded normal domain over an algebraically closed field ${}K$. Let ${}C=\operatorname {Proj} {\left(R\right)}$ be the corresponding smooth projective curve and let

${}I={\left(f_{1},\ldots ,f_{n}\right)}\,$

be an ${}R_{+}$-primary homogeneous ideal with generators of degrees ${}d_{1},\ldots ,d_{n}$. Then we get on ${}C$ the short exact sequence

$0\longrightarrow \operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}(m)\longrightarrow \bigoplus _{i=1}^{n}{\mathcal {O}}_{C}(m-d_{i}){\stackrel {f_{1},\ldots ,f_{n}}{\longrightarrow }}{\mathcal {O}}_{C}(m)\longrightarrow 0.$

Here ${}\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}(m)$ is a vector bundle, called the syzygy bundle, of rank ${}n-1$ and of degree

$((n-1)m-\sum _{i=1}^{n}d_{i})\operatorname {deg} \,(C).$

An element

${}f\in R_{m}=\Gamma (C,{\mathcal {O}}_{C}(m))\,$

defines a cohomology class

${}c=\delta (f)\in H^{1}(C,\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}(m))\,.$

With this notation we have

$f\in I^{*}{\text{ if and only if there exists }}z\neq 0$

(homogeneous of some degree ${}k$) such that ${}zF^{e*}(c)=0$ for all ${}e$. This cohomology class lives in

${}H^{1}(C,F^{e*}(\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}(m))\otimes {\mathcal {O}}(k))=H^{1}(C,\operatorname {Syz} {\left(f_{1}^{q},\ldots ,f_{n}^{q}\right)}(qm+k))\,.$

For the plus closure we have a similar correspondence:

${}f\in I^{+}$ if and only if there exists a curve ${}\varphi \colon C'\rightarrow C$ such that the pull-back ${}\varphi ^{*}(c)$ of the cohomology class vanishes.

Definition ((semistable and strongly semistable))

Let ${}{\mathcal {S}}$ be a vector bundle on a smooth projective curve ${}C$. It is called semistable, if ${}\mu ({\mathcal {T}})={\frac {\deg({\mathcal {T}})}{\operatorname {rk} ({\mathcal {T}})}}\leq {\frac {\deg({\mathcal {S}})}{\operatorname {rk} ({\mathcal {S}})}}=\mu ({\mathcal {S}})$ for all subbundles ${}{\mathcal {T}}$.

Suppose that the base field has positive characteristic ${}p>0$. Then ${}{\mathcal {S}}$ is called strongly semistable, if all (absolute) Frobenius pull-backs ${}F^{e*}({\mathcal {S}})$ are semistable.

Theorem ((Brenner))

Suppose that ${}\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}$ is strongly semistable. Then

$R_{m}\subseteq I^{*}{\text{ for }}m\geq {\frac {\sum d_{i}}{n-1}}{\text{ and (for almost all prime numbers) }}R_{m}\cap I^{*}\subseteq I{\text{ for }}m<{\frac {\sum d_{i}}{n-1}}.$

In general, there exists an exact criterion depending on $c=\delta (f)$ and the strong Harder-Narasimhan filtration of $Syz$.

If $K$ is finite, then the same criterion holds for plus closure.

A COUNTEREXAMPLE TO THE LOCALIZATION PROPERTY

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.

Example ((Monsky))

Let

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

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:

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

By the geometric interpretation of Hilbert-Kunz theory this means that the restricted cotangent bundle

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

is strongly semistable in the transcendental case, but not strongly semistable in the algebraic case. In fact, for $d=\deg(\alpha )$, $t\mapsto \alpha$, the $d$-th Frobenius pull-back destabilizes.

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

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

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! First, by strong semistability in the transcendental case we have

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

by the degree formula. If localization would hold, then $f$ 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 $f$ belongs never to the tight closure of $I$.

Lemma ((Monsky))

Let

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

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

Beweis

This is an elementary but tedious computation.

$\Box$

Theorem ((Brenner-Monsky))

Tight closure does not commute with localization.

Beweis

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

$\Box$

Corollary

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

Beweis

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 ${}Y\to C_{{\mathbb {F} }_{2}(t)}$ which annihilates the cohomology class ${}c$ and this would extend to a morphism of relative curves almost everywhere.

$\Box$