Kurs:Vector bundles and ideal closure operations (MSRI 2012)/Lecture 3/kontrolle

In the last lecture we will focus on the question when are the torsors given by a forcing algebras over a two-dimensional ring affine? We will look at the graded situation to be able to work on the corresponding projective curve.

Geometric interpretation in dimension two

We will restrict now to the two-dimensional homogeneous case in order to work on the corresponding projective curve. We want to find an object over the curve which corresponds to the forcing algebra or its induced torsor.

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

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

Thus a homogeneous element ${}f$ of degree ${}m$ defines a cohomology class ${}\delta (f)\in H^{1}(C,\operatorname {Syz} _{}^{}{\left(f_{1},\ldots ,f_{n}\right)}(m))$, so this defines a torsor over the projective curve.

Semistability of vector bundles

In the situation of a forcing algebra of homogeneous elements, this torsor ${}T$ can also be obtained as ${}\operatorname {Proj} {\left(B\right)}$, where ${}B$ is the (not necessarily positively) graded forcing algebra. In particular, it follows that the containment ${}f\in I^{*}$ is equivalent to the property that ${}T$ is not an affine variety. For this properties, positivity (ampleness) properties of the syzygy bundle are crucial. We need the concept of (Mumford-) semistability.

Definition Referenznummer erstellen

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.

An important property of a semistable bundle of negative degree is that it can not have any global section ${}\neq 0$. Note that a semistable vector bundle need not be strongly semistable, the following is probably the simplest example.

Example

Let ${}C$ be the smooth Fermat quartic given by ${}x^{4}+y^{4}+z^{4}$ and consider on it the syzygy bundle ${}\operatorname {Syz} {\left(x,y,z\right)}$ (which is also the restricted cotangent bundle from the projective plane). This bundle is semistable. Suppose that the characteristic is ${}3$. Then its Frobenius pull-back is ${}\operatorname {Syz} {\left(x^{3},y^{3},z^{3}\right)}$. The curve equation gives a global non-trivial section of this bundle of total degree ${}4$. But the degree of ${}\operatorname {Syz} {\left(x^{3},y^{3},z^{3}\right)}(4)$ is negative, hence it can not be semistable anymore.

The following example is related to Beispiel *****.

Example

Let ${}R=K[x,y,z]/{\left(x^{3}+y^{3}+z^{3}\right)}$, where ${}K$ is a field of positive characteristic ${}p\neq 3$, ${}I={\left(x^{2},y^{2},z^{2}\right)}$, and

${}C=\operatorname {Proj} {\left(R\right)}\,.$

The equation ${}x^{3}+y^{3}+z^{3}=0$ yields the short exact sequence

$0\longrightarrow {\mathcal {O}}_{C}\longrightarrow \operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}(3)\longrightarrow {\mathcal {O}}_{C}\longrightarrow 0.$

This shows that ${}\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}$ is strongly semistable.

For a strongly semistable vector bundle ${}{\mathcal {S}}$ on ${}C$ and a cohomology class ${}c\in H^{1}(C,{\mathcal {S}})$ with corresponding torsor we obtain the following affineness criterion.

Theorem

Let ${}C$ denote a smooth projective curve over an algebraically closed field ${}K$ and let ${}{\mathcal {S}}$ be a strongly semistable vector bundle over ${}C$ together with a cohomology class ${}c\in H^{1}(C,{\mathcal {S}})$.

Then the torsor ${}T(c)$ is an affine scheme if and only if ${}\operatorname {deg} _{}^{}{\left({\mathcal {S}}\right)}<0$ and ${}c\neq 0$ (${}F^{e}(c)\neq 0$ for all ${}e$ in positive characteristic).

This result rests on the ampleness of ${}{\mathcal {S}}'^{\vee }$ occuring in the dual exact sequence ${}0\rightarrow {\mathcal {O}}_{C}\rightarrow {\mathcal {S}}'^{\vee }\rightarrow {\mathcal {S}}^{\vee }\rightarrow 0$ given by ${}c$ (this rests on work of Gieseker and Hartshorne). It implies for a strongly semistable syzygy bundle the following degree formula for tight closure.

Theorem

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

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

If we take on the right hand side ${}I^{F}$, the Frobenius closure of the ideal, instead of ${}I$, then this statement is true for all characteristics. As stated, it is true in a relative setting for ${}p$ large enough.

We indicate the proof of the inclusion result. The degree condition implies that ${}c\in \delta (f)=H^{1}(C,{\mathcal {S}})$ is such that ${}{\mathcal {S}}=\operatorname {Syz} {\left(f_{1},\ldots ,f_{n}\right)}(m)$ has non-negative degree. Then also all Frobenius pull-backs ${}F^{*}({\mathcal {S}})$ have non-negative degree. Let ${}{\mathcal {L}}={\mathcal {O}}(k)$ be a twist of the tautological line bundle on ${}C$ such that its degree is larger than the degree of ${}\omega _{C}^{-1}$, the dual of the canonical sheaf. Let ${}z\in H^{0}(Y,{\mathcal {L}})$ be a non-zero element. Then ${}zF^{e*}(c)\in H^{1}(C,F^{e*}({\mathcal {S}})\otimes {\mathcal {L}})$, and by Serre duality we have

${}H^{1}(C,F^{e*}({\mathcal {S}})\otimes {\mathcal {L}})\cong H^{0}(F^{e*}({\mathcal {S}}^{\vee })\otimes {\mathcal {L}}^{-1}\otimes \omega _{C})^{\vee }\,.$

On the right hand side we have a semistable sheaf of negative degree, which can not have a non-trivial section. Hence

${}zF^{e*}(c)=0\,$

and therefore ${}f$ belongs to the tight closure.

Harder-Narasimhan filtration

In general, there exists an exact criterion depending on ${}c$ and the strong Harder-Narasimhan filtration of ${}{\mathcal {S}}$. For this we give the definition of the Harder-Narasimhan filtration.

Definition

Let ${}{\mathcal {S}}$ be a vector bundle on a smooth projective curve ${}C$ over an algebraically closed field ${}K$. Then the (uniquely determined) filtration

${}0={\mathcal {S}}_{0}\subset {\mathcal {S}}_{1}\subset \ldots \subset {\mathcal {S}}_{t-1}\subset {\mathcal {S}}_{t}={\mathcal {S}}\,$

of subbundles such that all quotient bundles ${}{\mathcal {S}}_{k}/{\mathcal {S}}_{k-1}$ are semistable with decreasing slopes ${}\mu _{k}=\mu ({\mathcal {S}}_{k}/{\mathcal {S}}_{k-1})$, is called the Harder-Narasimhan filtration of ${}{\mathcal {S}}$.

The Harder-Narasimhan filtration exists uniquely (by a Theorem of Harder and Narasimhan). A Harder-Narasimhan filtration is called strong if all the quotients ${}{\mathcal {S}}_{i}/{\mathcal {S}}_{i-1}$ are strongly semistable. A Harder-Narasimhan filtration is not strong in general, however, by a Theorem of A. Langer, there exists some Frobenius pull-back ${}F^{e*}({\mathcal {S}})$ such that its Harder-Narasimhan filtration is strong.

Theorem Theorem 3.7 ändern

Let ${}C$ denote a smooth projective curve over an algebraically closed field ${}K$ and let ${}E$ be a vector bundle over ${}C$ together with a cohomology class ${}c\in H^{1}(C,{\mathcal {S}})$. Let

${}E_{1}\subset E_{2}\subset \ldots \subset E_{t-1}\subset E_{t}=F^{e*}(E)\,$

be a strong Harder-Narasimhan filtration. We choose ${}i$ such that ${}E_{i}/E_{i-1}$ has degree ${}\geq 0$ and that ${}E_{i+1}/E_{i}$ has degree ${}<0$. We set ${}{\mathcal {Q}}=F^{e*}(E)/E_{i}$.

Then the following are equivalent.

1. The torsor ${}T(c)$ is not an affine scheme.
2. Some Frobenius power of the image of ${}F^{e*}(c)$ inside ${}H^{1}(X,{\mathcal {Q}})$ is ${}0$.

Plus closure in dimension two

Let ${}K$ be a field and let ${}R$ be a normal two-dimensional standard-graded domain over ${}K$ with corresponding smooth projective curve ${}C$. A homogeneous ${}{\mathfrak {m}}$-primary ideal with homogeneous ideal generators ${}f_{1},\ldots ,f_{n}$ and another homogeneous element ${}f$ of degree ${}m$ yield a cohomology class

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

Let ${}T(c)$ be the corresponding torsor. We have seen that the affineness of this torsor over ${}C$ is equivalent to the affineness of the corresponding torsor over ${}D({\mathfrak {m}})\subseteq \operatorname {Spec} {\left(R\right)}$. Now we want to understand what the property ${}f\in I^{+}$ means for ${}c$ and for ${}T(c)$. Instead of the plus closure we will work with the graded plus closure ${}I^{+{\text{gr}}}$, where ${}f\in I^{+{\text{gr}}}$ holds if and only if there exists a finite graded extension ${}R\subseteq S$ such that ${}f\in IS$. The existence of such an ${}S$ translates into the existence of a finite morphism

$\varphi \colon C'=\operatorname {Proj} {\left(S\right)}\longrightarrow \operatorname {Proj} {\left(R\right)}=C$

such that ${}\varphi ^{*}(c)=0$. Here we may assume that ${}C'$ is also smooth. Therefore we discuss the more general question when a cohomology class ${}c\in H^{1}(C,{\mathcal {S}})$, where ${}{\mathcal {S}}$ is a locally free sheaf on ${}C$, can be annihilated by a finite morphism

$C'\longrightarrow C$

of smooth projective curves. The advantage of this more general approach is that we may work with short exact sequences (in particular, the sequences coming from the Harder-Narasimhan filtration) in order to reduce the problem to semistable bundles which do not necessarily come from an ideal situation.

Lemma

Let ${}C$ denote a smooth projective curve over an algebraically closed field ${}K$, let ${}{\mathcal {S}}$ be a locally free sheaf on ${}C$ and let ${}c\in H^{1}(C,{\mathcal {S}})$ be a cohomology class with corresponding torsor ${}T\rightarrow C$. Then the following conditions are equivalent.

1. There exists a finite morphism

   $\varphi \colon C'\longrightarrow C$

   from a smooth projective curve ${}C'$ such that ${}\varphi ^{*}(c)=0$.

2. There exists a projective curve ${}Z\subseteq T$.

Proof

If (1) holds, then the pull-back ${}\varphi ^{*}(T)=T\times _{C}C'$ is trivial (as a torsor), as it equals the torsor given by ${}\varphi ^{*}(c)=0$. Hence ${}\varphi ^{*}(T)$ is isomorphic to a vector bundle and contains in particular a copy of ${}C'$. The image ${}Z$ of this copy is a projective curve inside ${}T$.

If (2) holds, then let ${}C'$ be the normalization of ${}Z$. Since ${}Z$ dominates ${}C$, the resulting morphism

$\varphi \colon C'\longrightarrow C$

is finite. Since this morphism factors through ${}T$ and since ${}T$ annihilates the cohomology class by which it is defined, it follows that

${}\varphi ^{*}(c)=0$.

$\Box$

We want to show that the cohomological criterion for (non)-affineness of a torsor along the Harder-Narasimhan filtration of the vector bundle also holds for the existence of projective curves inside the torsor, under the condition that the projective curve is defined over a finite field. This implies that tight closure is (graded) plus closure for graded ${}{\mathfrak {m}}$-primary ideals in a two-dimensional graded domain over a finite field.

Annihilation of cohomology classes of strongly semistable sheaves

We deal first with the situation of a strongly semistable sheaf ${}{\mathcal {S}}$ of degree ${}0$. The following two results are due to Lange and Stuhler. We say that a locally free sheaf is étale trivializable if there exists a finite étale morphism ${}\varphi \colon C'\rightarrow C$ such that ${}\varphi ^{*}({\mathcal {S}})\cong {\mathcal {O}}_{C'}^{r}$. Such bundles are directly related to linear representations of the étale fundamental group.

Lemma

Let ${}K$ denote a finite field (or the algebraic closure of a finite field) and let ${}C$ be a smooth projective curve over ${}K$. Let ${}{\mathcal {S}}$ be a locally free sheaf over ${}C$.

Then ${}{\mathcal {S}}$ is étale trivializable if and only if there exists some ${}n$ such that ${}F^{n*}{\mathcal {S}}\cong {\mathcal {S}}$.

Theorem

Let ${}K$ denote a finite field (or the algebraic closure of a finite field) and let ${}C$ be a smooth projective curve over ${}K$. Let ${}{\mathcal {S}}$ be a strongly semistable locally free sheaf over ${}C$ of degree ${}0$.

Then there exists a finite morphism

$\varphi \colon C'\longrightarrow C$

such that ${}\varphi ^{*}({\mathcal {S}})$ is trivial.

Proof

We consider the family of locally free sheaves ${}F^{e*}({\mathcal {S}})$, ${}e\in \mathbb {N}$. Because these are all semistable of degree ${}0$, and defined over the same finite field, we must have (by the existence of the moduli space for vector bundles) a repetition, i.e.

${}F^{e*}({\mathcal {S}})\cong F^{e'*}({\mathcal {S}})\,$

for some ${}e'>e$. By Lemma 3.9 the bundle ${}F^{e*}({\mathcal {S}})$ admits an étale trivialization ${}\varphi \colon C'\rightarrow C$. Hence the finite map ${}F^{e}\circ \varphi$ trivializes the bundle.

$\Box$

Theorem

Let ${}K$ denote a finite field (or the algebraic closure of a finite field) and let ${}C$ be a smooth projective curve over ${}K$. Let ${}{\mathcal {S}}$ be a strongly semistable locally free sheaf over ${}C$ of nonnegative degree and let ${}c\in H^{1}(C,{\mathcal {S}})$ denote a cohomology class.

Then there exists a finite morphism

$\varphi \colon C'\longrightarrow C$

such that ${}\varphi ^{*}(c)$ is trivial.

Proof

If the degree of ${}{\mathcal {S}}$ is positive, then a Frobenius pull-back ${}F^{e*}({\mathcal {S}})$ has arbitrary large degree and is still semistable. By Serre duality we get that ${}H^{1}(C,F^{e*}({\mathcal {S}}))=0$. So in this case we can annihilate the class by an iteration of the Frobenius alone.

So suppose that the degree is ${}0$. Then there exists by Theorem 3.10 a finite morphism which trivializes the bundle. So we may assume that ${}{\mathcal {S}}\cong {\mathcal {O}}_{C}^{r}$. Then the cohomology class has several components ${}c_{i}\in H^{1}(C,{\mathcal {O}}_{C})$ and it is enough to annihilate them separately by finite morphisms. But this is possible by the parameter theorem of K. Smith (or directly using Frobenius and Artin-Schreier extensions).

$\Box$

The general case

We look now at an arbitrary locally free sheaf ${}{\mathcal {S}}$ on ${}C$, a smooth projective curve over a finite field. We want to show that the same numerical criterion (formulated in terms of the Harder-Narasimhan filtration) for non-affineness of a torsor holds also for the finite annihilation of the corresponding cohomomology class (or the existence of a projective curve inside the torsor).

Theorem

Let ${}K$ denote a finite field (or the algebraic closure of a finite field) and let ${}C$ be a smooth projective curve over ${}K$. Let ${}{\mathcal {S}}$ be a locally free sheaf over ${}C$ and let ${}c\in H^{1}(C,{\mathcal {S}})$ denote a cohomology class. Let ${}{\mathcal {S}}_{1}\subset \ldots \subset {\mathcal {S}}_{t}$ be a strong Harder-Narasimhan filtration of ${}F^{e*}({\mathcal {S}})$. We choose ${}i$ such that ${}{\mathcal {S}}_{i}/{\mathcal {S}}_{i-1}$ has degree ${}\geq 0$ and that ${}{\mathcal {S}}_{i+1}/{\mathcal {S}}_{i}$ has degree ${}<0$. We set ${}{\mathcal {Q}}=F^{e*}({\mathcal {S}})/{\mathcal {S}}_{i}$.

Then the following are equivalent.

1. The class ${}c$ can be annihilated by a finite morphism.
2. Some Frobenius power of the image of ${}F^{e*}(c)$ inside ${}H^{1}(C,{\mathcal {Q}})$ is ${}0$.

Proof

Suppose that (1) holds. Then the torsor is not affine and hence by Theorem 3.7 also (2) holds.

So suppose that (2) is true. By applying a certain power of the Frobenius we may assume that the image of the cohomology class in ${}{\mathcal {Q}}$ is ${}0$. Hence the class stems from a cohomology class ${}c_{i}\in H^{1}(C,{\mathcal {S}}_{i})$. We look at the short exact sequence

$0\longrightarrow {\mathcal {S}}_{i-1}\longrightarrow {\mathcal {S}}_{i}\longrightarrow {\mathcal {S}}_{i}/{\mathcal {S}}_{i-1}\longrightarrow 0,$

where the sheaf of the right hand side has a nonnegative degree. Therefore the image of ${}c_{i}$ in ${}H^{1}(C,{\mathcal {S}}_{i}/{\mathcal {S}}_{i-1})$ can be annihilated by a finite morphism due to Fakt *****. Hence after applying a finite morphism we may assume that ${}c_{i}$ stems from a cohomology class ${}c_{i-1}\in H^{1}(C,{\mathcal {S}}_{i-1})$. Going on inductively we see that ${}c$ can be annihilated by a finite morphism.

$\Box$

Theorem

Let ${}C$ denote a smooth projective curve over the algebraic closure of a finite field ${}K$, let ${}{\mathcal {S}}$ be a locally free sheaf on ${}C$ and let ${}c\in H^{1}(C,{\mathcal {S}})$ be a cohomology class with corresponding torsor ${}T\rightarrow C$.

Then ${}T$ is affine if and only if it does not contain any projective curve.

Proof

Due to Theorem 3.7 and Theorem 3.12, for both properties the same numerical criterion does hold.

$\Box$

These results imply the following theorem in the setting of a two-dimensional graded ring.

Theorem

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

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

This is also true for non-primary graded ideals and also for submodules in finitely generated graded submodules. Moreover, G. Dietz has shown that one can get rid also of the graded assumption (of the ideal or module, but not of the ring).

Geometric deformations - A counterexample to the 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 ${}R$?

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 ${}D$.

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 field ${}{\mathbb {F} }_{p}[t]\subseteq K$)

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

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

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

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

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

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. 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 ${}f$ 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

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 the previous Lemma shows that ${}f\notin I^{*}$.

$\Box$

In terms of affineness (or local cohomology) this example has the following properties: 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 ${}Y\to C_{{\mathbb {F} }_{2}(t)}$ which annihilates the cohomology class ${}c$ and this would extend to a morphism of relative curves almost everywhere.

$\Box$

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