Fermat-Kubik/z^2 in tight closure von (x,y)/Beispiel/en

We consider the Fermat cubic ${}R=K[X,Y,Z]/{\left(X^{3}+Y^{3}+Z^{3}\right)}$, the ideal ${}I=(X,Y)$ and the element ${}Z^{2}$. We claim that in positive characteristic ${}\neq 3$ the element ${}Z^{2}$ does belong to the tight closure of ${}I$. Equivalently, the open subset

${}D(X,Y)\subseteq \operatorname {Spec} {\left(R[S,T]/(XS+YT+Z^{2})\right)}\,$

is not affine. The element ${}Z^{2}$ defines the cohomology class

${}c={\frac {Z^{2}}{XY}}\in H^{1}(D(X,Y),{\mathcal {O}}_{X})\,$

and its Frobenius pull-backs are

${}F^{e*}(c)={\frac {Z^{2q}}{X^{q}Y^{q}}}\in H^{1}(D(X,Y),{\mathcal {O}}_{X})\,.$

This cohomology module has a ${}\mathbb {Z}$-graded structure (the degree is given by the difference of the degree of the numerator and the degree of the denominator) and, moreover, it is ${}0$ in positive degree (this is related to the fact that the corresponding projective curve is elliptic). Therefore for any homogeneous element ${}t\in R$ of positive degree we have ${}tF^{e*}(c)=0$ and so ${}Z^{2}$ belongs to the tight closure.

From this it follows also that in characteristic ${}0$ the element ${}Z^{2}$ belongs to the solid closure, because affineness is an open property in an arithmetic family.