Fermat cubic/xyz in (x^2,y^2,z^2)^*/Cohomological proof/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 ${}f=xyz$. We consider the short exact sequence

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

and the cohomology class

${}c=\delta (xyz)=H^{1}{\left(U,\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}\,.$

We want to show that ${}zF^{e*}(c)=0$ for all ${}e\geq 0$ (here the test element ${}z$ equals the element ${}z$ in the ring). It is helpful to work with the graded structure on this syzygy sheaf (or to work on the corresponding elliptic curve ${}\operatorname {Proj} R$ directly). Now the equation ${}x^{3}+y^{3}+z^{3}=0$ can be considered as a syzygy (of total degree ${}3$) for ${}x^{2},y^{2},z^{2}$, yielding an inclusion

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

Since this syzygy does not vanish anywhere on ${}U$ the quotient sheaf is invertible and in fact isomorphic to the structure sheaf. Hence we have

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

and the cohomology sequence

$\longrightarrow H^{1}{\left(U,{\mathcal {O}}_{U}\right)}_{s}\longrightarrow H^{1}{\left(U,\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}_{s+3}\longrightarrow H^{1}{\left(U,{\mathcal {O}}_{U}\right)}_{s}\longrightarrow 0,$

where ${}s$ denotes the degree-${}s$th piece. Our cohomology class ${}c$ lives in ${}H^{1}{\left(U,\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}_{3}$, so its Frobenius pull-backs live in ${}H^{1}{\left(U,F^{e*}\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}_{3q}$, and we can have a look at the cohomology of the pull-backs of the sequence, i.e.

$\longrightarrow H^{1}{\left(U,{\mathcal {O}}_{U}\right)}_{0}\longrightarrow H^{1}{\left(U,F^{e*}\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}_{3q}\longrightarrow H^{1}{\left(U,{\mathcal {O}}_{U}\right)}_{0}\longrightarrow 0.$

The class ${}zF^{e*}(c)$ lives in ${}H^{1}{\left(U,F^{e*}\operatorname {Syz} {\left(x^{2},y^{2},z^{2}\right)}\right)}_{3q+1}$. It is mapped on the right to ${}H^{1}{\left(U,{\mathcal {O}}_{U}\right)}_{1}$, which is ${}0$ (because we are working over an elliptic curve), hence it comes from the left, which is

${}H^{1}(U,{\mathcal {O}}_{U})_{1}=0\,.$

So ${}zF^{e*}(c)=0$ and ${}f\in {\left(x^{2},y^{2},z^{2}\right)}^{*}$.