Fermat cubic/z not in tight closure of (x,y)/Example

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 ${}Z$. We claim that for characteristic ${}\neq 3$ the element ${}Z$ does not belong to the solid closure of ${}I$. Equivalently, the open subset

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

is affine. For this we show that the extended ideal inside the ring of global sections is the unit ideal. First of all we get the equation

${}X^{3}+Y^{3}=(XS+YT)^{3}=X^{3}S^{3}+3X^{2}S^{2}YT+3XSY^{2}T^{2}+Y^{3}T^{3}\,$

or, equivalently,

${}X^{3}{\left(S^{3}-1\right)}+3X^{2}YS^{2}T+3XY^{2}ST^{2}+Y^{3}{\left(T^{3}-1\right)}=0\,.$

We write this as

${}{\begin{aligned}X^{3}{\left(S^{3}-1\right)}&=-3X^{2}YS^{2}T-3XY^{2}ST^{2}-Y^{3}{\left(T^{3}-1\right)}\\&=Y{\left(-3X^{2}S^{2}T-3XYST^{2}-Y^{2}{\left(T^{3}-1\right)}\right)},\end{aligned}}$

which yields on ${}D(X,Y)$ the rational function

${}Q={\frac {S^{3}-1}{Y}}={\frac {-3X^{2}S^{2}T-3XYST^{2}-Y^{2}{\left(T^{3}-1\right)}}{X^{3}}}\,.$

This shows that ${}S^{3}-1=QY$ belongs to the extended ideal. Similarly, one can show that also the other coefficients ${}3S^{2}T,3ST^{2},T^{3}-1$ belong to the extended ideal. Therefore in characteristic different from ${}3$, the extended ideal is the unit ideal.