Does there exist a point $ x $ of an affine toric variety $ U_{\sigma} $ such that $ x $ is not compatibly split?

Question

A variety $ X $ is $ F $-split if there exists an $ \mathcal{O}_{X} $-linear map $ \phi: F_{\ast}(\mathcal{O}_{X}) \to \mathcal{O}_{X} $ such that $ \phi \circ F^{\sharp} = \operatorname{id}_{\mathcal{O}_{X}} $. Such a map $ \phi $ is called a splitting. A closed sub-scheme $ Y $ of $ X $ is compatibly split if there exists a splitting $ \phi $ such that $ \phi(F_{\ast}(\mathcal{I}_{Y})) \subseteq \mathcal{I}_{Y} $. Note that $ \mathcal{I}_{Y} \subseteq \phi(F_{\ast}(\mathcal{I}_{Y})) $ already.

Toric varieties are $ F $-split. Does anyone know of an example of an affine, toric, variety $ U_{\sigma} $ and a point $ x \in U_{\sigma} $ such that $ x $ is not compatibly split? Here I am using Hartshorne's definition of a "variety" as an integral, separated, scheme of finite type over an algebraically closed field.

Answer 1

Actually, on an affine variety, if it is $F$-split, then it is compatibly $F$-split with every point (possibly changing the splitting). This is not true in the projective case (an ordinary projective elliptic curve is $F$-split, but not compatibly split with any point). It is also not true for non-closed points (for instance the generic point of a cusp in $\mathbb{A}^2$ cannot be compatibly split, since that would force the cusp itself to be $F$-split).

Lemma. If $R$ is a $F$-split $F$-finite ring, then for any maximal ideal $Q \subseteq R$, there is a Frobenius splitting compatibly split with $m$.

Proof. Suppose $R$ is an $F$-split ring and $Q \in Spec R$ is any closed point. Let $\phi : F_* R \to R$ be an $F$-splitting. If $\phi$ is compatible with $Q$ we are done, so suppose its not compatible with $Q$. Thus $\phi(F_* Q) \not\subseteq Q$. Since outside of $Q$, $\phi$ is surjective, we actually see that $\phi(F_* Q) = R$. Thus there exists $x_1 \in Q$ with $\phi(F_* x_1) = 1$. Let $\phi_1 = \phi \circ (\cdot F_* x)$. In other words, $\phi_1(F_* y) = \phi(F_* xy)$. Note $\phi_1$ is also a Frobenius splitting (since it also sends $1$ to $1$).

If $\phi_1$ is compatible with $Q$, we are also done, so we repeat the process. Continuing in in this way, we eventually come to a Frobenius splitting $\phi_{n}$ where $n > p d + 1$ where $d$ is the number of generators of the ideal $Q = (f_1, \dots, f_d)$. Now, $$1 = \phi_n(F_* 1) = \phi(F_* x_1 x_2 \dots x_n).$$ By construction, $x_1 x_2 \dots x_n \in Q^{pd + 1} \subseteq (f_1^p, \dots, f_d^p) =: Q^{[p]}$, that containment is basically the pigeonhole principal. On the other hand, an easy computation shows that $\phi(F_* Q^{[p]}) \subseteq Q$ and so we see that $1 = \phi_n(F_* 1) \in Q$, a contradiction. Thus at some earlier point we must have had a Frobenius splitting $\phi_i$ that was compatible with $Q$.