# Frobenius splitting for an excellent, non $F$-finite, $F$-pure hypersurface

Let $$p$$ be an odd prime. Let $$k$$ be a field of characteristic $$p$$ such that $$[k:k^p]=\infty$$ (i.e. $$k$$ is not $$F$$-finite ) .

Also assume that $$-1$$ is not a square in $$k$$ . Consider the homogeneous polynomial $$f(x,y)=x^2+y^2\in k[x,y]$$ . Then $$f$$ is irreducible in $$k[x,y]$$ , hence $$R=k[x,y]/(f)$$ is an integral domain (of dimension $$1$$ , hence Cohen-Macaulay) . Moreover, $$R$$ is Excellent. (https://en.m.wikipedia.org/wiki/Excellent_ring)

Also, $$R$$ is not $$F$$-finite (as $$k$$ is not).

We also observe that by Fedder's criteria, $$R$$ is $$F$$-pure since $$f^{p-1}\notin (x^p,y^p)$$ .

Indeed, to see $$f^{p-1}\notin (x^p,y^p)$$, observe that $$f^{p-1}=(x^2+y^2)^{p-1}=x^{p-1}y^{p-1}+\sum_{0\le j\le p-1 , j\ne \dfrac {p-1}2 } (x^2)^{j}(y^2)^{p-1-j}$$.

Now $$x^{p-1}y^{p-1}\notin (x^p, y^p)$$.

Moreover, either $$j$$ or $$p-1-j$$ is $$\ge \dfrac{p-1}2+1=\dfrac {p+1}2$$ if $$j\ne \dfrac {p-1}2$$, hence $$(x^2)^{j}(y^2)^{p-1-j}\in (x^p, y^p)$$ when $$j\ne \dfrac {p-1}2$$. Thus $$f^{p-1} \notin (x^p, y^p)$$.

My question is: When can we say that $$k[x,y]/(x^2+y^2)$$ is $$F$$-split ?

Your ring is Frobenius split. I also don't think you need the condition that $$k$$ does not contain a square root of $$-1$$.

Proof. Denoting by $$\overline{k}$$ the algebraic closure of $$k$$, we see that $$\overline{R} := R \otimes_k \overline{k} \simeq \frac{\overline{k}[x,y]}{x^2+y^2}$$ is $$F$$-pure by Fedder's criterion, using the same computation you gave. Since $$\overline{k}$$ is algebraically closed, it is $$F$$-finite. Thus, $$\overline{R}$$ is also $$F$$-finite. Combining these two facts, we see that $$\overline{R}$$ is Frobenius split.

Now consider the commutative diagram $$\require{AMScd}\begin{CD} R @>F_R>> F_{R*}R @>\phi>\exists?> R\\ @VVV @VVV @VVV\\ \overline{R} @>F_{\overline{R}}>> F_{\overline{R}*}\overline{R} @>\bar{\phi}>> \overline{R} \end{CD}\tag{1}\label{eq:basechange}$$ where the composition in the bottom row is the identity on $$\overline{R}$$ and the vertical maps are obtained by extending the field extension $$k \subseteq \overline{k}$$ by scalars along $$k \to R$$.

We want to show that the map $$\phi$$ exists such that the composition in the top row is the identity on $$R$$. Since $$k \subseteq \overline{k}$$ splits as a map of $$k$$-vector spaces, the map $$R \to \overline{R}$$ splits as a map of $$R$$-modules. Let $$f\colon \overline{R} \to R$$ be such a splitting. Then, defining $$\phi$$ to be the composition of the three arrows in the right square of the diagram $$\begin{CD} R @>F_R>> F_{R*}R @>\phi>> R\\ & @VVV @AAfA\\ & & F_{\overline{R}*}\overline{R} @>\bar{\phi}>> \overline{R} \end{CD}$$ we see that the composition in the top row of the diagram \eqref{eq:basechange} is the identity on $$R$$. Thus, $$R$$ is Frobenius split. $$\blacksquare$$

I also want to point out that this is a special case of a result due to Rankeya Datta and myself. The statement for $$F$$-purity for complete local rings I believe first appeared in [Fedder, Lem. 1.2].

Theorem (Datta–M; see [M1, Thm. B.2.3]). Let $$R$$ be a ring essentially of finite type over a noetherian complete local ring $$(A,\mathfrak{m})$$ of prime characteristic $$p > 0$$. Then,

1. $$R$$ is $$F$$-pure if and only if $$R$$ is Frobenius split; and
2. $$R$$ is strongly $$F$$-regular if and only if $$R$$ is split $$F$$-regular.

Here, $$R$$ is strongly $$F$$-regular if every inclusion of modules is tightly closed, following [Hochster, Def. on p. 166], and $$R$$ is split $$F$$-regular if for every element $$c$$ avoiding every minimal prime of $$R$$, there exists an integer $$e > 0$$ such that the composition $$R \overset{F^e_R}{\longrightarrow} F^e_{R*}R \xrightarrow{F^e_*(-\cdot c)} F^e_{R*}R$$ splits as a map of $$R$$-modules. Split $$F$$-regularity is the original definition for strong $$F$$-regularity in the $$F$$-finite case [Hochster–Huneke, Def. 5.1], although the terminology comes from [Datta–Smith, Def. 6.6.1].

The proof of the theorem uses the gamma construction of Hochster–Huneke [Hochster–Huneke, (6.7) and (6.11)] and [M2, Thm. 3.4], but the idea is to construct a diagram like that in \eqref{eq:basechange}, and use the fact (communicated to Hochster by Auslander) that pure maps from complete local rings split.

In your case, you can set $$A$$ to be the ground field $$k$$.

Finally, it is worth mentioning that the theorem does not hold for excellent rings in general: Rankeya and I found excellent regular rings, and even DVR's, that are not Frobenius split [Datta–M]. The rings we consider are Tate algebras and rings of convergent power series over non-Archimedean valued fields. The specific field we use was provided to us by Gabber, but one can also use fields constructed by Blaszczok and Kuhlmann.

• Thanks for your answer ... I was indeed looking for an excellent, F-pure domain which is not F-split, so your last paragraph is right on target and I'll carefully take time reading your paper. Also, just wanted to make sure of one thing : $F_{\bar R*} \bar R \cong F_{R*} R\otimes_k \bar k$ in the proof, right ? Oh and btw, the point about $-1$ being not a square in $k$ is indeed just a little nitpicking for my own satisfaction to make sure $k[x,y]/(x^2+y^2)$ is indeed a domain ... – sde May 16 at 3:26
• Hi @sde, I'm glad you found those examples interesting! The isomorphism you state does not hold: using identifications of the form $F_{S*}S \cong S^{1/p}$ for reduced rings $S$, the left-hand side is $\overline{k}^{1/p}[x^{1/p},y^{1/p}]/(x^{2/p}+y^{2/p})$, while the right-hand side is $(k^{1/p} \otimes_k \overline{k})[x^{1/p},y^{1/p}]/(x^{2/p}+y^{2/p})$, which is not reduced. To see this last statement, note that the subring $k^{1/p} \otimes_k \overline{k}$ is non-reduced by [Stacks, Tag 030W], for instance. – Takumi Murayama May 16 at 4:54
• I see, thanks for explaining ... then do you think $F_{\bar R*}\bar R\cong F_{R*}R\otimes_R \bar R$ holds true ? Actually I'm trying to see why $\bar R$ would still remain $F$-pure via some general possibly characteristic free argument ... – sde May 16 at 6:50
• Also, just double checking, that when working with an Algebra, essentially of finite type over a field, there's no need to distinguish between F-purity and F-splitting by the result proved by you and Datta, right ? It's just that this is kind of a cool big leap (but very natural going by your proof) 'cause all the lecture notes by Schwede or Hochster that I've read so far always assumes $F$-finiteness when going from F-purity to F-split while applying Fedder's criteria ..., but as your result shows, there's no need to assume that – sde May 16 at 6:58
• Hi @sde, my previous comment says that your isomorphism cannot hold. Also, even if $R$ is $F$-pure, $\overline{R}$ is sometimes not even reduced; see this example of Chevalley. Finally, yes, our result says that when working with essentially of finite type algebras over a field, there is no need to distinguish between $F$-purity and Frobenius splitting, although there are other reasons why one would want $F$-finite hypotheses. I should also mention that Fedder proved they coincide in the complete local case. – Takumi Murayama May 16 at 15:59