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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 ?

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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.

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  • $\begingroup$ 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 ... $\endgroup$ – sde May 16 at 3:26
  • $\begingroup$ 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. $\endgroup$ – Takumi Murayama May 16 at 4:54
  • $\begingroup$ 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 ... $\endgroup$ – sde May 16 at 6:50
  • $\begingroup$ 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 $\endgroup$ – sde May 16 at 6:58
  • $\begingroup$ 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. $\endgroup$ – Takumi Murayama May 16 at 15:59

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