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I am currently taking a short course on p-adic geometry, and coming up with the following exercise:

Exercise: Let $f: A \rightarrow B$ be a ring homomorphism of rings of characteristic $p>0$. It induces a homomorphism $$ f^{\mathrm{perf}}: A^{\mathrm{perf}} \rightarrow B^{\mathrm{perf}}. $$ Then

  • $f^{\mathrm{perf}}$ is injective if and only if $\bigcap_{n} (\ker f)^{p^n} = 0$.
  • $f^{\mathrm{perf}}$ is surjective if and only if $\mathrm{im}(f)[\mathrm{perf}] = B[\mathrm{perf}]$, where for a ring $R$ of char $p>0$, $R[\mathrm{perf}]$ is the set of perfect elements in $R$, that is, elements with arbitrary $p$-power roots.

I can figure out the first conclusion by straightforward definitions. But I am getting stuck on the if direction of the surjectivity, i.e. the direction to show the surjectivity of $f^{\mathrm{perf}}$.

My attempts: Let $b=(b_0, \ldots) \in B^{\mathrm{perf}}$, then I am hoping to find an $a = (a_0, \ldots) \in A^{\mathrm{perf}}$ that maps to $b$.

As $b_0$ is a perfect element in $B$ (with arbitrary $p^{n}$-th power root $b_n$), by assumption $b_0$ lies in $\mathrm{im}(f)[\mathrm{perf}]$. Hence there exists $a_0, a_1, \ldots \in A$ such that $f(a_0) = b_0$ with $b_i^{\prime} = f(a_i)$ as a $p^n$-th power root of $b_0$. But then I got stuck on adjusting $a_i$ such that simultaneously

  1. $(a_i) \in A^{\mathrm{perf}}$, and
  2. $f(a_i) = b_i$.

Actually for (1), since $(b_i^{\prime})^{p^i} = b_0$, we see that $f(a_i^{p^i}) = f(a_0)$. To show $a_i^{p^i} = a_0$, it seems that we need some injectivity condition on $f$?

So I doubt whether the conclusion in the exercise is correct or not. I have read the post Isomorphism of the perfection of two ring, where the proof of the surjectivity part does rely on the nilpotency of $\ker f$.

And for (2), I am actually hoping to use the observation that not only $b_0 \in B[\mathrm{perf}]$, but every $b_i \in B[\mathrm{perf}]$ as well. So actually for each $i \geq 0$, we can do the same thing as above, and get $$ b_0= f(a_0^{(0)}), b_1^{(0)} = f(a_1^{(0)}), b_2^{(0)} = f(a_2^{(0)}), b_3^{(0)} = f(a_3^{(0)}), \ldots; $$ $$ b_0, b_1 = f(a_1^{(1)}), b_2^{(1)} = f(a_2^{(1)}), b_3^{(1)} = f(a_3^{(1)}), \ldots; $$ $$ b_0, b_1 , b_2, b_3^{(2)} = f(a_3^{(2)}), \ldots; \vdots $$ But it seems that this only complicates things, and I have no idea how to carry on.

So now I am getting stuck here. Thank you all for commenting and answering!

Hetong Xu
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  • Indeed, one really needs the kernel to be nilpotent in the question that you link (though maybe some other finiteness hypothesis can yield Mittag-Leffler and thus the conclusion). How confident are you that the statement is correct as is? – Aphelli Jul 05 '23 at 13:46
  • How about a surjection from $\mathbb{F}_p[T]$ to a finite field (strictly greater than $\mathbb{F}_p$)? – Aphelli Jul 05 '23 at 13:54

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The criterion for surjectivity seems false as is. It implies in particular that $f^{perf}$ is onto whenever $f$ is.

But consider a surjection $f: A:=\mathbb{F}_p[T] \rightarrow F$, where $F \neq \mathbb{F}_p$ is a finite field. Then $A^{perf}=\mathbb{F}_p$, so that $f^{perf}$ cannot be surjective.

Aphelli
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  • Thank you so much! Now I see that indeed the criterion is false. I will ask the lecturer about this tomorrow. Thank you for your counterexample! I realize how unfamiliar I am with such concrete examples of perfections. :( – Hetong Xu Jul 06 '23 at 01:03
  • Before I saw your counterexample, actually I was considering correcting the conclusion as $f(A[\mathrm{perf}])=B[\mathrm{perf}]$. As your example shows, this is also daydreaming, since it seems that $A[\mathrm{perf}] = \mathbb{F}_p$ and its image via $f$ is a copy of $\mathbb{F}_p$, while $B[\mathrm{perf}] = F \neq \mathbb{F}_p$. – Hetong Xu Jul 06 '23 at 01:09