Isomorphism of the perfection of two ring

I was working on Exercise 2.0.4 of Bhatt's notes, which are available here.

The exercise states:

Let $$f\colon R\to S$$ be a map of char $$p$$ rings that is surjective with nilpotent kernel. Then $$R^{perf}$$ and $$S^{perf}$$ are isomorphic. Here, $$R^{perf}:= \underleftarrow {\lim}R$$, the inverse limit of Frobenius automorphisms of $$R$$.

My solution: let N be the nilradical of R. We have an exact sequence $$0\to N\to R\to S\to 0.$$

Taking the inverse limit, we have

$$0 \to\underleftarrow {\lim}N\to R^{perf}\to S^{perf}$$ exact. So in order to show $$R^{perf}$$ and $$S^{perf}$$ are isomorphic, we have to prove that $$\underleftarrow {\lim}N$$ is zero, which implies that the Frobenius map on $$N$$ is surjective. However, this isn't always the case. (e.g $$R=\mathbb F_{5}[T]/(T^2)$$)

I think this exercise is elementary but I cannot figure out where is m mistake.

$$\newcommand{\perf}{\mathrm{perf}}$$ I'm not totally sure what you're trying to do. Can you add more detail?

If you're just looking for a solution, here's one I believe. By definition we have that

$$R^\perf=\{(x_i)\in R:x_{i+1}^p=x_i\}$$

Note then that the map $$f:R^\perf\to S^\perf$$ is merely the map $$f((x_i))=(f(x_i))$$. Let us set $$I:=\ker f$$ and suppose that $$I^{p^k}=0$$ for some fixed $$k\geqslant 0$$.

We first claim that $$f$$ is surjective. Suppose now that $$(y_0,\ldots)\in S^\perf$$. Let us then for all $$i\geqslant 0$$ take some $$x'_i$$ such that $$f(x'_i)=y_{i+k}$$. Let us then consider $$(x_i)$$ with $$x_i=(x'_i)^{p^k}$$. Note that $$f(x_i)=f(x'_i)^{p^k}=y_{i+k}^{p^k}=y_i$$. The only thing to verify is that $$x_{i+1}^p=x_i$$. To see this note that since

$$f((x'_{i+1})^p)=y_{i+k+1}^p=y_{i+k}=f(x'_i)$$

that $$(x'_{i+1})^p-x'_i\in I$$. This implies that

$$0=(x'_{i+1})^p-x'_i)^{p^k}=x_{i+1}^p-x_i$$

from where the conclusion follows.

To see that it's injective note that if $$f((x_i))=(0)$$ then $$x_i\in I$$ for all $$i$$. Note though that since $$x_i=x_{i+k}^{p^k}$$ this implies, since $$I^{p^k}=0$$, that $$x_i=0$$. The conclusion follows.

• Thank you very much! I have figured out where is my mistake.I misunderstood the condition "nilpotent kernel" and I thought it means the kernel is nilradical. I feel sorry for my stupid mistake. – Ff E Apr 27 at 10:26
• @FfE To be honest, it seems a bit ambiguous to me. The interpretation I took was that $(\ker f)^\ell=0$ for some $\ell$. Does that seem reasonable to you? – Alex Youcis Apr 27 at 10:28
• This is reasonable. In fact, we can show this is surjective by Mittag-Leffler condition(stacks.math.columbia.edu/tag/0594). This condition can be proved by the assumption that kernel is nilpotent. The injectivity is easy to see. – Ff E Apr 27 at 10:37
• @FfE Yeah, perhaps. It seems more natural to just do it by hand though. – Alex Youcis Apr 27 at 10:38