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.
 A: $\newcommand{\perf}{\mathrm{perf}}$ 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. 
