Computing $\Bbb F_p[t]^{perf}$

Question

This is the second example of 1. in Ex. 2.0.3 of Bhatt's notes in perfectoid space.

We define $R^{perf}:= \varprojlim ( \cdots R \xrightarrow{\phi} R)$ where $\phi$ is the Frobenius map.

He claims that $R=\Bbb F_p[t]$ we have $R^{perf}=\Bbb F_p$.

I don't see why.

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My thoughts: From the map $\Bbb F_p[t] \rightarrow \Bbb F_p$, $t \mapsto 0$. We get an induced map on limits. Since $\Bbb F_p$ is perfect, $$g:\varprojlim \Bbb F_p[t] \rightarrow \Bbb F_p$$

There is also a canonical $\Bbb F_p \rightarrow \Bbb F_p[t]$, inducing $$h:\Bbb F_p \rightarrow \varprojlim \Bbb F_p[t]$$

I'm guess that these two maps are inverses. Its clear that $hg=id$. However, it is less clear to me why $gh=id$.

Answer 1

The Frobenius map $\phi:\mathbb{F}_p[t]\to\mathbb{F}_p[t]$ is very simple: it's just the map that sends $t$ to $t^p$. So, identifying $\mathbb{F}_p[t]$ with $\mathbb{F}_p[t^p]$ by the obvious isomorphism, we can think of it as just the inclusion map $\mathbb{F}_p[t^p]\to\mathbb{F}_p[t]$. This means that the entire inverse system $$\dots\to \mathbb{F}_p[t]\stackrel{\phi}{\to} \mathbb{F}_p[t]\stackrel{\phi}{\to} \mathbb{F}_p[t]$$ is isomorphic to the inverse system $$\dots \to\mathbb{F}_p[t^{p^2}]\to \mathbb{F}_p[t^p]\to\mathbb{F}_p[t]$$ where the maps are inclusions. The inverse limit of a system of inclusions is just the intersection, so the inverse limit is $\bigcap_n\mathbb{F}_p[t^{p^n}]=\mathbb{F}_p$.

More generally, if $R$ is a ring of characteristic $p$ on which the Frobenius is injective, then a similar argument shows that $R^{perf}$ is the subring of $R$ consisting of elements that have $p^n$th roots for all $n$.

Answer 2

Your map $\Bbb F_p[t]\to\Bbb F_p$ isn't leading to a map $\varprojlim\Bbb F_p[t]\to\Bbb F_p$ by a universal property, rather you can get such a map from the composition $\varprojlim\Bbb F_p[t]\to\Bbb F_p[t]\to\Bbb F_p$ where the first map is a natural projection onto some factor. But let's ignore this point and define things in a different way.

As you pointed out, you have an inclusion map $\Bbb F_p\to\Bbb F_p[t]$, and this leads to a map $\Bbb F_p\to\varprojlim\Bbb F_p[t]$ by the universal property of limits (it works because the inclusion of $\Bbb F_p$ commutes with the Frobenius map $\Bbb F_p[t]\to\Bbb F_p[t]$, which boils down to the fact that $a^p=a$ for any $a\in\Bbb F_p$). Explicitly the map we have is $a\mapsto (\dots,a,a)$ and it is clearly injective.

Let's prove it is surjective: take some $(\dots,f_2,f_1)\in\varprojlim\Bbb F_p[t]$. Thus $f_{n+1}^p=f_n$ for each $n$. Using induction you can show that $f_k^{p^k}=f_1$ for any $k$. Now let $d:=\deg(f_1)$. Take some $k$ for which $p^k>d$, then we have $$d=\deg(f_1)=\deg(f_k^{p^k})=p^k\cdot\deg(f_k).$$

Since $p^k>d$, the only way this is not a contradiction is if $\deg(f_k)=0$, which implies $\deg(f_i)=0$ for all $i$, so $f_i\in\Bbb F_p$ for all $i$, and since Frobenius fixes elements of $\Bbb F_p$ we just conclude that all of the $f_i$ are the same element equal to some $a\in\Bbb F_p$, giving surjectivity.