Two Properties of Perfect Rings I came across Kunz's theorem about the characterization of regular rings in characteristic $p$. In the paper that I am reading, the author uses perfect rings to prove this result.
Perfect rings $R$ are defined to be rings of characteristic $p>0$ where the Frobenius homomorphism
$$
F:x \rightarrow x^p
$$
is an automorphism.
He states without explanation some properties of perfect rings and I don't understand some of them.

*

*If $R$ is perfect and local such that the maximal ideal $\mathfrak{m}$ is finitely generated, then $R$ is a field.
Why do we need that $\mathfrak{m}$ is finitely generated? We need to show that $\mathfrak{m}$ is the entire ring $R$. But how can we use that $R$ is perfect to show this?


*If $R$ is perfect and Noetherian, then it can be written as a finite product of perfect fields.
Could someone explain these two properties to me? I would also appreciate any hints.
 A: First of all, I'm pretty sure you're assuming commutativity, and that is how I will continue.  I'll also denote the Frobenius automorphism as $\alpha$
(in problem 1) $R$ is reduced
The perfect condition is going to eliminate nilpotent elements.  It is well known that a commutative ring has no nonzero nilpotent elements iff it has no nonzero elements squaring to zero.
That being the case, if $x^2=0$ for some $x\in R$, then $x^p=0$ too, but then by injectivity of the Frobenius map, $x=0$.  Therefore $R$ is reduced.
(in problem 1)  The maximal ideal is zero

We need to show that $\mathfrak{m}$ is the entire ring $R$.

No, you need to show $\mathfrak{m}=\{0\}$.  The trick is Nakayama's lemma, and this observation using the perfect condition:
Claim: $\mathfrak{m}^p=\mathfrak{m}$. The Frobenius automorphism has to map $\mathfrak{m}$ onto itself, and $\alpha(\mathfrak{m})$ is a subset of the ideal product $\mathfrak{m}^p$. So there is equality between the three.
Then to continue your problem, by Nakayama's lemma (since $\mathfrak{m}$ is finitely generated and $\mathfrak{m}^{p-1}$ is contained in the Jacobson radical) $\mathfrak{m}=\{0\}$.  This means simply that $R$ is a field.
(problem 2)
I do not have a complete argument for the claim at this time, but I can get almost all of it.

*

*It's well known that a reduced Noetherian ring has finitely many minimal primes, and that $R$ injects into $\prod \{R/P\mid P \text{ a minimal prime of } R\}$ via the usual map that sends $r$ to its residues mod the various prime ideals.  In fact this is known as a subdirect product since it has some more special properties than just injecting $R$ as a subring.


*$R/P$ is perfect as well when $P$ is a prime, since the Frobenius endomorphism is clearly still onto, and also injective since the ring is a domain.


*The Frobenius endomorphism of a perfect domain extends to its field of fractions, so that its field of fractions is perfect.
From the foregoing it's obvious that $R$ embeds in a finite product of perfect fields, but I haven't seen a reason (if there is one) that it itself is isomorphic to a product of perfect fields.
