I was thinking about when the Frobenius endomorphism is an automorphism. It is a well known fact that if you are taking the Frobenius endomorphism over a finite field such that $Char(F) = p$, then the mapping is also an automorphism.
Could the requirement be loosened to a finite integral domain with characteristic $p$ instead? The below argument and its mistakes (if any, which is why I posted this on MSE) are solely mine. Does it hold?
In a finite set, a injective map is also a surjective one, so it only suffices to prove one. We shall prove that the Frobenius endomorphism is a injective map by a proof by contradiction.
Let $R$ be a finite ring with characteristic $p$ with the property that it is an integral domain. The Frobenius endomorphism will be denoted by $\phi$.
If $\phi$ wasn't injective, then there $\exists$ $a,b \in R$ where $a \ne b$ and $ a, b \ne 0$ such that $\phi(a) = \phi(b)$. This would mean that $a^p$ = $b^p$.
For some $n$ where $1 < n <p $, $a^n = b^n$. For any number smaller than $n$, let's say $m$, $a^m \ne b^m$. The intuition in this step is that if you start out with $a \ne b$ and end up with $a^p = b^p$, then at some point you are going to wind up having $a^n = b^n$.
Since $p = kn + m$ where $m < n$, we have $a^{kn+m} = b^{kn+m}$, then $a^{kn}a^m = b^{kn}b^m $. Since $a^n = b^n$, $a^{kn} = b^{kn}$. Thus $a^{kn}a^m = a^{kn}b^m \rightarrow a^m = b^m$ in an integral domain, which is a contradiction!
Are there any errors in the proof? I am worried about the the claim that there is some $m < p$ where $a^m \ne b^m $ even though I think it's true.
I can't find any resources online on how the Frobenius endomorphism behaves in an integral domain. If anyone would also be kind enough to point me to more resources, I'd be grateful.