# Is this an automorphism of a commutative ring of characteristic $p$?

Let $R$ be a commutative ring of characteristic $p$ (prime). It is evident that $a\mapsto a^p$ is an endomorphism of $R$. But is it an automorphism?

This is the same as asking whether the equation $x^p = 0$ has only $x =0$ as a solution. It doesn't look true to me. If $R$ is a domain then clearly this is true. What about non-domains?

Maybe this is very easy but I'm a beginner and I don't have much examples of rings. All the rings I know either satisfy this or do not have characteristic $p$.

• This in general may be true or not. Even for domains this may not be surjective. – Crostul Jul 2 '17 at 21:14
• @Crostul ouch. I somehow overlooked surjectivity. – Cauchy Jul 2 '17 at 21:15

Let $F$ be a field of characteristic $p$ and consider the ideal $(X^p)$ generated by $X^p$ in the polynomial ring $F[X]$.

Then $R=F[X]/(X^p)$ has characteristic $p$ and contains an element $r$ such that $r\ne0$ and $r^p=0$.

If you consider $K=F(X)$ (the field of rational functions) and an indeterminate $t$ over $K$, then the endomorphism $f(t)\mapsto f(t)^p$ of $K(t)$ is not surjective, because $t^p-X$ cannot be a $p$-th power.

• Very nice example! Thanks – Cauchy Jul 2 '17 at 21:19

Even if R is a field, the question has no trivial answer. In that case, it is an automorphism iff R is a perfect field. The endomorphism you are looking for is called the Frobenius endomorphism. You can look for the internet and for the literature to discover several characterizing properties. Hope this helps

• It helps! Thanks – Cauchy Jul 2 '17 at 21:20

Consider the case $R = \mathbb{F}_p[x]$, then clearly, there is no polynomial $f$ such that $f^p = x$, thus $a \mapsto a^p$ is not surjective, although $R$ is a domain.

But you are right that is is injective if $R$ is domain.