Is the mapping an automorphism?

Question

$R$ is a commutative ring with unity and characteristic $P$. Where $P$ is a prime. $F$ is a function mapping each element of $R$ into its $P$-th multiplicative power. I have proven that $F$ is a homomorphism from $R$ to $R$. So far, I have proved that $F$ is injective when $R$ is an integral domain. Is it surjective too in this case? Is it an automorphism when $R$ is not an integral domain? Kindly provide me with some hints.

Answer 1

Consider the ring $F_p[X]$ of polynomial functions defined on the finite field $F_p$, $f(X)\rightarrow f(X)^p$ is not surjective.

Answer 2

The map $x \mapsto x^p$ is called the Frobenius endomorphism.

The map may be an automorphism even when $R$ is not an integral domain.

For instance, for $R=\mathbb F_p \times \mathbb F_p$, the map is the identity.

Answer 3

In the case of $\mathbb{Z}/p\mathbb{Z}$ the function $x \mapsto x^p$ is an automorphism since by Fermat's little theorem we have $a^p \equiv a \bmod p$ for any integer $a$ and prime $p$.