# Converse to freshman's dream

Suppose $x\to x^n$ is a ring homomorphism from $R$ to $R$, a commutative ring with unity. DoEs it imply that $R$ has characteristic $n$ (which is prime)?

No, a simple counterexample is with $n=3$ and $R=\mathbb{Z}_2$.
If this map is a ring homomorphism for some ring R whose characteristic $m$ does not divide $n$, then the restriction to Z/mZ is also a ring homomorphism. So it's enough to check if the identity holds in any such ring.