Is the Frobenius automorphism a polynomial?

Question

Let $p$ be a prime and $\Bbb{Z}/\Bbb{Z}_p$ the field of integers mod $p$, and since $\Bbb{Z}/\Bbb{Z}_p$ is a field we have the ring of polynomials in $X$ with coefficients in $\Bbb{Z}/\Bbb{Z}_p$ denoted $\Bbb{Z}/\Bbb{Z}_p[X]$ and we have the Frobenius automorphism $\varphi:x\mapsto{x}^p$ $\forall{x}\in{\Bbb{Z}/\Bbb{Z_p}}$ My question is this: is $\varphi\in{\Bbb{Z}/\Bbb{Z_p}}[X]$ ? My reasoning is that since any polynomial $q\in{\Bbb{Z}/\Bbb{Z_p}}[X]$ can be written as $$q=\sum_{i=0}^n{a_i{X}^i}$$ with $a_i\in{\Bbb{Z}/\Bbb{Z_p}}$ and since $\varphi:x\mapsto{x}^p$ can be written $$\varphi=\sum_{i=0}^pa_i{X}^i$$ where we set $a_1,...a_{p-1}=0$ and $a_p=1$ to obtain $1{X}^p={X}^p$ that $\varphi$ should be a polynomial but for some reason I am unsure. Any help or confirmation would be appreciated.

Answer 1

Your notation for the field with $p$ elements is weird. Write $\mathbb{Z}/{p\mathbb{Z}}$ or $\mathbb{F}_p$ instead.

By definition, the Frobenius automorphism is the polynomial function corresponding to the polynomial $X^p$ in $\mathbb{F}_p[X]$. Pay attention to distinguish between a polynomial and its associated polynomial function.

Exercise: Any function from $\mathbb{F}_p$ is a polynomial function.

Exercise: Compute the kernel of the map that to a polynomial associates its polynomial function.

Answer 2

The field $\mathbb{Z}_{p}$ (where $p$ is prime) does not have any nontrivial automorphisms. If you have a finite field $\mathbb{F}_{q}$ of order $q = p^{e}$ (with $e > 1$), then the automorphism $\varphi : a \mapsto a^{p}$ is the Frobenius automorphism of $\mathbb{F}_{q}$. While you can think of this as being a polynomial map, you would not say that the automorphism is a polynomial; a polynomial is a distinct object from the mapping it defines.