Automorphisms of $\mathbb Z_p[x]$

Question

I am trying to find all automorphisms of $\mathbb Z_p[x]$ (polynomials with coefficients from $\mathbb Z_p$ where $p$ is prime).

I know that automorphisms of $\mathbb Z[x]$ are $x\to x$ and $x\to -x$, but now when coefficients are in $\mathbb Z_p$, I am not entirely sure.

By $Z_p$, do you mean $\mathbb{Z}/p\mathbb{Z}$ (i.e. the integers mod $p$), or the $p$-adic numbers (i.e. the inverse limit of $(\mathbb{Z}/p^n\mathbb{Z})_{n=1}^{\infty}$)? — Nick Peterson, Jul 24 '13 at 01:47
@Daniel : Good point. These automorphisms are very important when one studies polynomial irreducibility for instance.
I'll go to sleep, I'm saying stupid stuff. -_- — Patrick Da Silva, Jul 24 '13 at 01:51

score 0 · Accepted Answer · answered Jul 24 '13 at 02:28

As you have probably noticed, an endomorphism of this ring is fully determined by its value on $x$ (because it must be identity on ${\bf Z}_p$, as $1\mapsto 1$ and others are multiples of $1$), and any choice of this value yields an endomorphism.

The question that remains is for what choices of a polynomial $P$ does $x\mapsto P$ give an automorphism.

Note that we always have all the constants in the range of an endomorphism, and ${\bf Z}_p$ is a field, so for an endomorphism to be surjective its enough for its range to contain some degree one polynomial. It's not hard to see that it is true if and only if $P$ has degree $1$ (this follows from the fact that ${\bf Z}_p$ is a domain).

On the other hand, a simple analysis of highest order terms can show that such an endomorphism has nontrivial kernel if and only if $P$ is constant (again, because ${\bf Z}_p$ is a domain).

In summary, $x\mapsto P$ yields an automorphism iff $P$ has degree $1$.

thank you for your elaborate answer. It is clear and pretty similar to the way I worked the $Z[x]$ case. If I am to describe these automorphisms, would it just be x -> x + k and x -> -x + k for any k in $Z_p$? — stella, Jul 24 '13 at 13:46
@stella: No, it will be $x\mapsto ax+b$ for $a\in {\mathbf Z}_p\setminus {0},b\in {\bf Z}_p$. This is the same as you've said for $p=2,3$ but for larger $p$ there's quite a bit more than that. — tomasz, Jul 24 '13 at 14:22
Oh, I see. I was misguided by the case of Z[x]. Thank you @tomasz — stella, Jul 24 '13 at 15:00

Automorphisms of $\mathbb Z_p[x]$

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