automorphisms of a finite field

Question

Let $F$ be a finite field of characteristic $p$ over $\mathbb{Z}_p$ with $p^r$ elements. Then I have proved that the map $\phi: x\mapsto x^p$ is a field automorphism of $F$. Moreover, $\phi(x)=x$ if and only if $x\in \mathbb{Z}_p$.

I am confused on:

(1). Is every field automorphism of $F$ fixing $\mathbb{Z}_p$ can be written in this form?

I do not know how to prove:

(2). $\phi$ is an invertible linear map on the $\mathbb{Z}_p$-vector space $F$ and determine the minimal polynomial of $\phi$ over $\mathbb{Z}_p$.

(3). Let $K$ be a finite field of characteristic $p$ over $\mathbb{Z}_p$ with $p^t$ elements. Then $K$ is a subfield of $F$ if and only if $t\mid r$.

(4). If $t\mid r$, then $\phi^t: x\mapsto x^{p^t}$ is a field automorphism of $F$ such that $\phi^t(x)=x$ if and only if $x\in K$. Moreover, how about the minimal polynomial of $\phi^t$ over $K$?

Answer 1

(1) Note that the "form" you are talking about, $x\mapsto x^p$, is just one single automorphism. It is not really a form at all. You are asking if this is the only automorphism. The answer is no. For one there is the trivial automorphism $x\mapsto x$, and by composing this so-called Frobenius automorphism with itself multiple times you also have $x\mapsto x^{p^2}$, $x\mapsto x^{p^3}$ etc. This sequence of automorphisms eventually repeats, since $x=x^{p^r}$ for all $x\in F$. (Exercise: why? Hint: every $x\in F^\times$ is a _-root of unity.) Note every field automorphism of $F$ automatically fixes $\Bbb Z/p\Bbb Z$ pointwise. (Exercise: why?)

(2) You have already proved $\phi:x\mapsto x^p$ is a field automorphism fixing $\Bbb Z/p\Bbb Z$ pointwise. Use this latter property to prove that $\phi$ is $\Bbb Z/p\Bbb Z$-linear, i.e. additive and $\phi(ax)=a\phi(x)$ for all $a\in\Bbb Z/p\Bbb Z$.

Every $x\in F$ satisfies $x^{p^r}=x$, which can be rewritten as $\phi^rx=x$, so we know $f(\phi)=0$ where $f(T)=T^r-T$. As $f$ is monic and degree $r$, it must be the characteristic polynomial of $\phi$. The minimal polynomial $m(T)$ must have degree $\le r$; suppose it is strictly $<r$. Write $m(\phi)x$ as a polynomial in $x$; what is its degree? Given its degree, how many roots can it have max? Can it be the zero map? What can we conclude?

(3) Hint ("only if"): if $F/K$ is an extension of fields then what is $|F|$ in terms of $|K|$ and $\dim_KF$?

Hint ("if" part): Use the fact that ${\Bbb F}_q$ must be a splitting field of $x^q-x$ (this fact follows from the discussion already above; do you see how?). It is an important field-theoretic fact that splitting fields are isomorphic, therefore a splitting field of a composite polynomial $f(T)g(T)$ will always contain a copy of a splitting field of $f(T)$ (why does this make sense?).

All of the reasoning put on the table now is sufficient to do (4). Try on your own.

Answer 2

Basically every statement on the list 1-4 is very simple but explaining it in isolation makes no sense. It is like solving a list of calculus problems without knowing that derivatives and integrals are related to each other.

You are probably missing several of the basic facts about finite fields, which are stated at Wikipedia, and make the answers (and maybe the proofs) to most of your questions relatively easy.

http://en.wikipedia.org/wiki/Finite_field

Let me quote some relevant parts

The finite fields are classified by size; there is exactly one finite field up to isomorphism of size $p^k$ for each prime p and positive integer k. Each finite field of size $q$ is the splitting field of the polynomial $x^q − x$, and thus the fixed field of the Frobenius endomorphism which takes x to $x^q$. Similarly, the multiplicative group of the field is a cyclic group.

The field $F_{p^n}$ contains a copy of $F_{p^m}$ if and only if m divides n.

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It may also be required to know a few simple facts about fields (not only finite ones), such as an extension of $F$ being a vector space over $F$, and the very first things from Galois theory, such as the fixed field (solutions of $\psi(x)=x$) of an automorphism being a subfield.