Uniqueness of prime-power fields

Question

I'm still stuck on the proof of the following theorem. I've asked two questions so far to get to where I am even at this point.

Theorem: Let $p$ be a prime and let $n\in\mathbb{Z}^{+}$. If $E$ and $E'$ are fields of order $p^{n}$, then $E\cong E'$.

Proof: Both $E$ And $E'$ have $\mathbb{Z}_{p}$ as prime fields (up to isomorphism). By Corollary 33.6, $E$ is a simple extension of $\mathbb{Z}_{p}$ of degree $n$, so there exists an irreducible polynomial $f(x)$ in $\mathbb{Z}_{p}[x]$ such that $$E\cong \mathbb{Z}_{p}[x] / \langle f(x)\rangle$$

The next line stumps me again. "Since the elements of $E$ and $E'$ are exactly the roots of $x^{p^{n}} - x$, $f(x)$ is a factor of $x^{p^{n}} - x$ in $\mathbb{Z}_{p}[x]$."

The rest of the proof:

"Because $E'$ also consists of zeros of $x^{p^{n}} - x$, we see that $E'$ also contains zeros of the irreducible $f(x)$ in $\mathbb{Z}_{p}[x]$. Thus because $E'$ also contains $p^{n}$ elements, $E'$ is also isomorphic to $\mathbb{Z}_{p}[x] / \langle f(x)\rangle$."

It follows from work already done that elements of $E$ and $E'$ are zeros of $x^{p^{n}} - x$. But I don't follow how we get $f(x)$ as a factor of $x^{p^{n}} - x$ from this. This is actually what prompted my Question about algebraic field extensions as I thought that might be relevant; but it turns out not to be the reason.

Thanks very much for any help you can give.

Answer 1

Are you missing the following bits?

1. The polynomial $f(x)\in \mathbb{Z}_p[x]$ of degree $n$ has a root in the field $E$. Because the field extension $E/\mathbb{Z}_p$ is Galois, and $f(x)$ is irreducible, all the roots of $f(x)$ are distinct and in $E$. So $$ f(x)=(x-a_1)(x-a_2)\cdots (x-a_n) $$ for some elements $a_1,a_2,\ldots,a_n\in E$.
2. The elements of $E$ are exactly the zeros of the polynomial $p(x)=x^{p^n}-x$. In other words $$ p(x)=\prod_{a\in E}(x-a). $$

The troubling claim follows from this. The zeros $a_i,i=1,2,\ldots,n,$ are among the zeros of $p(x)$, so $f(x)\mid p(x)$.

In particular, the polynomial $f(x)$ also has $n$ zeros in $E'$, because $p(x)$ has $p^n$ roots there, and the roots of $f(x)$ are among those.

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Edit: Proving my first claim. This depends heavily on the properties of the so called Frobenius homomorhpism $F:E\to E, x\mapsto x^p$. This is a homomorphism, because obviously $F(1)=1$ and $F(xy)=(xy)^p=x^py^p=F(x)F(y)$ for all $x,y\in E$. Less obvious is that $F$ respects addition as well, i.e. $$ F(x+y)=(x+y)^p=x^p+y^p=F(x)+F(y) $$ for all $x,y\in E$. This follows from the binomial formula together with the observation that the binomial coefficients ${p\choose i}$ are all divisible by $p$, when $1\le i\le p-1$.

From little Fermat it follows that $F(x)=x^p=x$ for all the elements $x$ of the subfield $\mathbb{Z}_p$. We need to also make the observation that $x^p=x$ only when $x\in\mathbb{Z}_p$. This is because the polynomial equation $x^p-x=0$ can have at most $p$ solutions in the field $E$, and we already found $p$ solutions.

So we assume that $f(x)=x^n+f_{n-1}x^{n-1}+f_{n-2}x^{n-2}+\cdots+f_1x+f_0\in \mathbb{Z}_p[x]$ is irreducible, and has a root $a_1$ in $E$ (=the coset of $x$ in $\mathbb{Z}_p[x]/\langle f(x)\rangle$). In other words $a_1\notin\mathbb{Z}_p$ and $$ a_1^n+f_{n-1}a_1^{n-1}+\cdots f_1a_1+f_0=0. $$ Let's apply the mapping $F$ to this equation. Remember that $F(f_i)=f_i$ for all $i$. We get $$ a_1^{pn}+f_{n-1}a_1^{p(n-1)}+\cdots f_1a_1^p+f_0=0, $$ or, upon inspection, $f(a_1^p)=0$. Because $a_1\notin\mathbb{Z}_p$, $a_1^p\neq a_1$. Therefore we have found another zero $a_2=a_1^p$ of $f(x)$ in $E$.

We can repeat the argument and keep finding roots of $f(x)$: $a_3=a_2^p$, $a_4=a_3^p$ et cetera. Because $f(x)$ can have at most $n$ roots in $E$, this sequence of roots will have to start repeating at some point. Because $F$ is injective (its kernel is trivial), the repetition must start from $a_1$, in other words $a_1=a_1^{p^k}$ for some $k, 2\le k\le n$.

The polynomial $$ g(x)=(x-a_1)(x-a_1^p)\cdots (x-a_1^{p^{k-1}}) $$ is stable under $F$, so its coefficients are in $\mathbb{Z}_p$. Furthermore, $g(x)\mid f(x)$. But $f(x)$ was irreducible, so we must have $g(x)=f(x)$, and $k=n$. But all the roots of $g(x)$ are distinct and in $E$ by construction. Therefore the same holds for $f(x)=g(x)$.

Answer 2

If $E$ is a finite field with $q=p^f$ elements, then we know that its multiplicative group is cyclic (of order $q-1$). Thus the elements of $E$ are exactly the roots of $X^q-X$ in a (fixed) algebraic closure of $\Bbb F_p$.

Note that the polynomial $X^q-X$ has distinct roots (also) in characteristic $p$.

Answer 3

Note: This answer may very well be incorrect, but I cannot yet see why. It looks messy only because I am insisting on being explicit about identifications as I am still uncomfortable with ignoring them at this point.

Since $E$ and $E'$ have $p^{n}$ elements, they are degree $n$ finite extensions of appropriate subfields which are isomorphic to $\mathbb{Z}_p$ (any other degree would yield a cardinality different from $p^{n}$).

Roughly speaking, they are isomorphic because they are both $n$ dimensional vector spaces over isomorphic scalar fields.

To make this more precise, I will take a typical element of $E$ and and by converting coefficients and basis elements (via given isomorphisms) I will map it to an appropriate element of $E'$.

Proof: $E$ and $E'$ are degree $n$ finite extensions of appropriate subfields which are isomorphic to $\mathbb{Z}_p$. Denote these subfields by $E_p$ and $E_p'$. Then letting $\phi:E_{p}\to \mathbb{Z}_{p}$ and $\phi':E_{p}'\to \mathbb{Z}_{p}$ be the respective isomorphisms.

Then let $B = \{b_1, ..., b_n\}\subset E$ and $B' = \{b_1', ..., b_n'\}\subset E'$ be $n$-element bases for $E$ as a vector space over $E_p$ and $E'$ as a vector space over $E_p'$, respectively.

For each $x\in E$, write $x = \sum_{i=1}^{n}a_{i}b_{i}$ for some choice of $a_{i}\in E_{p}$. Then define the map $f:E\to E'$ by $f(x) = \sum_{i=1}^{n}a_{i}'b_{i}'$, where $a_{i}' = [\phi'^{-1}\circ \phi](a_{i})$.