proof of characterization of perfect fields

Proposition 1

The following are equivalent:

1. Every algebraic extension of $K$ is separable.

2. Either $\mathrm{char}K=0$ or $\mathrm{char}K=p$ and the Frobenius map is surjective.

Proof.Suppose (a) and not (b). Then we must have $\mathrm{char}K=p>0$, and there must be $a\in K$with no $p$-th root in $K$. Let $L$ be a splitting field over $K$ for the polynomial $X^p-a$,and let $\alpha \in L$ be a root of this polynomial.Then ${(X-\alpha )}^p=X^p-{\alpha }^p=X^p-a$, which has coefficients in $K$.This means that the minimum polynomial for $\alpha$ over $K$ must be a divisor of${(X-\alpha )}^p$ and so must have repeated roots. This is not possible since $L$ is separableover $K$.

Conversely, suppose (b) and not (a).Let $\alpha$ be an element which is algebraic over $K$ but not separable.Then its minimum polynomial $f$ must have a repeated root, and by replacing $\alpha$ by thisroot if necessary, we may assume that $\alpha$ is a repeated root of $f$.Now, $f^{\prime }$ has coefficients in $K$ and also has $\alpha$ as a root. Since it is of lowerdegree than $f$, this is not possible unless $f^{\prime }=0$,whence $\mathrm{char}K=p>0$ and $f$ has the form:

$$f=x^{pn}+a_{n-1}{x}^{p(n-1)}+\mathrm{\dots }+a_1{x}^p+a_0.$$

with $a_o\ne 0$.By (b), we may choose elements $b_i\in K$, for $0\le i\le n-1$ such that $b_i{}^p=a_i$.Then we may write $f$ as:

$$f={(x^n+b_{n-1}{x}^{n-1}+\mathrm{\dots }+b_{1}x+b_0)}^p.$$

Since $f(\alpha )=0$ and since the Frobenius map $x\mapsto x^p$ is injective, we see that

$${\alpha }^n+b_{n-1}{\alpha }^{n-1}+\mathrm{\dots }+b_1\alpha +b_0=0$$

But then $\alpha$ is a root of the polynomial

$$x^n+b_{n-1}{x}^{n-1}+\mathrm{\dots }+b_{1}x+b_0$$

which has coefficients in $K$, is non-zero (since $b_o\ne 0$), and has lower degree than $f$.This contradicts the choice of $f$ as the minimum polynomial of $\alpha$.