Characterizing algebraic extensions

Question

Show that $K$ is an algebraic extension over $F$ if and only if for every intermediate field $E$, every monomorphism $\sigma: E\rightarrow E$ which is the identity on $F$ is in fact an automorphism of $E$.

The $\Rightarrow$ direction seems fairly easy, and doesn't really use the fact that $K$ is algebraic: any arbitrary intermediate field $E$ is an $F$-vector space, so a linear map from $E$ to itself is injective if and only if it is surjective. Thus any $\sigma: E\rightarrow E$ which is a monomorphism and fixes $F$ is, among other things, an injective $F$-linear map, and so must be surjective. Thus it's a bijective field homomorphism, i.e. an automorphism.

The other direction is stumping me. I can figure out how to begin: let $u\in K-F$, and we want to find a polynomial over $F$ which $u$ is a root of. Since we know something about every intermediate field, let's consider $F(u)$. Now we know something about monomorphisms from $F(u)$ to $F(u)$. My problem is that I can't see any obvious monomorphism to pick. But I think if I can pick the right map, then surjectivity will allow me to find a polynomial which $u$ satisfies.

Any hints are appreciated!

Answer 1

$\Rightarrow$: Assume $K$ is algebraic and $E$ an intermediate field and $\sigma\colon E\to E$ a monomorphism. Let $u\in E$, $f$ its minimal polynomial, $U\subset E$ the set of roots that $f$ has in $E$. Then $\sigma$ maps $U\to U$ and as $U$ is finite and $\sigma$ injective, it in fact permutes $U$. Hence for suitable $m>1$, $\sigma^m$ is the identity on $U$. In particular, $\sigma$ maps $\sigma^{m-1}(u)$ to $u$. As $u$ was arbitrary, $\sigma$ is surjective and finally it is an automorphism.

$\Leftarrow$: Assume $u$ is transcendental over $F$ and let $E=F(u)$. Then $u\mapsto u^2$ induces a monomorphism $\sigma\colon E\to E$. By assumption, it is onto, hence there exists an element $\frac{f(u)}{g(u)}\in E$ that maps to $u$. We conclude $f(u^2)=ug(u^2)$. As the terms cannot cancel (and at least $g\ne 0$), we obtain a polynomial equation for $u$, contraditing the assumption that $u$ is transcendental.