Showing a homomorphism of a field algebraic over $\mathbb{Q}$ to itself is an isomorphism.

Suppose $F$ is algebraic over $\mathbb{Q}$ and $\varphi : F\to F$ is a homomorphism. Prove $\varphi$ is an isomorphism.

Showing injectivity follows from the fact that the only ideals in a field are $(0)$ and $F$. But how do you show surjectivity?

Solution 1:

Let $\alpha$ be an element of $F$. Let $f(X)$ be the minimal polynomial of $\alpha$. Let $S$ be the set of all the roots of $f(X)$ in $F$. $\varphi$ induces an injective map $S\to S$. Since $S$ is a finite set, this map is surjective. Hence $\varphi$ is surjective.

Solution 2:

The possible images of $\alpha \in F$ under $\varphi$ are the conjugates of $\alpha$ in $F$. This is a finite set $A$ because $\alpha$ is algebraic. Since $\varphi$ is injective and takes $A$ into $A$, it must be surjective on $A$. In particular, $\alpha$ is in the image of $\varphi$. Thus, $\varphi$ is surjective on $F$.

Solution 3:

If $\varphi$ isn't surjective, then since it's injective, it's isomorphic onto its image which must be a field. So this would mean $\varphi$ is an isomorphism of $F$ onto a proper subfield of $F$, but this can't happen for dimension reasons.

EDIT: As Dylan pointed out, this doesn't work unless our extension is finite, so it's probably best to look at the other answers.