TITLE: Showing a homomorphism of a field algebraic over $\mathbb{Q}$ to itself is an isomorphism.
QUESTION [10 upvotes]: 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?

REPLY [20 votes]: 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.

REPLY [5 votes]: 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 $F$.