TITLE: Field homomorphism into itself
QUESTION [5 upvotes]: Suppose $F$ is algebraic over $\mathbb{Q}$ and $\varphi \colon F \to F$ is a homomorphism. Prove that $\varphi$ is an isomorphism.

I came across this question here but I don't quite get why $F$ has to be algebraic over a field.
I know $\phi$ is injective since $\ker(\phi) = 0$. Shouldn't it then follow that $\phi$ is an isomorphism since $F$ and $F$ have the same cardinality?
Can anyone give a counterexample?
Edit: I also don't get why $\phi$ is surjective. 

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.

Since $\phi$ maps $S$ onto $S$, how does that imply $\phi$ is surjective in the whole field?

REPLY [2 votes]: Let $\theta\in F$. Restriction  $\psi$ of $\varphi$ on $\mathbb{Q}(\theta)$ is a homomorphism of finite extensions of $\mathbb{Q}$. You've mentioned that it's injective. But $\psi$ is a linear mapping of finite dimensional $\mathbb{Q}$-spaces and so has a determinant. It does not vanish due to injectivity. Then $\psi$ has an inverse. So $(\theta\psi^{-1})\varphi = \theta$