# Field homomorphism into itself

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?

• Your proof only works when $F$ is finite (which is not the case). But notice that the actual proof of the statement looks at a finite set of roots of some minimal polynomial and then uses the fact that every injective endomorphism of a finite set is an isomorphism. – Martin Brandenburg Jun 8 '14 at 17:35

Let $$F=\mathbb Q(T)$$. Then there is a homomorphism $$F\to F$$ given by $$T\mapsto T^2$$. It is not onto.
Edit: $$\phi\colon F\to F$$ is surjective because you can pick $$\alpha\in F$$ arbitrarily, then consider $$S$$ as in your quote, notice that $$\phi$$ is a permutation of the finite set $$S$$, hence there exists $$\alpha'\in S$$ with $$\phi(\alpha')=\alpha$$. In summary, for arbitrary $$\alpha\in F$$ there exists $$\alpha'\in F$$ such that $$\phi(\alpha')=\alpha$$.
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 $$(\theta\psi^{-1})\varphi = \theta$$