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?

  • 2
    $\begingroup$ 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. $\endgroup$ – 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 $(\theta\psi^{-1})\varphi = \theta$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.