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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?

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Of course $\mathbb{Q}$ may be replaced by an arbitrary ground field. – Martin Brandenburg Jun 1 '12 at 15:44
up vote 17 down vote accepted

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.

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Oh, you beat me to it. – lhf Jun 1 '12 at 15:32
Sorry about that. – Makoto Kato Jun 1 '12 at 16:15
Why $\varphi$ induces an injective map $S \mapsto S$? – user124697 Apr 18 '14 at 1:39

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.

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There is no assumption on the dimension though. – M Turgeon Jun 1 '12 at 15:23
Are you assuming that $F$ is finite over $\mathbf Q$? In the general case the argument still boils down to dimension, but there is slightly more to say. – Dylan Moreland Jun 1 '12 at 15:23
I am not assuming dimension is finite. What is the dimension argument? I don't see it. – Galois Jun 1 '12 at 15:27
@Galois I didn't claim that you were. The argument is something you know: if $V$ is a finite-dimensional vector space over a field and $f\colon V \to V$ is an injective linear map, then $f$ is necessarily surjective. – Dylan Moreland Jun 1 '12 at 15:33
You could push this through by taking $\alpha \in F$ and looking at the extension of $\mathbf Q$ generated by all the conjugates of $\alpha$ lying in $F$. – Dylan Moreland Jun 1 '12 at 15:40

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 $F$.

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