9
$\begingroup$

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?

  • 4
    Of course $\mathbb{Q}$ may be replaced by an arbitrary ground field.2012-06-01

3 Answers 3

17

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.

  • 0
    Why $\varphi$ induces an injective map $S \mapsto S$?2014-04-18
4

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 $F$.

3

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.

  • 1
    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 $F$.2012-06-01