Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
Of course $\mathbb{Q}$ may be replaced by an arbitrary ground field. –  Martin Brandenburg Jun 1 '12 at 15:44

3 Answers 3

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.

share|improve this answer
    
Oh, you beat me to it. –  lhf Jun 1 '12 at 15:32
3  
Sorry about that. –  Makoto Kato Jun 1 '12 at 16:15
    
Why $\varphi$ induces an injective map $S \mapsto S$? –  user124697 Apr 18 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.

share|improve this answer
    
There is no assumption on the dimension though. –  M Turgeon Jun 1 '12 at 15:23
3  
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
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 $F$. –  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 $F$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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