0
$\begingroup$

I seem to be stuck on the following problem about field extensions from an old prelim exam in algebra. Let $K$ be an algebraic field extension of a field $F$ and let $L$ be a subfield of $K$ such that $F \subseteq L$ and $L$ is normal (but not necessarily finite) over $F$. Show that if $\sigma$ is any automorphism of $K$ over $F$, then $\sigma(L)=L$.

I think the finite case should be pretty straightforward (is it?), but I'm not so sure about the infinite case. How should I approach this problem? Might trying to get some sort of contradiction be of any use here? In the problem, I think by definition $L$ is the normal closure of the extension $K$ of $F$ but I'm not so certain if that will help.

$\endgroup$
1
$\begingroup$

It suffices to prove $\sigma (L)\subseteq L$. To do this you can combine two facts: first, an automorphism of $K/F$ maps an element $x$ to another root of the minimal polynomial of $x$ over $F$, and second, a polynomial $p(x)\in F[x]$ that has a root in a normal extension of $F$ has all of its roots there.

$\endgroup$
  • $\begingroup$ Is $x \in K$ or $x \in F$? $\endgroup$ – Libertron Jan 17 '15 at 20:29
  • 1
    $\begingroup$ In general $x\in K$; of course $x\in F$ is allowed, but then $\sigma (x)=x$ and there is nothing to prove. $\endgroup$ – Hagen Knaf Jan 19 '15 at 7:46

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.