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.


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.

  • $\begingroup$ Is $x \in K$ or $x \in F$? $\endgroup$ – Libertron Jan 17 '15 at 20:29
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    $\begingroup$ 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

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