0
$\begingroup$

Let $K \subset L$ be a finite Galois extension. Assume that there is $a \in L$ such that $\sigma(a) \not= a$ for every $\sigma \in$ Gal$(L/K)$. Let $P$ be the minimal polynomial of $a$ over $K$.

(a) Show that deg $P$ = |Gal $(L/K)$|.

(b) Show that $L = K(a).$

I can see how (b) will follow (a). Any help with the first part would be appreciated - I know that deg $P$ will divide|Gal $(L/K)$|. I tried looking at the roots of the minimal polynomial but did not get anywhere useful.

$\endgroup$
  • 1
    $\begingroup$ For a Galois extension, $\sigma(a)=a$ for every $\sigma$ means that $a \in K$. $\endgroup$ – Crostul May 15 '16 at 14:43
  • 1
    $\begingroup$ are you really really REALLY sure it doesn't say "$\sigma(a) \neq a$" instead ? $\endgroup$ – mercio May 15 '16 at 14:46
  • $\begingroup$ Sorry, yes that is what it should say @mercio ! I have edited it now. $\endgroup$ – abc May 15 '16 at 14:48
  • $\begingroup$ $\sigma$ not identity! $\endgroup$ – mich95 May 15 '16 at 14:49
1
$\begingroup$

Consider $K(\alpha)$. It's clear that $K(\alpha)$ is a subfield of $L$. Let's look for the subgroup of Gal$(L/K)$ that fixes $K(\alpha)$. It's enough to check what are the automorphisms of $L/K$ fixing $\alpha$. By assumption, there is only the identity fixing $\alpha$ and hence the subgroup fxing $K(\alpha)$ is the trivial subgroup of Gal$(L/K)$. By fundamental theorem of Galois theory, $K(\alpha)=L$. Part $a$ then follows!

$\endgroup$

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.