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.

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    $\begingroup$ For a Galois extension, $\sigma(a)=a$ for every $\sigma$ means that $a \in K$. $\endgroup$ – Crostul May 15 '16 at 14:43
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    $\begingroup$ 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

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!


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