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Suppose $L$ is a splitting field of $F$ and let $f(x)$ be an separable irreducible polynomial in $F[x]$. Let $\sigma \in Aut(L/F)$ then we know that if $\alpha$ is a root of $f(x)$ then $\sigma(\alpha)$ is also a root of $f(x)$. Does it imply that degree of minimal polynomial of $\alpha$ is same as degree of minimal polynomial of $\sigma(\alpha)$?

Or $L$ is required to be Galois Extension of $F$?

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  • $\begingroup$ You mean $L$ is a finite extension of $F$ and is the splitting field of some separable polynomial $P \in F[x]$. In that case $L/F$ is Galois. $\endgroup$
    – reuns
    Feb 1, 2017 at 3:41

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I'm not sure what happens when you don't require everything to be separable. So assume it is

In general $\alpha \in \overline{F}$ but $\alpha \not \in L$, so with $\sigma \in Aut(L/F)$, $\sigma(\alpha)$ only makes sense once you extended it to an automorphism $\sigma \in Aut(\overline{F}/F)$ (view $\overline{F}$ as a field extension of $L$, so as a $L$-algebra, onto which the action of $\sigma$ is obvious)

Once this is done $f(\sigma(a)) = \sigma(f(a))= 0$, and $h \in F[x], h(\sigma(a)) = 0 \implies h(a) = h(\sigma^{-1}(\sigma(a))) = \sigma^{-1}(h(\sigma(a)) = 0$ so that $deg(h) < deg(f)$ would contradict the minimality of $f$

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  • $\begingroup$ You have not used separability anywhere. And presumably you are meant to assume $\alpha\in L$. $\endgroup$
    – Eric Wofsey
    Feb 1, 2017 at 4:11
  • $\begingroup$ @EricWofsey Yes, indeed I wanted to write something about $\prod_{\sigma \in Gal(L/F)} (x-\sigma(a)) $ but I deleted it. $\endgroup$
    – reuns
    Feb 1, 2017 at 4:55

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