# degree of minimal polynomial of $\alpha$ is same as degree of minimal polynomial of $\sigma(\alpha)$

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 $F$?

• 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. – reuns Feb 1 '17 at 3:41

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 $f$
• You have not used separability anywhere. And presumably you are meant to assume $\alpha\in L$. – Eric Wofsey Feb 1 '17 at 4:11
• @EricWofsey Yes, indeed I wanted to write something about $\prod_{\sigma \in Gal(L/F)} (x-\sigma(a))$ but I deleted it. – reuns Feb 1 '17 at 4:55