# If every polynomial of odd degree is reducible in $\mathbb{F}$ then $[E:\mathbb{F}]$ is not odd

$\mathbb{F}$ is a field of characteristic zero, and every odd-degree polynomial over $F$ has a root in $\mathbb{F}$. now I have two questions:-

1. If $E$ is a finite extension of $F$ show that $[E:\mathbb{F}]$ is not odd.
2. If $E$ is a finite Galois extension of $\mathbb{F}$, Show that $[E:\mathbb{F}]$ is a power of two.

My attempt for the first one: In case the degree of extension is odd its mean we have a minimal polynomial of degree odd such that irreducible over $\mathbb{F}$ and this is not happen since at least we have one root.

For second one I have not idea, but from part one I know it will multiple of 2.

Your supposition about part $1$ is correct. For part $2$, note that if $[E:F]$ is not a power of $2$, then $[E:F]=2^km$ with $m=2j+1$ an odd, natural number.
By Sylow's theorem, there exists a subgroup of $\text{Gal}(E/F)$ of order $2^k$ and index $m$. But then there is a fixed sub-field $E'/F$ of degree $m$, namely the one fixed by this Sylow-$2$ subgroup, and therefore $E'/F$ would have odd degree, a contradiction.