Field extension with Galois group $C_2$ Given a field extension $E \subseteq F$ with the Galois automorphism group of $F:E$ having order 2, must $F:E$ necessarily be a normal extension?
I remembered seeing this claim somewhere (I can't remember where) and I think it is true, but I can't seem to work out a proof for this. I'm not sure if we need the hypothesis that $[F:E]$ is finite - if so then we can perhaps use the criterion that $F:E$ is normal iff $F$ is the splitting field of some polynomial in $E[x]$?
Edit: I should clarify that we are not assuming $[F:E] = 2$.
 A: No. Here is how you can see this by carefully interrogating the Galois correspondence.
Suppose the extension $E \to F$ is not normal but is finite and separable, so we can embed it into a finite Galois extension $E \to F \to K$ (for example, its normal closure). Let $G = \text{Gal}(K/E)$. Under the Galois correspondence, $F$ corresponds to some non-normal subgroup $H$ of $G$, and its automorphism group as an extension is $\text{Aut}_G(G/H)$, the automorphism group of $G/H$ as a $G$-set. Explicitly, this is $N_G(H)/H$, where $N_G(H)$ denotes the normalizer
$$N_G(H) = \{ g \in G \mid gHg^{-1} = H \}.$$
($H$ is normal iff its normalizer is all of $G$, in which case we get that the automorphism group is $G/H$ as expected.)
So it remains to find some nontrivial non-normal subgroup $H$ of some finite group $G$ which has index $2$ in its normalizer, then find an extension with Galois group $G$. The latter is always possible if $G$ is finite so we'll concentrate on the former. Actually it will be cleanest to construct the $G$-set $X = G/H$ first; we'll try to write down a set and some permutations acting transitively on it such that the group of permutations commuting with all of them has order $2$, and then we'll take $G$ to be the group generated by these permutations and $H$ to be the stabilizer of the corresponding action.
So take $X = \{ 1, 2, 3, 4 \}$ and take the permutations $(1234), (13)$. These generate $D_4$ inside $S_4$; thinking of the elements of $X$ as being the vertices of a square, these generators are a rotation and a reflection. The only nontrivial permutation commuting with all of them is $(13)(24)$, a rotation by $\pi$ (which is central). Hence we can take $G = D_4$ and $H$ to be the non-normal copy of $C_2$ generated by any reflection, say $(13)$. The normalizer of $H$ is then a copy of $C_2 \times C_2$ generated by the reflection we picked and by the central rotation $(13)(24)$. 
Going back to field extensions, we can take $E = \mathbb{Q}, K = \mathbb{Q}(i, \sqrt[4]{2})$ (a dihedral extension with Galois group $D_4$), and $F = \mathbb{Q}(\sqrt[4]{2})$. Here the nontrivial automorphism of order $2$ is $\sqrt[4]{2} \mapsto - \sqrt[4]{2}$. 
A: Suppose that $[F:E]=2$, let $a\in F$ which is not in $E$, $1,a$ is a basis of the $E$-vector space $F$. This implies that $a^2+ba+c=0, b,c\in E$. Thus $a$ is a root of $X^2+bX+c$, if you divides $X^2+bX+c$ by $X-a$ you have $(X-a)(X-d)$, thus $ad=c\in E$ this implies that $d=ca^{-1}\in F$ thus $F$ is the splitting field of $X^2+bX+c$.
