Let $L/K$ be a Galois extension, and let $R\subseteq L$ be a subring such that $\tau(R)=R$ for every $\tau\in\text{Gal}(L/K)$.
Let $\alpha\in R$. How would I show that $H=\{\tau\in\text{Gal}(L/K):\tau(\alpha)=\alpha\}$ a normal subgroup of $\text{Gal}(L/K)$?
I've tried using the Fundamental Theorem of Galois Theory, by which it would be equivalent to show that $L^H/K$ is a Galois extension. I would think that $L^H=K(\alpha)$, but haven't been able to show this, or that assuming this the extension is Galois.
Best Answer
Unless I misunderstood something the claim is false. Take the field extension $\mathbb{Q} \subset \mathbb{Q}(\sqrt[4]{2},i)$, which is a Galois extension, being the splitting field of $x^4-2$
Now take $R = \mathbb{Q}(\sqrt[4]{2},i)$ and obviously the condition is satisfied. Now pick $\alpha = \sqrt[4]{2}$. Then it's not hard to show that $L^H = \mathbb{Q}(\sqrt[4]{2})$. But this is obviously not a Galois extension of $\mathbb{Q}$, so H can't be a normal group of $\text{Gal}(L/K)$
As you have mentioned it's true that $L^H = K(\alpha)$. To prove this first note that $K(\alpha) \subseteq L^H$, as obviously $K(\alpha)$ is fixed pointwise by $H$. For the other inclusion, by the Fundamental Theorem of Galois Theory we have $K(\alpha) = L^{H'}$ for some subgroup $H'$ of $\text{Gal}(L/K)$. Now as $H'$ fixes $\alpha$ we have that $H' \le H$. From this $L^{H} \subseteq L^{H'} = K(\alpha)$. Hence $L^H = K(\alpha)$.
Furthermore, the reason why the statement fails is because $R$ isn't necessarily a subring of $L^H$. This means that although the conjugates of $\alpha$ are in $R$, they might not be in $K(\alpha)$, which has to be the case if it were a Galois extension of $K$.