Fundamental Thm of Galois Theory: Showing that $L^H$ is normal over $F$ iff $H$ is normal in $G$

Question

I am trying to understand the proof of the following Lemma (from this source) preceding the proof of the Fundamental Theorem of Galois Theory:

We assume L/F to be a finite-dimensional Galois extension of fields with Galois group $G=\textrm{Gal}⁡(L/F).$ (...)

Lemma 3. Let $H$ be a subgroup of $G$. Then the following are equivalent:

1. $L^H$ is normal over $F$.

2. $\sigma⁢(L^H)=L^H$ for all $\sigma\in G$.

3. $\sigma ⁢H⁢ \sigma^{-1} = H$ for all $\sigma \in G$.

In that source the implication $3 \Rightarrow 1$ is proven as follows.

Let $\alpha \in L^H$, and let $f$ be the minimal polynomial of $\alpha$ over $F$. Since $L/F$ is normal, $f$ splits into linear factors in $L⁢[X]$. Suppose $\alpha′ \in L$ is another zero of $f$, and let $\sigma \in G$ be such that $\sigma⁢(\alpha′)=\alpha$ (such a $\sigma$ always exists). By assumption, for all $\tau \in H$ we have $\tau′:=\sigma \tau\sigma^{-1} \in H$, so that $$\tau(\alpha′)=\sigma^{-1}⁢\tau′⁢\sigma⁢(\alpha′)=\sigma^{-1}⁢\tau′⁢(\alpha)=\sigma^{-1}⁢(\alpha)=\alpha′.$$ This shows that $\alpha′$ lies in $L^H$ as well, so $f$ splits in $L^H⁢[X]$. We conclude that $L^H$ is normal over $F$.

I am fine with everything up to the last sentence, which I think should elaborated a bit further on. Probably the fact that a field extension is normal iff it is the splitting field of some collection of polynomials is used here. First I thought that it remains to show that $L^H$ is the smallest extension of $F$ such that $f$ splits in it (and hence a splitting field). This might, however, not be the case (e.g. if $H$ is not the trivial group and $\alpha \in F$). So, what I think is happening here is that $L^H$ is the smallest extension such that for all $f \in M := \{min_\alpha|\alpha \in L^H\}$, where $min_\alpha$ is the minimal polynomial for $\alpha$ over $F$, $f$ splits in $L^H$. The extension is smallest, for if there would exist any smaller such extension $S$, then there would exist some $f \in M$ such that $f$ has a root $\beta \in L^H$ with $\beta \notin S$ (hence $f$ does not split in $S$).

Do I understand this correctly and is this the most straightforward way to think about it or is there something more obvious that I am missing?

Answer 1

Just to have an actual answer here, the test uses the following characterisation of normal field extensions. A field extension $K/F$ is normal if for every $\alpha\in K$ the minimal polynomial of $\alpha$ over $F$ (which by definition has $\alpha$ as a root) splits into linear factors in the ring$~K[X]$. In the application of this characterisation $K=L^H$, and it is shown that for any $\alpha\in K$ and any linear factor of its minimal polynomial, a factor a priori in $L[X]$ where this polynomial splits, this linear factor already lies in $K[X]$ (so that the minimal polynomial splits in $K[X]$). This is obtained by showing that the root $\alpha'$ of that linear factor lies in $K=L^H$, in other words it is fixed under $H$.