TITLE: Composite of fields of the form $\sigma(K)$ is precisely $L$
QUESTION [6 upvotes]: I'm trying to understand the proof of lemma 38 in chapter 14 of Dummit and Foote.

Suppose $K/F$ is a root extension and $L$ is the Galois closure of $K$ over $F$. Then the composite field of $\sigma(K)$ for all $\sigma \in \operatorname{Gal}(L/F)$ is precisely $L$.

Could someone please explain to me why this is true?
Thanks in advance!

REPLY [3 votes]: This is similar to a proof in Lang's book algebra page 242.
Let $G=Gal(L/F)=\{\sigma_{1},\cdots,\sigma_{n}\}$. We show that $L=\sigma_{1}(K)\cdots\sigma_{n}(K)$. Let $L'=\sigma_{1}(K)\cdots\sigma_{n}(K)$. 
$L\supset L'$: since each $\sigma_{i}(K)\subset L$, we have $L'\subset L$.
$L\subset L'$: we show that $L'/F$ is Galois. Since $F\subset L'\subset L$, we have $L'/F$ is finite and separable. Let $\overline{F}$ be an algebraic closure of $F$ such that $L'\subset \overline{F}$. Let $\tau$ be an embedding of $L'$ in $\overline{F}$ over $F$. Then $\tau\circ\sigma_{i}$ is an embedding of $L$ in $\overline{F}$ over $F$. Since $L/F$ is normal, we have $\tau\circ\sigma_{i}\in Gal(L/F)$. Also, since $\tau$ is injective, the $\tau\circ\sigma_{i}$'s are distinct. So $\{\tau\circ\sigma_{1},\cdots,\tau\circ\sigma_{n}\}=\{\sigma_{1},\cdots,\sigma_{n}\}$. Hence $\tau(L')=(\tau\circ\sigma_{1})(K)\cdots(\tau\circ\sigma_{n})(K)=L'$. So $\tau$ induces an automorphism of $L'$. Hence $L'/F$ is normal. So $L'/F$ is Galois. 
Let $\overline{L}$ be an algebraic closure of $L$. Since $K\subset L'$, we have $L'$ is a Galois extension of $F$ containing $K$. By corollary 23, $L'\supset L$.