Composite of fields of the form $\sigma(K)$ is precisely $L$

Question

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!

Answer 1

I suppose the thing you're trying to prove is this:

Let $K/F$ a finite field extension, and let $L$ be the normal closure of $K/F$. Then the composite of the fields $\sigma(K)$ (with $\sigma$ ranging over all automorphisms in ${\rm Aut}(L/F)$) is precisely $L$.

One proof that comes to mind is the following. Suppose $\{ \alpha_1, \dots, \alpha_n \}$ is an $F$-basis for $K$, and suppose that $m_1(X), \dots, m_n(X) \in F[X]$ are the minimal polynomials of $\alpha_1, \dots, \alpha_n$ respectively. Then the normal closure $L$ is precisely the splitting field of the polynomial $m_1(X) \times \dots \times m_n(X)$.

Now, for each $i \in \{1, \dots, n\}$, denote the roots of $m_i(X)$ in $L$ by $\beta_{i,1}, \beta_{i,2}, \dots, \beta_{i, k_i}$ (where $\beta_{i,1} = \alpha_i$). Since $\beta_{i,1}, \beta_{i,2}, \dots, \beta_{i, k_i}$ are all roots of the same irreducible polynomial $m_i(X) \in F[X]$, and since $L/F$ is a normal extension, the automorphism group $ {\rm Aut}(L/F)$ acts transitively on the roots $\beta_{i,1}, \beta_{i,2}, \dots, \beta_{i, k_i}$. In particular, for each $j \in \{1, \dots, k_i\}$, there exists a $\sigma \in {\rm Aut}(L/F)$ that maps $\alpha_i = \beta_{i,1}$ to $\beta_{i,j}$. Since $\alpha_i \in K$, this proves that $\beta_{i,j} \in \sigma(K)$.

Applying this argument for every $i \in \{ 1 , \dots n \}$ and for every $ j \in \{1, \dots, k_i \} $, we learn that every root $\beta_{i,j}$ is in $\sigma(K)$ for some $\sigma \in {\rm Aut}(L/F)$. It follows that the composite of the $\sigma(K)$ (with $\sigma$ ranging over all automorphisms in ${\rm Aut}(L/F)$) contains all of the $\beta_{i,j}$'s. But the field $L$ is generated over $F$ by the $\beta_{i,j}$'s! So it must be the case that the composite of the $\sigma(K)$'s is the whole of $L$.

Answer 2

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, $L'\supset L$.