Exercise 12.2.10

Proof.

(a)

Since $F\subset L$ is a Galois extension, there is a polynomial $f\in F[x]$ such that $L=F({\alpha }_1,\dots ,{\alpha }_n)$, where ${\alpha }_1,\dots ,{\alpha }_n$ are the roots of $f$ in $L$.

By Exercise 4, we can take $M=F({\alpha }_1,\dots ,{\alpha }_n,{\beta }_1,\dots ,{\beta }_m)$ the splitting field of $\mathit{fg}$, where ${\beta }_1,\dots ,{\beta }_m$ the roots in $M$ of the common minimal polynomial $g$ of $\beta ,{\beta }^{\prime }$. If we replace the initial fields by the new fields, written $K,L$, and $\beta ,{\beta }^{\prime }$ by ${\beta }_1,{\beta }_2$, then

$$F\subset K=F(\beta )\subset M,F\subset K^{\prime }=F({\beta }^{\prime })\subset MF\subset L\subset M.$$

We must suppose $g$ separable. Then $F\subset M$ is separable (Theorem 5.3.15 (a)), so there exists a Galois closure $F\subset M^{\prime }$ of the extension $F\subset M$.

Since $M\subset M^{\prime }$, $L,K$, and $K^{\prime }$ are subfields of $M^{\prime }$.

(b)

$F\subset M^{\prime }$ is a Galois extension (therefore $M^{\prime }$ is a splitting field over $F$ of some polynomial), and $\beta ,{\beta }^{\prime }$ have the same minimal polynomial. By Proposition 5.1.8 there exists an $F$-automorphism $\tau$ of $M^{\prime }$ which sends $\beta$ on ${\beta }^{\prime }$. Since $\tau (\beta )={\beta }^{\prime }$, $\tau (K)=\tau (F(\beta ))=F(\tau (\beta ))=F({\beta }^{\prime })=K^{\prime }$.

(c)

Since $F\subset L$ is normal, $\tau (L)=L$, so $\tau {\text{|}}_L\in \mathrm{Gal}(L∕F)$, and by part (b), $\tau (K)=K^{\prime }$. Therefore $\tau (K\cap L)=K^{\prime }\cap L$, so $\tau {\text{|}}_L$ sends $K\cap L$ on $K^{\prime }\cap L$. Thus $K\cap L$ and $K^{\prime }\cap L$ are conjugate subfields of $L$.

(d)

Since $K^{\prime }\cap L=\sigma (K\cap L)$, where $\sigma =\tau {\text{|}}_L\in \mathrm{Gal}(L∕F)$, by Lemma 7.2.4, $$\mathrm{Gal}(L∕K^{\prime }\cap L)=\mathrm{Gal}(L∕\sigma (K\cap L))=\sigma \mathrm{Gal}(L∕K\cap L){\sigma }^{-1}.$$

Since the isomorphisms of Theorem 7.2.4 send $\mathrm{Gal}(\mathit{KL}∕K)$ on $\mathrm{Gal}(L∕K\cap L)$ and $\mathrm{Gal}(K^{\prime }L∕K^{\prime })$ on $\mathrm{Gal}(L∕K^{\prime }\cap L)$, $\mathrm{Gal}(\mathit{KL}∕K)$ and $\mathrm{Gal}(K^{\prime }L∕K^{\prime })$ map to conjugate subgroups of $\mathrm{Gal}(L∕F)$.