I have a question about an argument used in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 124):
My question is why does $L_0 \cap K \bar{k}=KL$ hold. One inclusion is obviously since $L_0$ and $K \bar{k}$ contain $KL$. But what about the "$\subset$" inclusion?
My ideas: By consruction we have $L_0= KL[x]/(f)$ so $L_0=KL(\alpha)$ with $f(\alpha)=0$.
Let $d \in L_0 \cap K \bar{k}$. Then $d= \sum_{i=0} ^{deg(f)-1}b_i \alpha^i$ with $b_j \in KL$. We want to show that $b_j=0$ iff $j \neq 0$.
The naive abbroach would be the embedd $L_0$ canonically to $K_0=L_0 \bar{k}$. This works since the coefficients $b_j \in KL \subset K\bar{k}$. But then we obtain following problem: the $\alpha^j$ are not more linear independ so we cannot use a "coefficient comparison" argument...
There's the following general theorem in Galois theory: Let $L/K$ be a field extension, and $E,F$ intermediate fields. If $E/K$ is Galois, then $EF/F$ is Galois and $$ \mathrm{Gal}(EF/F) \cong \mathrm{Gal}(E/E\cap F) \leq \mathrm{Gal}(E/K). $$ Applying this in your setting, so to the field extension $K_0/KL$, and the intermediate fields $L_0$ and $K\bar k$. Now $L_0K\bar k=K_0$, so the theorem tells us that $$ G = \mathrm{Gal}(K_0/K\bar k) \cong \mathrm{Gal}(L_0/L_0\cap K\bar k) \leq \mathrm{Gal}(L_0/KL) = G, $$ so necessarily $L_0\cap K\bar k=KL$.