Let $ K $ be a field and $ f \in K[X] $ a separable polynomial which factorizes over $ K[X] $ in irreducible factors as $ f(x)=g(x)h(x) $. Let $ K_{f}, K_{g} $ and $ K_{h} $ be the splitting fields of $ f,g $ and $ h $ over $ K $ and $ G_{f}=Gal(K_{f}/K), G_{g}=Gal(K_{g}/K), G_{h}=Gal(K_{h}/K) $. I am asked to show that $ G_{g} \cap G_{h} =\{1 \} $ and that if $ K_{g} \cap K_{h}=K $, then $ G_{f}=G_{g}G_{h} $.
Suppose $ K_{g}=K(\alpha_{1}, \dots , \alpha_{k}) $ and $ K_{h}=K(\alpha_{k+1}, \dots , \alpha_{n}) $, where $ \alpha_{1}, \dots , \alpha_{n} $ are the roots of $ f $. Then for $ \sigma \in G_{g} \cap G_{h} $, we must have that $ \sigma (x)=x \,\forall x \in K_{g} $ and $ \forall \, x \in K_{h} $. How do I conclude that $ \sigma $ must be the identity? Also, how do I prove the second statement?
I see now that the first statement actually follows trivially. I am still stuck proving the second one.
I would appreciate any help. Thank you!
Consider the map from $Gal(K_f/K_g)$ to $Gal(K_h,K)$ that takes $\sigma \rightarrow \sigma_{K_h}$. The map is well defines since $K_h$ is Galois over $K$. The map is injective because $\sigma_{K_h}=id$ implies fixed field of $\sigma$ is $K_gK_h=K_f$ and hence $\sigma=id$. Now consider the image of the map which is a subgroup of $Gal(K_h/K)$. The fixed field of the image is $K_h\cap K_g=K$ and hence by the fundamental theorem of Galois theory the image is precisely $Gal(K_h/K)$. Thus we have $G_f=G_g.G_h$