Lecture 21
Galois Groups of Composite Extensions
 


Theorem 21.4  

Let $ E/K$ and $ F/K$ be finite Galois extensions so that $ E, F \subset \overline {K}.$ Then the homomorphism

$\displaystyle \psi: G(EF/K) \longrightarrow G(E/K) \times G(F/K), \; \psi(\sigma)=(\sigma\vert _E, \sigma\vert _F)$
is injective. If $ E \cap F =K$ then $ \psi$ is an isomorphism.


Proof.
It is clear that $ \psi$ is a group homomorphism. The kernel of $ \psi$ consists of $ \sigma \in G(EF/K)$ so that $ \sigma(a)=a$ for all $ a \in E$ and for all $ a \in F.$ Hence such $ \sigma=id_{EF}.$ Thus $ \psi$ is injective.

Suppose that $ E \cap F=K.$ Then by Corollary (21.3),
$\displaystyle \vert G(EF/K)\vert=[EF:E \cap F]=[F:K][E:K]=\vert G(E/K)\vert\vert G(F/K)\vert.$

This shows that $ \psi$ is an isomorphism.


The Normal Closure of an Algebraic Extension

Let $ K/F$ be an algebraic extension and $ K \subset \overline {F}.$ The normal closure of $ K/F$ in $ \overline {K}$ is the splitting field $ N$ over $ F$ of the polynomials $ \{\;$irr$ \;(a,F) \mid a \in K\}.$ It is clear that $ N$ is a normal extension of $ F$ containing $ K.$ Moreover any normal extension $ N' \subset \overline {F}$ of $ F$ containing $ K$ must contain the splitting fields of $ \{\;$irr$ \;(a,F) \mid a \in K\}.$ Hence $ N =N'.$ If $ K=F(a_1,a_2, \ldots,a_n)$ then $ N$ is the splitting field of the polynomials $ \;$irr$ \;(a_i,F)$ for all $ i=1,2, \ldots,n.$

If $ K/F$ is separable then $ N/F$ is a separable extension as it is obtained by adjoining roots of separable polynomials over $ F.$ Hence the normal closure of $ K/F$ when $ K/F$ is separable, is a Galois extension.

Let $ K/F$ be a separable extension that is not normal. Let $ N$ be a normal closure of $ K/F.$ Put $ H=G(N/K).$ Then $ K=N^H.$ Let $ H' < H$ be a normal subgroup of $ G=G(N/F).$ Then $ N^{H'} > N^H=K$ and $ N^{H'}/F$ is a normal extension of $ K.$ Thus $ N^{H'}=N$ by minimality of $ N.$ Hence $ H'=(id).$