Intersection of subgroups and the one of intermediate subfields

\mathbf{Theorem 1:} Show that in the fundamental theorem of Galois theory, the subfield corresponding to the intersection of two subgroups H_1 and H_2 is the subfield generated by the corresponding intermediate subfields \text{Inv}H_1 and \text{Inv}H_2.
\mathbf{Proof:} Suppose K is the subfield generated by \text{Inv}H_1 and \text{Inv}H_2. Then K=\displaystyle \bigcap L here L is intermediate subfield embarcing \text{Inv}H_1 and \text{Inv}H_2.
For any such L, \text{Gal }E/L\leq H_1 and \text{Gal }E/L\leq H_2, so \text{Gal }E/L\leq H_1\cap H_2.
Thus K\subset \text{Inv}(H_1\cap H_2).
Conversely, \text{Inv}(H_1\cap H_2)\supset \text{Inv}H_1, \text{Inv}(H_1\cap H_2)\supset \text{Inv}H_2. So \text{Inv}(H_1\cap H_2)\supset K.
\text{Q.E.D}\hfill \square
\mathbf{Theorem 2:} The intersection of two intermediate fields K_1 and K_2 corresponds to the subgroup generated by the corresponding subgroups \text{Gal }E/K_1 and \text{Gal } E/K_2.
The proof is very similar to the previous one.

\mathbf{Corollary:} Suppose we also know K_1 is Galois over the groud field and K_2 is another intermediate field. Then \text{Gal }K_1/K_2\cap K_2 is isomorphic to \text{Gal }K_1K_2/K_2.

\mathbf{Proof:} By assumption, H is normal. From group theory,

\displaystyle H_1H_2/H_1\cong H_2/H_1\cap H_2

Theorem 2 says that the corresponding group to K_1\cap K_2 is \langle H_1,H_2\rangle. Since H_1 is normal then \langle H_1,H_2\rangle=H_1H_2. And \text{Gal }K_1/K_1\cap K_2=H_1H_2/H_1.

Theorem 1 says that the corresponding group to intermediate field K_1K_2 is H_1\cap H_2. And \text{Gal }K_1K_2/K_2=H_2/H_1\cap H_2.

So the conclusion holds.

\text{Q.E.D}\hfill \square

In fact this corollary is valid even K_2 is transcedental extension, namely

\mathbf{Theorem 3:} Let K be a Galois extension of k, let F be an arbitrary extension and assume that K,F are subfields of some other field. Then KF is Galois over F and K is Galois over K\cap F. Let H be the Galois group of KF over F, and G the Galois group of K over k. If \sigma\in H then the restriction of \sigma to K is in G and the map

\sigma\to \sigma|_K

gives an isomorphism of H on the Galois group of K over K\cap F.

\mathbf{Remark:} Jacobson Algebra I. p243

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