Let M/F and K/F be Galois extensions with Galois groups G = Gal(M/F) and H =...

Let M/F and K/F be Galois extensions with Galois groups G = Gal(M/F) and H = Gal(K/F). Since M/F is Galois, and K/F is a field extension, we have the composite extension field K M.

Show that σ → (σ|_M , σ|_K) is a homomorphism from Gal(K M/F) to G × H, and that it is one-to-one. [As in the notes, σ|_X means the restriction of the map σ to the subset X of its domain.]

Solutions

Expert Solution

A composite of Galois extensions is Galois, so K M /F is Galois.

Any σ ∈ Gal(K M /F) restricted to M or K is an automorphism, since M and K are both Galois over F. So we get a function R: Gal(K M /F) → Gal(M/F)×Gal(K/F) = G×H by R(σ) = (σ|M , σ|K ).

We will show R is an injective homomorphism.

To show R is a homomorphism, it suffices to check the separate restriction maps σ → σ|M and σ → σ|K are each homomorphisms from Gal(K M/F) to Gal(M/F) and Gal(K/F). For σ and τ in Gal(K M/F) and any α ∈ M (στ )|M (α) = (στ )(α) = σ(τ (α)), and τ (α) ∈ M since M/F is Galois, so σ(τ (α)) = σ|M (τ | M(α)) = (σ|M ◦ τ|M) Thus (στ )|M (α) = (σ|M ◦ τ |M )(α) for all α ∈ M, so (στ )|M = σ|M ◦ τ |M .

The proof that (στ )|K = σ|K ◦ τ |K is the same.

The kernel of R is trivial, since if σ is the identity on M and K then it is the identity on K M . Thus R embeds Gal(K M/F) into the direct product of the Galois groups of M and K over F. If the groups Gal(M/F) and Gal(K/F) are abelian, their direct product is abelian. Therefore the embedded subgroup Gal(K M/F) is abelian.

Colby Messinger answered 1 year ago

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