2
$\begingroup$

Assume that $M$ is a finite abelian extension of $\mathbb{Q}$, such that $M\subset \mathbb{Q}(\zeta_m)$ is the smallest cyclotomic field which contains the $M$ via the Kronecker-Weber theorem.

Is the restriction of the Galois group $Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})$ to subfield $M$ simply the Galois group $Gal(M/\mathbb{Q})$?

$\endgroup$
  • 1
    $\begingroup$ Yes, this is true in any tower of Galois extensions. $\endgroup$ – Lord Shark the Unknown Aug 28 '17 at 15:27
1
$\begingroup$

We have a tower of normal extensions of $\mathbb Q$: $$ \mathbb Q \subset M \subset \mathbb Q(\zeta_m).$$

In this situation, we have a surjective group homomorphism: $$ \phi : Gal(\mathbb Q(\zeta_m) : \mathbb Q) \to Gal(M : \mathbb Q),$$ which sends each $\sigma \in Gal(\mathbb Q(\zeta_m) : \mathbb Q)$ to its restriction $\sigma|_M \in Gal(M : \mathbb Q)$.

[Since $M : \mathbb Q$ is a normal extension, we have $\sigma(M) = M $, so this definition makes sense. And since $\mathbb Q(\zeta_m) : \mathbb Q$ is a normal extension, every $\mathbb Q$-automorphism of $M$ extends to a $\mathbb Q$-automorphism of $\mathbb Q(\zeta_m)$, which means that $\phi$ is surjective.]

Anyway, the kernel of the group homomorphism $\phi$ is $Gal(\mathbb Q(\zeta_m) : M)$. Hence $\phi$ induces a group isomorphism: $$ \frac{Gal(\mathbb Q(\zeta_m) : \mathbb Q)}{Gal(\mathbb Q(\zeta_m) : M)} \cong Gal(M : \mathbb Q),$$ which is the isomorphism you are looking for.

$$ $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.