Let $K(\alpha)/K$ and $K(\beta)/K$ algebraic field extensions so that their respective Galois groups are abelian.

Prove that the Galois group of the field extension $K(\alpha + \beta)/K$ is also abelian.

My attempt:

I've tried considering the towers

$$K(\alpha,\beta)/K(\alpha)/K \qquad K(\alpha,\beta)/K(\beta)/K$$

Are somehow related to

$$K(\alpha,\beta)/K(\alpha+\beta)/K \qquad K(\alpha,\beta)/K(\alpha - \beta)/K$$

But I don't know how to relate this to the fact that the quotient is abelian or whether this statement is true or not.


$K(\alpha, \beta)$ is Galois over $K$ and the corresponding Galois group is a subgroup of $\text{Gal}(K(\alpha)/K)\times \text{Gal}(K(\beta)/K)$. So $K(\alpha, \beta)/K$ is abelian. Thus, any intermediate Galois extension must be abelian since the corresponding Galois group would be a quotient of $\text{Gal}(K(\alpha,\beta)/K)$

  • $\begingroup$ Sounds good, but I'm having trouble seeing that $\text{Gal}(K(\alpha, \beta)/K)$ is a subgroup of $\text{Gal}(K(\alpha)/K)\times \text{Gal}(K(\beta)/K)$. $\endgroup$ – Darth Geek Jul 13 '15 at 16:39
  • $\begingroup$ There are restriction maps $\text{Gal}(K(\alpha,\beta)/K) \to \text{Gal}(K(\alpha)/K)$ and to $\text{Gal}(K(\beta)/K)$. These maps give you the required injection. (Just google "compositum of galois extensions") $\endgroup$ – Prahlad Vaidyanathan Jul 14 '15 at 1:38

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