This is part of an exam preparation:


  1. If $L_1/K$ and $L_2/K$ are not Galois, then $L_1L_2/K$ is not Galois.
  2. If $L_1/K$ and $L_2/K$ are not solvable Galois extensions, then $L_1L_2/K$ is not a solvable Galois extension.

For 1, I think that the answer is false. I thought of "splitting" the splitting field of a polynomial. For example, the splitting field of $X^3-2$ over $\mathbb{Q}$ is $\mathbb{Q}(2^{1/3},\omega)$ and then taking $\mathbb{Q}(2^{1/3})$ and $\mathbb{Q}(\omega)$. The problem is that $\mathbb{Q}(w)/\mathbb{Q}$ is Galois and I can't think of a different kinds of example.

For 2, I think that the answer is true. We have $$G_1 = Gal(L_1L_2/K) \leq Gal(L_1/K) \times Gal(L_2/K) = G_2$$

Then I thought of making some claim about a subgup of a solvable group but Im not really sure how to proceed.


  • $\begingroup$ i think for the second part, one has to assume that $L_1$ and $L_2$ are not the same extension (otherwise one of the 2 conditions implies the result). When the $L_1$, $L_2$ are different extensions the new extension $L_1L_2$ would have a Galois group that contains the $G_1$ and $G_2$ as (proper) subgroups and since none of them is solvable the $G_{12}$ is also not solvable $\endgroup$ – Nikos M. Jul 5 '14 at 13:31
  • $\begingroup$ in other words the series of galois extensions for the solvable group $G_{12}$ would have to contain at some steps the series of galois extensions of $G_1$ and $G_2$ (at different steps in general) thus since none of them is solvable, the resulting galois series of simple extensions is also not solvable (i may use slightly different terminology, but i think the main point of the argument comes accross) $\endgroup$ – Nikos M. Jul 5 '14 at 13:58

Promoting comments to an answer.

  1. If you pick $K=\Bbb{Q}$, $L_1=K(\root3\of2)$, $L_2=K(\omega\root3\of2)$. Neither of those is Galois, but $L_1L_2$ is the splitting field of $x^3-2$ over $K$.
  2. Here the answer depends on the exact meaning of the question. If NOT applies to "solvable and Galois", then the counterexample in (1) applies, because neiter $L_1$ nor $L_2$ is "solvable and Galois" for they are not Galois. But that smacks of sophistry. More likely the intended meaning is "Galois but not solvable". In this case we know that the compositum $L_1L_2$ is also Galois over $K$. Let's denote $G=Gal(L_1L_2/K)$, $G_i=Gal(L_i/K)$, $i=1,2$. Here the mappings $f_i:G\to G_i$, $f(\sigma)=\sigma\vert_{L_i}$ gotten by restricting the domain of a $K$-automorphism are homomorphisms of groups. By basic Galois theory they are both surjective. Thus $G_1$ and $G_2$ are both quotient groups of $G$. If $G$ were solvable, what would that tell you about the solvability of $G_i$?

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