The translation theorem of Galois theory (as I learned it) roughly states that two Galois extensions $L,K$ of a ground field $k$ that are "far enough" (i.e. $L \cap K = k$) inside the same algebraic closure $\bar k$ of $k$ give rise to an extension $M = LK$ (the field compositum) of $k$ having both, $L$ and $K$ as subextensions with Galois group

$$Gal(M/k) = Gal(L/k) \times Gal(K/k)$$

(I know that there are some generalisations that not both extensions have to be Galois and that the Galois group $Gal(L/k)$ is isomorphic to the Galois group $Gal(M/L)$ but that should not play a role for my question)

Is there a converse statement to the translation theorem of Galois theory?

I'm thinking of a statement that whenever one can express the Galois group $H$ of an extension $M$ over a field $k$ as a direct product of groups, then those groups have to be Galois groups of certain extensions of $k$ and one can realize them? Or is this just what normal Galois theory gives?

EDIT And if we start from a Galois extension $M$ over $L \supseteq k$, under which conditions do we get a Galois extension $M'$ over $k$ which has the same Galois group $Gal(M'/k) \cong Gal(M/L)$

Thank you :-)

  • $\begingroup$ Do you mean $L \cap K = k$? $\endgroup$
    – JSchlather
    Oct 28, 2013 at 17:45
  • $\begingroup$ Yes thank you :-) $\endgroup$
    – BIS HD
    Oct 28, 2013 at 17:46
  • 1
    $\begingroup$ If $G=H_1\times H_2$ where $G=Gal(F/k)$ then the fixed fields $K_1,K_2$ of $H_1$ and $H_2$ are normal (the subgroups being normal) with Galois groups over $k$ $H_2$ and $H_1$, and Galois correspondence gives $K_1\cap K_2=k$, $K_1K_2=F$. I.e. no rocket science :) $\endgroup$
    – user8268
    Oct 28, 2013 at 17:59
  • $\begingroup$ @user8268 Thank you, feel free to post that as an answer ;-) I added a second question to the same theorem. Maybe you could have a look?? $\endgroup$
    – BIS HD
    Oct 28, 2013 at 18:52

1 Answer 1


As user8268 pointed out: If the Galois group $G$ of a Galois extension $F/k$ is the direct product $$G = H_1 \times H_2$$ of two normal subgroups, then the main theorem of Galois theory is the converse of the translation theorem I've looked for: Each fixed field $K_i = F^{H_i}$ has $Gal(K_i / F) = H_i$ for $i = 1,2$ and the compositum equals $F = K_1 K_2$.

My second question can be answered in the following way: If the group $Gal(L/k)$ acts on $Gal(M/L)$ trivially and $M/k$ is Galois, then we can write $$Gal(M/k) = Gal(M/L) \times Gal(L/k)$$ and the above statement gives a Galois extension $M'/k$ with Galois group $Gal(M'/k) = Gal(M/L)$. The trivial action could come from (different) prime group orders.

Thank you for your comments :-) That helped me a lot!


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