How can I show that every finite group is the Galois group of an extension $K/F$ where $F$ is itself a finite extension of $\mathbb Q$?

I know the following:

  1. Every finite group is contained in $S_p$ for a large enough prime $p$.

  2. Every irreducible polynomial in $\mathbb Q[x]$ of degree $p$ having exactly $p-2$ real roots has a Galois group $S_p$ over $\mathbb Q$.

  3. For any $n$ there is an irreducible polynomial in $\mathbb Q[x]$ of degree $n$ having exactly $n-2$ real roots.

Does this have something to do with the inverse Galois problem?


1 Answer 1


Well this is "a" inverse Galois problem in some sense but significantly easier than "the" inverse Galois problem.

Your three results basically solve the problem already.

Take your finite group $G$ and embed it in $S_p$ for some prime $p$ (via (1)). Take some irreducible polynomial $f$ over $\mathbb{Q}$ with exactly $p-2$ real roots (via (3)). Let $K$ be the splitting field of $f$ (over $\mathbb{Q}$) then $K$ has Galois group $S_p$ (via (2)). Set $F$ the fixed field of $G$ (considered as a subgroup of $S_p$). Then by the main theorem of Galois theory $G$ is the Galois group of $K/F$.

  • 1
    $\begingroup$ How is this different from the inverse Galois problem? Just set $F=\mathbb{Q}$? I know there is something I'm missing here, but what is it? $\endgroup$ Jul 7, 2019 at 22:47
  • $\begingroup$ You can't set $F=\mathbb{Q}$ because then the Galois group of $K/F$ is $S_p$ and not $G$. $F$ has to be the fixed field of $G$... $\endgroup$ Jul 8, 2019 at 14:23
  • $\begingroup$ Aaah, I see, you get to choose who the field $F$ is $\endgroup$ Jul 8, 2019 at 15:24

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