1
$\begingroup$

I'm confused about the distinction between these two:

For a field $F$ and a finite group $G$, we can find a galois extension $E/F$ with galois group $G$.

Given a finite group $G$, there exists a field $F$ and a galois extension $E/F$ with galois group $G$.

What's the distinction between these statements and are they both true?

$\endgroup$
  • 2
    $\begingroup$ @rogerl The second question is quite easy to answer in the affirmative. The first question is unsolved for $F=\Bbb Q$. $\endgroup$ – Lord Shark the Unknown Jan 15 '18 at 2:22
  • $\begingroup$ Indeed, @LordSharktheUnknown, as I know you know, for certain fields $F$ (like the finite fields), the first statement is resoundingly false. $\endgroup$ – Lubin Jan 15 '18 at 2:26
  • $\begingroup$ Could you give me an example why it fails with $F$ finite fields or $/mathbb{Q}$? $\endgroup$ – nyim Jan 15 '18 at 3:01
2
$\begingroup$

The first statement is a generalized statement of the Inverse Galois problem. For this, the field $F$ is fixed and cannot be chosen. The second statement removes the restriction: $F$ can be any field.

If we set $F = \mathbb{Q}$, the first statement is unknown save for a few choices of $G$. On the other hand, this statement fails when $F$ is a finite field because every Galois extension will have a cyclic Galois group: only $\mathbb{Z}_n$ will appear as a Galois group (for any $n \in \mathbb{N})$.

However, the second statement is true for any finite group $G$. To see this, first recall Cayley's theorem, which states that any finite group $G$ appears as a subgroup of $S_n$ for $n \geq |G|$. Next, Hilbert proved that $S_n$ appears as a Galois group over $\mathbb{Q}$; let $K$ be a Galois extension with $\text{Gal}(K/\mathbb{Q}) \cong S_n$. Finally, the Galois correspondence tells us that, for each subgroup $H$ of $S_n$, there will exist an intermediate field $\mathbb{Q} \subset E_H \subset K$ such that the subgroup is isomorphic to $\text{Gal}(K/E_H)$. This is to say, $G$ will appear as the Galois group for the extension $K/E_G$.

$\endgroup$

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.