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?

  • 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

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 $K/E_G$.


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