# For a given finite group G, is there a cover of P^1 over Q s.t. over C it's G-Galois?

For any finite group, G, we can find a cover of ℙ1 which is G-Galois. The regular inverse Galois problem is equivalent to there existing such a cover that descends with action to ℚ. My question is about the easier problem: given a finite group G, can we find a cover of ℙ1 such that it descends to ℚ as a mere cover (meaning not necessarily with group action)?

From the results that I know, I would be really surprised if this is solved. But what is known? And where is it written?

## 1 Answer

I don't know about every finite group $$G$$ (I'll guess no), but there are definitely infinitely many finite groups $$G$$ for which the situation you describe obtains: the extension $$K/\mathbb{C}(t)$$ has a model over $$\mathbb{Q}$$ but is not Galois over $$\mathbb{Q}$$. (And for most of these groups, we do not know how to realize them as Galois groups over $$\mathbb{Q}$$, regularly or otherwise.)

For instance, this is the situation in a work in progress of John Voight and me:

http://alpha.math.uga.edu/~pete/triangle-091309.pdf

In our slightly different language, there are plenty of situations where the covering itself is defined over $$\mathbb{Q}$$ but the field of definition of the automorphism group $$G$$ is strictly larger. (This is equivalent to what you're asking, right? Please let me know.)

[Warning: recently, with the help of Noam Elkies, John and I realized that our arguments as given only work when (in our notation) $$a = 2$$. This is still a generalization of the setting in which I began this work some years ago: I had $$a = 2$$, $$b = 3$$, so I know for sure that there are infinitely many examples of this form.]

• Yes, this is what I asked. Thank you! I'll wait a bit to see if people have other results before picking an answer. Commented Feb 25, 2010 at 2:37