For any finite group, G, we can find a cover of ℙ1ℂ 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 ℙ1ℂ 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?



For a given finite group G, is there a cover of P^1 over Q s.t. over C it's G-Galois?
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](http://alpha.math.uga.edu/%7Epete/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.]


