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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?

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up vote 3 down vote accepted

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:

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.]

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Yes, this is what I asked. Thank you! I'll wait a bit to see if people have other results before picking an answer. – H. Hasson Feb 25 '10 at 2:37

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