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Conjecturally, every finite group is the Galois group of some extension of the rationals. This question made me wonder what is known about infinite simple groups occurring as Galois groups.

What are the infinite simple groups that are expected to be Galois groups, i.e., profinite? Are they classified? Are there any examples of such extensions?

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I'm not sure I understand. Infinite Galois groups can never be simple, right? –  Qiaochu Yuan Jan 29 '11 at 14:02
@Qiaochu: right. (As long by Galois group we mean automorphism group of a normal, separable algebraic extension. I have some notes in which I suggest a definition of Galois transcendental extensions, and then $\mathbb{C}/\overline{\mathbb{Q}}$ is Galois and its automorphism group is indeed a huge simple group. But it's not a profinite group, of course.) –  Pete L. Clark Jan 29 '11 at 14:58
The problem with good answers to stupid questions is that you can't delete the whole stuff -). Any irrational element of the field in question lieves in the normal extension of the field it generates, and produces a normal subgroup. Thanks, everyone. –  Franz Lemmermeyer Jan 30 '11 at 18:28
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3 Answers

up vote 13 down vote accepted

Any profinite simple group is finite, since it has nontrivial finite quotients (the conjugates of a finite index subgroup intersect in a finite index subgroup).

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Th notion of simple is not very interesting for profinite groups. The "right" concept is just infinite. We say that a profinite group is just infinite if all its non-trivial normal closed subgroups are of finite index. $SL_2(\mathbb{Z}_p)$ is an example of a just infinite profinite group. Another example coming from Galois theory is the Nottingham group which is an open subgroup of index p-1 in the automorphism group of the field $\mathbb{F}_p((t))$.

There are many examples of just infinite profinite groups. There isn't much general theory except Wilson's dichotomy that they are either Branch Groups (e.g. Grigorchuk group or the Gupta-Sidki group) or they contain an open subgroup which is direct sum of hereditarily just infinte profinite groups (i.e. every open subgroup is also profinite) and some results of Colin Reid.

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@Yiftach: actually $\operatorname{SL}_2(\mathbb{Z}_p)$ is not just infinite: consider the subgroup $\pm 1$ (of scalar matrices). This same comment applies to JSE's answer: because of the existence of $\operatorname{PSL}_2(\mathbb{Z})$, $\operatorname{SL}_2(\mathbb{Z})$ is not a simple $p$-adic Lie group according to the definition given. (But close enough. Anyway, I certainly agree that it's interesting to try to get these groups as Galois groups.) –  Pete L. Clark Jan 29 '11 at 21:50
(In my last comment, where you see $\mathbb{Z}$ it should be $\mathbb{Z}_p$. I am too lazy to tex up the comment again.) –  Pete L. Clark Jan 29 '11 at 21:51
By the way, if you are mentioning "just infinite profinite groups" and Galois extensions of number fields, do be sure to give the example of $\mathbb{Z}_p$! –  Pete L. Clark Jan 29 '11 at 21:54
Pete, you are of course right. I tend to work with the first congruence subgroup which is pro-$p$, so I am not used to think about $SL_2(\Z_p)$. Regarding, $\Z_p$ it is indeed hereditarily just infinite, but similar to the way a cyclic group of prime order is simple, not very interesting from group theoretic point of view. –  Yiftach Barnea Jan 29 '11 at 22:12
@Yiftach: but $\mathbb{Z}_p$ extensions are extremely interesting from a number theoretic point of view, enough so to cancel out their group theoretic banality, IMO. (More objectively, the just infinite property of $\mathbb{Z}_p$ is part of the reason why $\mathbb{Z}_p$-extensions are so nice to study, more so than $\hat{\mathbb{Z}}$-extensions, for instance.) –  Pete L. Clark Jan 29 '11 at 23:05
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One way to intepret the question is: if G is a simple p-adic Lie group, like SL_2(Z_p), do we expect there to be an extension of Q with Galois group G? (Here the point is that G is not literally simple as a group, but it has no positive-dimenional p-adic analytic group quotient.)

GL_2 is easy -- adjoin the Tate module of an elliptic curve -- but is not simple. Maybe you can get SL_2(Z_p) via Shih's construction (as described e.g. in Serre's "Topics in Galois Theory"?)

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How about: weight 3 modular form gives a Galois rep, generically surjective or almost so, and then make a finite extension to kill the Dirichlet character, and now det is cyclo squared so twist by inverse of cyclo? Doesn't that give $SL(2,Z_p)$? Maybe up to a finite group... –  Kevin Buzzard Jan 29 '11 at 16:11
See my comment to Yiftech's answer. It is not wholly nitpicky, because Shih's construction indeed gives groups of the form $\operatorname{PSL}_2$ rather than $\operatorname{SL}_2$, although IIRC Serre gives an $\operatorname{SL}_2$-variant in his book. –  Pete L. Clark Jan 29 '11 at 21:53
I like Guzdek's comment, especially given that there's already a large body of literature about constructing extensions of G whose Galois groups are FINITE groups of Lie type. Re Buzzard's comment -- good point -- so now I'm confused, why is it a big deal to generate PSL_2(F_p) - exts for almost all p if you can do this with one wt-3 form? –  JSE Jan 29 '11 at 22:33
As someone who has worked in this area of IGP: if the "programme" in Buzzard's comment can really be worked out to get $\operatorname{SL}_2(\mathbb{Z}_p)$ as a Galois group over $\mathbb{Q}$ for all primes $p$, then someone should drop everything and write that up: that would be an exciting advance. I see two issues with it: (i) as Kevin says, general open image stuff seems to tell you that the image of the Galois rep is...open, rather than surjective.... –  Pete L. Clark Jan 29 '11 at 23:10
(ii) Kevin seems to be passing to a finite extension. It is certainly no problem to get $\operatorname{SL}_2(\mathbb{F}_p)$ as a Galois group over some number field depending on $p$. It's not completely obvious to me how to do the same for $\operatorname{SL}_2(\mathbb{Z}_p)$ (we do want a finite degree extension of $\mathbb{Q}$, right?) but to do this would be less impressive. Getting things uniformly over $\mathbb{Q}$ is much of the problem. –  Pete L. Clark Jan 29 '11 at 23:12
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