Let $A$ be complete noetherian local ring with maximal ideal $m$ and residue field $A/m$ a finite field (in other words, $A$ is a noetherian compact local ring)

For which $A$ as above is there a subgroup of $GL_2(A)$ containing $SL_2(A)$ which is the Galois group of a Galois extension of $\mathbb Q$ unramified outside a finite set of primes $S=S(A)$?

(The italic part of the question above has been edited: the first, ill-formulated, version of this question asked if $GL_2(A)$ itself was a Galois group over $\mathbb Q$ and has been answered by Will Savin below).

For example, we know since the work of Serre on the points of torsion of elliptic curves that when $A=\mathbb Z_p$, the answer is yes. On the other hand, the answer should not be always yes: when $A$ has Krull dimension $5$ or more for instance, such a group would define a deformation to $A$ of the representation $G_{\mathbb Q,S} \rightarrow GL_2(A/m)$, which would have to be a quotient of the universal deformation ring of that representation, and these deformation ring are expected to have dimension at most $4$ (see Mazur's article in Galois Groups over A).

Obviously, I would be very very surprised if I received a complete answer to this question, but is it possible to give a conjectural answer, based on the most optimistic conjectures on the inverse Galois problems (e.g. Shafarevich's conjecture that $Gal(\mathbb Q/ \mathbb Q^{ab})$ is a free profinite group of countable rank), or at least, some reasonable guess?

  • $\begingroup$ We may not even know a conjectural answer. Do we even know the answer to first order, i.e. what is the maximum over all representations $V$ of $\operatorname{Gal}(\mathbb Q)$ on two-dimensional vector spaces over $\mathbb F_l$ of the rank over $\mathbb F_\ell$ of $H^1(\mathbb Q, ad(V))$? $\endgroup$ – Will Sawin Aug 12 '15 at 22:41

Claim: Given a representation from the Galois group to $GL_2(\mathbb F_q)$ (maybe $p>2$ to be safe) whose image contains $SL_2(\mathbb F_q)$, if $R$ is any quotient of the deformation ring of that representation (actually, the ring parameterizign deformations with fixed determinant character), then the image of the induced map to $GL_2(R)$ contains $SL_2(R)$.

Proof: By a limit we may reduce to finite length rings. Then by induction we may reduce to an extension by $\mathbb F_q$.

So let $R$ be a quotient of the deformation ring with maximal ideal $m$ and an ideal $I$ that is isomorphic as an $R$-module to $R/m$. Assume the representation to $GL_2(R/I)$ contains $SL_2(R/I)$. We must show that the representation to $GL_2(R)$ contains $SL_2(R)$.

It's sufficient to show that it contains the kernel of $SL_2(R)$ to $SL_2(R/I)$ or by conjugation sufficient to show that it contains any nontrivial element of that kernel.

If $I$ is contained in $m^2$, we can take $a,b \in m$ with $ab$ generating $I$ and take the commutator of the two matrices:

$$ \left[ \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & 1+b \end{pmatrix} \right] =\begin{pmatrix} 1 & ab \\ 0 & 1 \end{pmatrix} $$

By surjectivity in $R/I$, we have two matrices congruent mod $I$ to these, and their commutator will be the same. So that handles that case.

If $I$ is contained in $(p)$, you can do the same thing with

$$\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}^p = \begin{pmatrix} 1 & pa \\ 0 & 1 \end{pmatrix}$$

Otherwise $I$ corresponds to an element of $m/(p,m^2)$, which comes from a nontrivial extension class of the representation with itself. It's sufficient to show that for such a nontrivial extension there are some nontrivial elements that fix the sum and quotient. But otherwise you would get a nontrivial extension of the standard representation of $SL_2(\mathbb F_q)$ as a representation of $SL_2(\mathbb F_q)$, which there probably isn't by modular representation theory.

OK so this proof is not completely rigorous. But I think it can be made so with a bit more care.

Anyway, this would show that the question is basically equivalent to Passerby's version about quotients of deformation rings.

Very rarely. If $GL_2(A)$ is a Galois group then $GL_1(A)$ is a Galois group.

So in particular $Hom ( GL_1(A), \mathbb Z/p)$ must be finite, because there are finitely many $\mathbb Z/p$-extensions ramified outside a given finite set of primes.

If the residue field of $A$ has characteristic $p$, then $(A/p)^\times$ had better also have finite maps to $\mathbb Z/p$, which I think implies that it has Krull dimension $0$. Then it will be finite, and $A$ will be a finite extension of $\mathbb Z_p$ by Nakayama's lemma.

If $A$ is the ring of integers of a degree $n$ extension of $\mathbb Z_p$, then $GL_1(A)$ maps to $\mathbb Z_p^n$, so $\mathbb Z_p^n$ is also a Galois group over $\mathbb Q$. But $\mathbb Q$ has a unique $\mathbb Z_p$-extension, the cyclotomic extension.

So I think just $\mathbb Z_p$ works here.

  • $\begingroup$ Thank Will. That answers the question as asked (initially). Unfortunately, the question I asked is not exactly the one I wanted to ask, which was "modulo the center". I tried to simplify it at the last minute when asking it and messed it up. I will change the question accordingly. $\endgroup$ – Joël Aug 12 '15 at 20:13

This isn't an answer, just some idle thoughts about the following closely related problem:

Which complete local Noetherian rings $A$ can occur as the quotient of the universal deformation ring $R$ of a representation $\rho: G_{\mathbb{Q},S} \rightarrow \mathrm{GL}_2(\mathbb{F}_q)$?

Such a universal ring $R$ should always have dimension $4$ and be a complete intersection, at least conjecturally.

So straight away the following question occurs:
is the set $\mathcal{X}$ of isomorphism classes of complete, local, Noetherian, integral, $4$-dimensional $W(\mathbb{F}_q)$-algebras actually uncountable? (Conditions slightly edited)

I don't know the answer but if indeed $\mathcal{X}$ is uncountable, that seems to be a problem: Only finitely many of the elements of $\mathcal{X}$ could occur as quotients of each of the countably many universal $R$s (namely, one for each minimal prime ideal of $R$).

On the other hand, I think one might be able to verify that the set of universal deformation rings is dense in $\mathcal{X}$ (for some reasonable topology). For example, I believe that any Artin local $A$ is a quotient of an $R$ as above. This should follow from the method of Taylor and Wiles: they show that you can get a formal power series ring, in as many variables as you want, as a limit of such deformation rings $R$.

  • $\begingroup$ Thanks for these interesting reflexions. My guess is that $\mathcal X$ is not countable but I don't see how to prove it. That should be an easy question for specialists of classifications of singularities. (BTW, I don't really believe in the conjecture that all universal deformation rings in our context should be complete intersection. That seems wishful thinking to me.) $\endgroup$ – Joël Aug 13 '15 at 18:21
  • $\begingroup$ Yes, $\mathcal X$ is uncountable. From the ring $\mathbb Z_p[x,y,z,w]/(f(x,y,z,w))$ for a homogeneous polynomial $f$ we can recover the projective variety $f(x,y,z,w)$ by blowing up, but there are uncountably many non-isomorphic hypersurfaces. In fact this trick works with any projective variety. $\endgroup$ – Will Sawin Aug 14 '15 at 22:24

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