# Galois groups and prescribed ramification

What is known about finite groups $G$ for which there exists a Galois extension $K$ of $\mathbb{Q}$ ramified only at $2$ such that $\text{Gal}(K/\mathbb{Q}) \cong G$ ? More generally, which groups can be realized over $\mathbb{Q}$ with no ramification outside a given (finite) set of primes?

I am thus interested in results of two kinds:

Realization of specific groups.

Examples of groups which are not realizable in this manner, and restrictions on groups which are.

Let me first note that there is a slight ambiguity when one says "ramified only at 2". Strictly speaking, that means that the extension is unramified at every place of $\mathbb Q$ except 2, including infinity. The latter mean that the extension is totally real. Often, however, "ramified only at 2" means "ramified only possibly at 2 and $\infty$", and it is probably what you mean. Here, to remove ambiguity, for $S$ a finite set of places of $\mathbb Q$, I will use "ramified only at $S$" in the strict sense.

That being said, the short answer is that whatever the finite set $S$, there are strong restrictions on the finite groups $G$ that can appear as the Galois group of an extension of $\mathbb Q$ unramified outside at $S$, but that in general, to "describe" all these restrictions (we may mean different things by that) is in general an open problem. To see what kind of restrictions can appear, let us consider several situation, from the very particular to the general.

If $S=\emptyset$ or $S=\{\infty\}$, then Minkowski's theorem tell that there is no nontrivial extension unramified outside $S$, so the only possible Galois group $G$ is the trivial one. A very strong restriction indeed.

If $S$ is any set, but you try to determine what abelian $G$ may appear, then the answer is given by class field theory. Precisely, if $S=\{\ell_1,\dots,\ell_k,\infty\}$, then the abelian group which appear are the ones that are quotient of $\mathbb Z_{ell_1}^\ast \times \dots \mathbb Z_{\ell_k}^\ast$, and it is obvious that many abelian groups are not of this type (e.g. $\mathbb Z/\ell \mathbb Z$ for $\ell > \ell_1,\dots,\ell_k$). If $S=\{\ell_1,\dots,\ell_k\}$, replace $\mathbb Z_{ell_1}^\ast \times \dots \mathbb Z_{\ell_k}^\ast$ by its quotient by the diagonal subgroup $\{1,-1\}$.

If $S$ is any set, and you're interested in groups $G$ that are $p$-groups (for a $p$ that may or may nor be in $S$), then again, class field theory can help you because of Frattini's argument saying that a $p$-group is generated by any set that maps subjectively to the maximal abelian $p$-torsion quotient of $G$. So by the above, the $p$-groups $G$ appearing this way have a system of generators with less elements than the dimension over $\mathbb F_p$ of the maximal $p$-torsion quotient of $\mathbb Z_{ell_1}^\ast \times \dots \mathbb Z_{\ell_k}^\ast$. Is that all? No. We can describe in some cases all the $p$-group $G$ appearing, but not in all. For instance, with $S=\{2,\infty\}$, and $p=2$, the groups $G$ appearing are exactly the $2$-groups having a system of two generators, one of them being of square 1. For $S=\{2\}$, the only $2$-groups appearing are the cyclic groups. In general, the case $p \in S$ is better understood than the case $p \not \in S$. In the latter case, it is a folklore conjecture, on which not much is known, that only finitely many $p$-groups $G$ appearing (one can check readily that the conjecture is true for abelian $p$-groups, by the above paragraph).

If now we consider general finite groups $G$, well, the above shows that there are restrictions, but determining them all is largely open. For instance, a conjecture of Shafarevich states that there is an $n=n(S)$ such that every group $G$ which is the Galois group of an extension unramified outside $S$ has a system of generators with less than $n$ elements. But this is open in every non-trivial case.

Let me also mention a simple but too little known result of Serre, even if it is not strictly speaking part of your question: [Edit: Sorry, I messed up the statement of the result. Here is the right version]. If every finite group is a Galois group over $\mathbb Q$ (if the inverse galois problem has a positive solution) then every finite group is the Galois group of an extension of $\mathbb Q$ unramified at infinity -- that is, a real extension. In other words, according to the inverse Galois problem, there should need no restriction of $G$ in the case where $S$ is the set containing all finite places, but not the place at infinity.

• Is the situation for $\mathbb{F}_p[t]$ instead of $\mathbb{Q}$ more understood? – Pablo Sep 25 '14 at 6:13
• Pablo: I don't know. – Joël Sep 25 '14 at 18:25

Concerning the question of Pablo that follows Joël's answer:

If $k$ is an algebraically closed field, then the situation is completely understood, thanks to work of Grothendieck (in characteristic 0) completed by the proof of Abhyankar's conjecture by Raynaud and Harbater.

Precisely, let $C/k$ be a smooth affine curve, complement of $r\geq1$ points in a projective smooth connected curve of genus $g$, over the alg. closed field $k$. Let $F=k(C)$ be the field of rational functions on $C$. Extensions of $C$ that are unramified on $C$ correspond to connected étale covers of $C$.

If $\mathop{\rm char}(k)=0$, then Grothendieck showed (this is the main result of SGA 1) that a finite group $G$ is the Galois group of a connected étale cover of $C$ if and only if it is generated by $2g+r-1$ elements. More precisely, the Galois group of the maximal extension of $k(C)$ unramified on $C$ is a free profinite group on $2g+r-1$ elements. It is worth observing the fundamental group of a compact connected Riemann surface of genus $g$ deprived from $r\geq1$ points is a free group on $2g+r-1$ generators. (The case $r=0$ is also treated by Grothendieck, $G$ then needs to be generated by $2g$ elements $a_1,b_1,\dots,a_g,b_g$ satisfying the relation $(a_1,b_1)(a_2,b_2)\dots(a_g,b_g)=1$.)

If $\mathop{\rm char}(k)=p>0$, then $C$ has « more » étale covers. For example, if $C=\mathbf A^1_k$, then $f\colon C\to C$ given by $f(t)=t-t^p$ is a connected étale cover of degree $p$. Raynaud (when $C=\mathbf A^1_k$) and Harbater (in general) showed that a finite group $G$ is the Galois group of a connected étale cover of $C$ if and only if the quotient of $G$ by the (normal) subgroup generated by its $p$-Sylow subgroups is generated by $2g+r-1$ elements.

Specific examples had been exhibited by many mathematicians, such as Abhyankar itself (he has a long series of papers in this style) and Nori (when $G$ is a finite group of Lie type).

• So, was the conjecture on the bounded number of generators of the galois group of any extension of $\mathbb{F}_p[t]$ unramified outside of a finite set of primes settled? (the function field analogue of Shafarevich's conjecture mentioned in Joel's answer) – Pablo Sep 26 '14 at 6:26
• I'm not really sure of the following argument, but let's try something. If $X$ is an affine curve over a field $k$, and $\overline X$ is the base change to an algebraic closure $\overline k$, there is an exact sequence of fundamental groups, which is split if $X$ has a $k$-rational point: $1\to \pi_1(\overline X)\to \pi_1(X) \to \mathop{\rm Gal}(\overline k/k)\to 1$. It follows that $\pi_1(X)$ is the direct product of $\widehat{\mathbf Z}$ with a profinite free group on $2g+r-1$ generators. – ACL Sep 26 '14 at 19:06
• (followed) : It thus seems that the Galois group of any extension of $k(X)$ which is unramified above $X$ is generated by $2g+r$ elements. – ACL Sep 26 '14 at 19:07
• Do you really need the sequence to split? its exactness should be enough to bound the number of generators of the middle group by the sum of the number of generators of the other two. – Pablo Sep 27 '14 at 6:16
• You're right. Knowing that it splits allows to understand precisely those extensions which are regular, is contain no nontrivial extension of the finite ground field. – ACL Sep 28 '14 at 7:09