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So I was having tea with a colleague immensely more talented than myself and we were discussing his teaching algebraic number theory. He told me that he had given a few examples of abelian and solvable extensions unramified everywhere for his students to play with and that he had find this easy to construct with class field theory in the back of his head. But then he asked me if I knew how to construct an extension of number fields with Galois group $A_{5}$ and unramified everywhere. All I could say at the time (and now) is:

  1. There are Hilbert modular forms unramified everywhere.
  2. There are Hilbert modular forms whose residual $G_{{F}_{v}}$-representation mod $p$ is trivial for all $v|p$.
  3. There are Hilbert modular forms whose residual $G_{F}$-representation mod $p$ has image $A_{5}$ inside $\operatorname{GL}_{2}(\mathbb F_{p})$.

Suppose there is a Hilbert modular form satisfying all three conditions. Then the Galois extension through which its residual $G_{F}$-representation factors would have Galois group $A_{5}$ and would be unramified everywhere.

Can this be made to work?

Regardless of the validity of this circle of idea, can you construct an extension of number fields unramified everywhere and with Galois group $A_{5}$?

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I've forgotten the standard example, but I think there's a standard example. You write down some explicit polynomial of degree 5 with rational coefficients (it's something like $x^5+x+1$ but it's not exactly that) and its Galois group over $\mathbf{Q}$ is $S_5$, and then you let the base be the quadratic subextension. –  Kevin Buzzard Nov 4 '10 at 10:47
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The problem with using Hilbert modular forms is that it is very common that the associated mod $p$ representation is ramified at $p$, making the problem computationally very hard. For example if you want to compute in parallel weight 2 then even the determinant of the mod $p$ representation will be ramified at $p$ unless $p=2$. This rules out using, say, standard tables of elliptic curves of conductor 1. –  Kevin Buzzard Nov 4 '10 at 10:50
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I don't know whether this is relevant here, but since $A_5$ is isomorphic to $\mathrm{PSL}_2(\mathbf{F}_5)$, one could try to start with an elliptic curve over some number field with good reduction everywhere, and look at the Galois representation on the 5-torsion points. –  François Brunault Nov 4 '10 at 11:00
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You'll need the base to contain $Q(\zeta_5)$ or something, to stop it being ramified at 5. –  Kevin Buzzard Nov 4 '10 at 11:04
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4 Answers

up vote 15 down vote accepted

If you take the splitting field of $x^5+ax+b$ and consider it as an extension of its quadratic subfield, then it will be unramified with Galois group contained in $A_5$ whenever $4a$ and $5b$ are relatively prime. This is a result of Yamamoto. For almost all $a$ and $b$ (specifically, on the complement of a thin set), the group is $A_5$.

You might also enjoy this preprint of Kedlaya, which I found very readable. A note on Kedlaya's webpage, dated May 2003, says that he will not be publishing this because it has been superseded by a recent result of Ellenberg and Venkatesh. I assume he is referring to this paper, but I can't figure out why that one supersedes his.

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Surely one needs the hypothesis that the splitting field of $x^5+ax+b$ has Galois group $S_{5}$. –  Olivier Nov 4 '10 at 13:08
    
You're right, I'll edit. –  David Speyer Nov 4 '10 at 13:29
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Here's the standard example. I found it in Lang's Algebraic Number Theory where he attributes it to Artin. Let $K$ be the splitting field of $X^5-X+1$ over $\mathbb{Q}$. Then $K$ has Galois group $S_5$ over $\mathbb{Q}$ and $A_5$ over $L=\mathbb{Q}(\sqrt{2869})$. Also $K$ is unramified over $L$.

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Thanks Robin. I knew I was close :-) –  Kevin Buzzard Nov 4 '10 at 11:37
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Is $2869$ something like the discriminant of $X^5-X+1$ ? –  Denis Serre Nov 4 '10 at 11:57
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@Denis Serre: is there no computer algebra package on your computer? :-) Yes, 2869 is the discriminant. –  Kevin Buzzard Nov 4 '10 at 12:04
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@François: I think any quintic $f$ over ${\mathbb Q}$ with Galois group $S_5$ and square-free discriminant $D$ defines an unramified $A_5$-extension of $Q(\sqrt D)$. This is because exactly two roots of $f$ coincide modulo any prime divisor of $D$, so the inertia at $p$ must be exactly $C_2$. E.g. $f(x)=x^5-x^4-x^3+x^2-1$ has $D=1609$, but I don't know whether there are any smaller examples. –  Tim Dokchitser Nov 15 '10 at 16:34
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Per the philosophy of Bhargava (and indeed it seems plausible that this could be proved via his techniques) the number of S_5-extensions with squarefree discriminant between -N and N should be about (13/120)*(6/pi^2)*2N; i.e. each squarefree discriminant has 13/120 A_5-extensions on average. –  JSE Dec 15 '10 at 2:08
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Oh, I know how I would try and build examples. First I would write down a random $A_5$ extension $K$ of $\mathbf{Q}$, ramified at some primes (in fact I would look in a table, e.g. in Buhler's thesis or the Frey et al LNM on Artin's conjecture, to find an example ramified at only one prime, probably). I would then try to kill this ramification by making an extension $L$ of $\mathbf{Q}$ which is disjoint from $K$ globally but exactly the same as it locally for the ramified primes (e.g. by going up to a splitting field of a polynomial which is highly $p$-adically congruent to the original poly whose splitting field was the $A_5$ field, for all $p$ where $K$ ramified). I'd then look at the extension $LK/L$ which I think now should be totally split at the primes above the prime where $K/\mathbf{Q}$ ramified and hence unramified everywhere. I think this stands a fair chance of working, and indeed becoming a general machine which turns an extension of number fields with group $G$ into an extension of number fields unramified everywhere with group $G$. Or have I missed something? If not then this is undoubtedly a well-known technique.

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Here's an explicit worked example in a much easier case. Consider $K=\mathbf{Q}(\sqrt{2})/\mathbf{Q}$. It's the splitting field of $x^2-2$. It's ramified at 2. So let's choose a polynomial highly congruent to $x^2-2$, for example $x^2-34$, and set $L=\mathbf{Q}(\sqrt{34})$. Now $KL/L$ should be unramified everywhere and if my calculations are right, it is. –  Kevin Buzzard Nov 4 '10 at 11:42
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If I understand correctly, the approach of Yamammoto (in the specific case of $G=A_{n}$) is somewhat dual to what you suggest. Instead of killing the ramification by composing with an auxiliary extension, he realizes $G$ as a normal subgroup of a larger group $G'$ and ensures that the ramification can only occur in the quotient $G'/G$. When $G=A_{n}$, this works nicely because $G'=S_{n}$ and it is easy to build polynomials with at most one double roots modulo $p$ for all $p$. –  Olivier Nov 4 '10 at 13:13
    
@Kevin: this certainly sounds good to me. Indeed, the basic technique (choosing global extensions which have prescribed behavior at finitely many places) is one that I have used many times in my own work... –  Pete L. Clark Nov 4 '10 at 19:21
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Although the question was only about unramified $\mathfrak{A}_5$-extensions, and has been completely answered, it might not be superfluous to mention the following paper which I happened to come across today:

MR0819826 (87e:11122)

Elstrodt, J.(D-MUNS); Grunewald, F.(D-BONN); Mennicke, J.(D-BLF)

On unramified $A_m$-extensions of quadratic number fields.

Glasgow Math. J. 27 (1985), 31–37.

An explicit description is given of unramified extensions $S/k$ with Galois group equal to the alternating group $A_n$, where $k$ is a quadratic number field. The authors prove that if $f(x)\in {\bf Z}[x]$ is a monic, irreducible polynomial of degree $n$ with square-free discriminant and Galois group $S_n$, then $S/k$ is an unramified $A_n$-extension. Here $S$ denotes the splitting field for $f(x)$ over ${\bf Q}$ and $k={\bf Q}(\sqrt{\Delta})$, where $\Delta$ is the determinant of $f$. The proof involves a series of calculations which show that $S/k$ has relative different 1.

In the final section, 84 examples of unramified $A_5$-extensions of quadratic fields are given. In 15 of the cases the quadratic field is real and in 69 cases it is imaginary. This list contains an example (with real quadratic field) due to E. Artin, which was mentioned by S. Lang [Algebraic number theory, see p. 121, Addison-Wesley, Reading, Mass., 1970].

Reviewed by Charles J. Parry

Addendum (2011/03/30) Kedlaya's preprint mentioned by Speyer is now available on the arXiv. A corollary is that for each $n\geq3$, infinitely many quadratic number fields admit everywhere unramified degree-$n$ extensions whose normal closures have Galois group $\mathfrak{A}_n$.

Addendum Kedlaya's paper has now appeared in the Proceedings of the AMS 140 (2012), 3025--3033

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That preprint of Kedlaya is exactly the same as the one in Speyer's answer (well, different web links, but the same content). –  KConrad Mar 30 '11 at 6:52
    
I'm really sorry not to have realised this. The Addendum has been edited accordingly. –  Chandan Singh Dalawat Mar 30 '11 at 7:21
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