Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is probably common knowledge, alas I have to confess my ignorance.

In simpler more abstract language, does $\mathcal{O}_K$ being simply connected (having trivial etale $\pi_1$) imply $\mathcal{O}_K=\mathbb{Z}$?

share|improve this question
    
I think the Gaussian integers also work. –  S. Carnahan May 30 '10 at 19:44
7  
In fact it's an exercise in Washington's Cyclotomic fields to show that it's true for all the imaginary quadratic number fields with class number 1. –  dke May 30 '10 at 20:50
4  
A variant of this question is to find a number field $K$ such that the fundamental group of $O_K$ is finite and non-trivial. I don't myself know an example, but I haven't tried too hard to find one either. –  Minhyong Kim May 30 '10 at 23:23
2  
For the record, Cam McLeman gives such an example below. –  JBorger Jun 1 '10 at 7:16
1  
@Minhyong: I found the following example in Serre (BAMS, 2003). Let $f=T^3-T-1$, $E=\mathbb{Q}[T]/f$, and $L$ the galoisian closure of $E$. Then $\operatorname{Gal}(L|\mathbb{Q})=S_3$, the symmetric group on three letters, and $L$ is an unramified cubic cyclic extension of $K=\mathbb{Q}(\sqrt{-23})$. As $h(-23)=3$, it is the maximal unramified abelian extension of $K$. Serre says that $L$ is in fact the maximal unramified extension --- abelian or not --- of $K$, as follows from Oldyzko's discriminant bounds (Martinet, Petits discriminant... , Journées arithmétiques 1980). –  Chandan Singh Dalawat Jun 4 '10 at 4:25
add comment

3 Answers

up vote 22 down vote accepted

Yes, there are examples and Minkowski's proof for ${\mathbf Q}$ can be adapted to find a few of them. Some examples of this kind among quadratic fields $F$, listed in increasing size of discriminant (in absolute value), are $$ {\mathbf Q}(\sqrt{-3}), \ \ {\mathbf Q}(i), \ \ {\mathbf Q}(\sqrt{5}), \ \ {\mathbf Q}(\sqrt{-7}), \ \ {\mathbf Q}(\sqrt{2}), \ \ {\mathbf Q}(\sqrt{-2}). $$ A cubic and quartic field $F$ that will come out of the method I describe below are ${\mathbf Q}(\alpha)$ where $\alpha^3 - \alpha - 1 = 0$ and ${\mathbf Q}(\zeta_5)$.

Now for the details. I suggest when reading this through for the first time that you keep a concrete example in mind, like $F = {\mathbf Q}(i)$. (That's what I did the first time I worked this out.)

Over the rationals, Minkowski showed a number field with degree larger than 1 must have a discriminant whose absolute value is larger than 1. Over other number fields $F$ besides the rationals, the goal is to find sufficient conditions on $F$ so that any finite extension $E/F$ with $[E:F] > 1$ has its discriminant ideal ${\mathfrak d}_{E/F}$ not equal to the unit ideal, and then a prime ideal factor will ramify in $E$.

Rather than show ${\mathfrak d}_{E/F} \not= (1)$, we will look for a sufficient condition on $F$ which assures us that the norm of this ideal is not 1. That means absolutely the same thing, but it's easier to work with ideal norms since they are positive integers rather than ideals, and moreover it lets us express the problem in terms of discriminants of number fields: the discriminants of $E$ and $F$ are related by $$|d_E| = {\rm N}({\mathfrak d}_{E/F})|d_F|^{[E:F]}.$$ So aiming to show ${\mathfrak d}_{E/F} \not= (1)$ is the same as avoiding $|d_E| = |d_F|^{[E:F]}$, which is the same as avoiding $$|d_E|^{1/[E:{\mathbf Q}]} = |d_F|^{1/[F:{\mathbf Q}]}.$$ We want sufficient conditions on $F$ to guarantee this equation for any proper finite extension $E/F$ can't take place.

For any number field $K$, the quantity $|d_K|^{1/[K:{\mathbf Q}]}$ is called the root discriminant of $K$. When $n = [K:{\mathbf Q}]$, Minkowski's lower bound on $|d_K|$ is $$ |d_K| \geq \left(\frac{\pi}{4}\right)^n\frac{n^{2n}}{n!^2}, $$ so we get the root discriminant lower bound $$ |d_K|^{1/n} \geq \left(\frac{\pi}{4}\right)\frac{n^{2}}{n!^{2/n}}. $$ Call the right side $f(n)$, so the Minkowski bound says $|d_K|^{1/[K:{\mathbf Q}]} \geq f([K:{\mathbf Q}])$.

For $n = 1, 2, 3, 4$ the values of $f(n)$ are $.785, 1.570, 2.140, 2.565$, so we guess $f(n)$ is increasing and it's left as an exercise to prove that. (Hint: use the one-sided Stirling estimate $n! > (n/e)^n\sqrt{2\pi{n}}$.)

Returning to the extension $E/F$, let $m = [F:{\mathbf Q}]$, so $[E:{\mathbf Q}] = [E:F][F:{\mathbf Q}] \geq 2m$ since $E$ is a larger field than $F$. Now if $E/F$ had trivial discriminant ideal, we would have $$ |d_F|^{1/[F:{\mathbf Q}]} = |d_E|^{1/[E:{\mathbf Q}]} \geq f([E:{\mathbf Q}]) \geq f(2m). $$ This hypothetical lower bound on the root discriminant of $F$ is larger than the proved Minkowski bound of $f(m)$. Suppose $F$ is a number field of degree $m$ whose root discriminant is less than $f(2m)$. If $E/F$ is unramified at all primes in $F$ then $E$ and $F$ have equal root discriminants, so the root discriminant of $E$ is less than $f(2m)$. However, we saw above that the root discriminant of $E$ is $\geq f(2m)$ when $[E:F] \geq 2$, so the only choice is $E = F$, i.e., no proper finite extension of $F$ can be unramified at all primes in $F$.

Our goal now is to find examples of number fields $F$ with degree $m$ whose root discriminant is smaller than $f(2m)$: $|d_F|^{1/m} < f(2m)$. This is the same as $$ |d_F| < f(2m)^m = \frac{\pi^mm^{2m}}{(2m)!}. $$ Any $F$ which fits this condition will be an example.

As a reality check, let $F$ be the rationals, so $m = 1$. We have $f(2) = \pi/2$ and $|d_{\mathbf Q}| = 1 < \pi/2$, so $\mathbf Q$ has no unramified extensions. We're on the right track.

Taking $m = 2$ we want $|d_F| < f(4)^2 = 6.57$ and the fields ${\mathbf Q}(i)$, ${\mathbf Q}(\sqrt{-3})$, and ${\mathbf Q}(\sqrt{5})$ all work.

If the root discriminant of $F$ is not below $f(2m)$, where $m = [F:{\mathbf Q}]$, we can still squeeze out some information, namely an upper bound on the degree of an everywhere (= at finite places) unramified extension of $F$. For instance, ${\mathbf Q}(\sqrt{2})$ has discriminant 8, which is not below 6.57, so this argument doesn't show on its own that every proper extension of ${\mathbf Q}(\sqrt{2})$ is ramified at some prime in ${\mathbf Q}(\sqrt{2})$. However, we can bound the degree of such an extension. The root disc. of ${\mathbf Q}(\sqrt{2})$ is $\sqrt{8} \approx 2.828$, which lies between $f(4)$ and $f(5)$, so any proper unramified extension of ${\mathbf Q}(\sqrt{2})$ must be a quadratic extension of ${\mathbf Q}(\sqrt{2})$. Quadratic extensions are automatically abelian, so we would have an abelian extension of ${\mathbf Q}(\sqrt{2})$ unramified at no finite places, and there's no such thing by class field theory since ${\mathbf Q}(\sqrt{2})$ has narrow class number 1 (the class number not involving ramification at infinity). Therefore you can add ${\mathbf Q}(\sqrt{2})$ to the list of number fields with no proper extension unramified at all finite places. The same arguments apply to ${\mathbf Q}(\sqrt{-2})$ and ${\mathbf Q}(\sqrt{-7})$, whose root discriminants are also between $f(4)$ and $f(5)$.

Unfortunately, this method is really limited because although $f(n)$ is increasing, it's actually bounded. By Stirling's formula, $f(n)$ has limit $\pi e^2/4 \approx 5.803$. So when a number field $F$ has root discriminant exceeding this value, this method won't give us any examples at all (you can't find an $m$ for which the root discriminant is below $f(2m)$). [Edit: Using Odlyzko-type lower bounds on discriminants, one can produce some more examples as Torsten points out.]

This kind of argument using Minkowski's bound does work for a few cubic fields and quartic fields: for cubic fields it works as long as the discriminant of the field (in absolute value) is less than 31.39, and there are two such fields: ${\mathbf Q}(\alpha)$ and ${\mathbf Q}(\beta)$ where $\alpha^3 - \alpha - 1 = 0$ (discriminant -23) and $\beta^3 + \beta + 1 = 0$ (discriminant -31). The next smallest absolute value of a discriminant of a cubic field is 44, which is above the bound. For quartic fields we need the discriminant to be less than 158.32, and I know of three fields which work: ${\mathbf Q}(\gamma)$ where $\gamma^4 + 2\gamma^3 + 3\gamma + 1 = 0$ (discriminant 117), ${\mathbf Q}(\zeta_5)$ has discriminant 125, and ${\mathbf Q}(\zeta_{12})$ has discriminant 144.

share|improve this answer
3  
I think Odlyzko's bounds on the root discriminants (On conductors and discriminants) can be used to go a bit further as they are better than the Minkowski bounds. –  Torsten Ekedahl May 30 '10 at 21:22
1  
As a concrete example, let F be splitting field of x^3 - 2. Its disc. is 34992. The Minkowski bound implies that if [K:Q]=n then log|disc(K)| > 1.39n - 1.93. If K were unram. extension of F then |disc(K)| = 34992^(n/6), so log|disc(K)| = 1.74n, which is already above the Minkowski bound, so we learn nothing. On the other hand, using the Odlyzko bound (as in Washington's book, noting K is tot. cpx) with sigma = 2.8 gives: log|disc(K)| > 2.178n - 4.945, and putting 1.74n on the left we get n < 11.28. Since 6|n, we get n = 6, so K = F. Thus F has no proper unram. extns. Way to go, Odlyzko. –  KConrad May 31 '10 at 3:52
    
Too many symbols... Can you, please, append a summary to your post (the answer, not the comment? –  Victor Protsak May 31 '10 at 4:51
    
Victor, I don't understand what you mean. Are the symbols not coming out properly for you? Where do you want a summary: at the start, end,...? –  KConrad May 31 '10 at 21:40
1  
I rewrote the beginning to make the plan of attack clearer (I hope) and listed at the start some of the fields to which the method applies. You should just set F = Q(i) all the way through to see what is happening. –  KConrad Jun 1 '10 at 6:29
show 1 more comment

Yes, there are many such fields.

(Edit: Let me put up here that "many" is still finite, and only in the hundreds if my list below represents most of the known examples. Note that "state-of-the-art" is still pretty primitive here -- infinitely many examples would give infinitely many number fields with class number one, which we don't know to exist (but strongly strongly suspect to!).)

As Torsten brings up (and what is behind the exercise in Washington that dke mentions), the Odlyzko bounds are really the key here. Briefly, the remarkable result that comes from these bounds is:

The ring of integers in any number field with sufficiently small root discriminant has trivial fundamental group, i.e., admits no non-trivial unramified extensions.

The "sufficiently small" is large enough, with the Odlyzko bounds, to generate fairly large examples of what you're looking for:

  • All quadratic imaginary number fields with class number 1 (of which there are 9)
  • All cyclotomic fields with class number 1 (of which there are 30)
  • All real quadratic fields with prime-power conductor $\leq 67$ (I count 9).
  • The top of the class field tower over a quadratic imaginary number field $K$ for many many $K$, including at least all such number fields with conductor $\leq 1000$ (a couple hundred).

The last two of these are work of Yamamura (results spread out, but see "Maximal Unramified Extensions of Imaginary Quadratic Fields of Small Conductors", parts I and II).

While I'm here, this last one answers affirmatively the question brought up by Minhyong in the comments. For example, $\mathbb{Q}(\sqrt{-771})$ has a ring of integers with fundamental group $S_4\times \mathbb{Z}/3\mathbb{Z}$, for this is precisely the Galois group of the Hilbert class field tower over this base. Finite but non-trivial. Many other examples of fundamental groups that can show up in this way are found in Yamamura.

I think the natural follow-up (and incredibly interesting) question is to classify exactly which groups can show up in this fashion. This is completely intractable at the moment, modulo some easy partial results in both directions. Edit: For example, by class group considerations, it is impossible to have a quadratic imaginary number field with etale fundamental group isomorphic to $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$.

share|improve this answer
    
Here is a specialisation of the question at the end. The polynomial (an example "dear to Artin" according to Lang, if I remember the comment in his book correctly) $x^5-x+1$ gives an unramified $A_5$-extension $K$ of $\mathbb Q[\sqrt{19\cdot151}]$. Is this the maximal unramified extension of $\mathbb Q[\sqrt{19\cdot151}]$? The class group of $K$ is trivial so a further solvable extension is out. On the other hand Odlyzko's bounds are not strong enough to give anything it seems. –  Torsten Ekedahl Jun 2 '10 at 5:17
    
You only can expect a positive answer if unramified means unramified everywhere. If you allow ramification at infinity, you get a larger tower by adjoining a square root of -19 and then continuing with the Hilbert class field of ${\mathbb Q}(\sqrt{-151})$. BTW, how do you prove that K has class number 1 if Odlyko's bounds are too weak? –  Franz Lemmermeyer Jun 2 '10 at 11:07
    
Well it is even worse I thought about the distinction for $K$ where I noted that it doesn't matter as $K$ is totally complex but forget that it matters for $\mathbb Q[\sqrt{19\cdot151}$. What is much worse I however is that when I asked Magma (and also Kant to doublecheck) about the class number I did for the field obtained by adjoining one root of the polynomial and not all of them. Hence I got class number (which is probably OK) but for the wrong field... Back to the drawing board. –  Torsten Ekedahl Jun 2 '10 at 15:04
add comment

Necessarily class number must be 1, then all exts with an abelian subextension must ramify. For 'small fields' discriminant estimates may help to establish the desired property, but I don't know to what generality this works, nor how to approach such a problem for 'large' fields. If I had had the possibility to leave a comment, I would have done that, for this is certainly not an 'answer'.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.