9
$\begingroup$

As an exercise I am trying to compute the class group of $\mathbb Q(\sqrt{-35})$.

We have $-35\equiv 1$ mod $4$, so the Minkowski bound is $\frac{4}{\pi}\frac12 \sqrt{35}<\frac23\cdot 6=4$. So we only need to look at the prime numbers $2$ and $3$.

$-35\equiv 5$ mod $8$, so $2$ is inert. Also, $-35\equiv 1$ mod $3$, so $3$ splits, i.e. $(3)=Q\overline Q$ with $Q=(3,1+\sqrt{-35})$, $\overline Q=(3,1-\sqrt{-35})$. The ideals $Q,\overline Q$ are not principal, because there are no solutions to $x^2+35y^2=12$, i.e. no elements of norm 3.

Now we know that there are at most $3$ elements (or do we?), namely $(1),Q,\overline Q$. Mathematica tells me that the class number is $2$, so $Q$ and $\overline Q$ must be in the same equivalence class and $Q^2$ has to be a principal ideal. But how can I show this?

$\endgroup$
  • $\begingroup$ I mean not to be rude, but why does this have to do with class-field theory? Per chance it is related to the defree of the Hilbert class field, otherwise, I see no connections. $\endgroup$ – awllower May 22 '13 at 15:21
  • $\begingroup$ I assumed "class groups" belong to the theory of "class-field-theory". I might be wrong, of course. $\endgroup$ – Phil-ZXX May 22 '13 at 15:23
  • $\begingroup$ Well, I cannot say they have no connections, but I believe that the question is not related to that theory, either the solution, or the concern. In fact, I think that the single tag algebraic-number-theory suffices in this case. Maybe there is somehow a way to solve the problem by the methods in CFT of course. So thanks for the attention then. :D $\endgroup$ – awllower May 22 '13 at 15:25
  • $\begingroup$ @awllower Sorry for not being too familiar with this subject matter, but aren't there bigger things to worry about? $\endgroup$ – Phil-ZXX May 22 '13 at 16:56
5
$\begingroup$

Note that the ring of integers is $\mathbb Z[(1+\sqrt{-35})/2]$.

If you compute $(3, 1 + \sqrt{-35})^2$, you get $$(9,3 + 3\sqrt{-35}, -34 + 2 \sqrt{-35} ) = (9, 1 + \sqrt{-35}) = ( \dfrac{1-\sqrt{-35}}{2} \dfrac{1 + \sqrt{-35}}{2}, 2 \dfrac{1+\sqrt{-35}}{2}) = ((1 + \sqrt{-35})/2 )$$ (because $\dfrac{1-\sqrt{-35}}{2}$ and $2$ are coprime, and so generate the ideal $1$).


I find the computation a bit easier by phrasing the factorization of $(3)$ in the following alternative way: $(3) = (3, (1 + \sqrt{-35})/2) (3,(1- \sqrt{-35})/2)$, and $$(3,(1+\sqrt{-35})/2)^2 = (9, 3(1+\sqrt{-35})/2,(-17+\sqrt{-35})/2) = ( (1+\sqrt{-35})/2 )$$ is principal.


As a consistency check, note that $ (9) = (3) (3) = Q \overline{Q} Q \overline{Q} = Q^2 \overline{Q}^2,$ but also $9 = ( (1+\sqrt{-35})/2) ( ( 1 - \sqrt{-35})/2),$ so we must have $Q^2$ equal to one of $( (1 \pm \sqrt{-35})/2).$

$\endgroup$
  • $\begingroup$ Does this mean that if $\mathbb Z[\frac{1+\sqrt{d}}{2}]$ is the ring of integers for $\mathbb Q(\sqrt{d})$ and a prime $p$ splits, then $(p)=(p,\frac{1+\sqrt{d}}{2})(p,\frac{1- \sqrt{d}}{2})$ and not $(p)=(p,1+\sqrt{d})(p,1- \sqrt{d})$ ? $\endgroup$ – Phil-ZXX May 16 '13 at 16:16
  • $\begingroup$ @Thomas: Dear Thomas, If a prime $p$ splits, the way it splits depends on the prime. E.g. if $p$ does not divide norm of $(1 +\sqrt{d})/2$ (and it won't for most $p$) then what you wrote is incorrect. In general, you need to find an element $\alpha$ whose norm is divisible by $p$ exactly once. Then $p = (p,\alpha)(p,\overline{\alpha}).$ You chose a different description, namely an element whose norm is divisible by $p^2$, but for the reason that it is divisible by $Q^2$ (in the notation of your question) rather than $Q \overline{Q})$. Regards, $\endgroup$ – Matt E May 16 '13 at 17:53
  • $\begingroup$ P.S. In writing the above, I realized that my claim that your factorization was wrong is incorrect. (It was written early in the morning; sorry about that!) I will edit. $\endgroup$ – Matt E May 16 '13 at 17:54
6
$\begingroup$

Let $\alpha=\frac{1+\sqrt{-35}}{2}$, then in fact, $(3)=Q\overline{Q}=(3,\alpha)(3,\alpha+2)$. In the class group then, $[Q]+[\overline{Q}]=[\mathcal{O}_K]$, so $[Q]$ and $[\overline{Q}]$ are inverses. So the class group is generated by $Q$, and we just need to show that $Q^2$ is principal.

The minimal polynomial of $\alpha$ is $f(x)=x^2-x+9$, and recall that $f(\beta)=\text{Nm}(\beta-\alpha)$. In particular, $f(0)=9=3^2$. The element $(\alpha)$ is sent to $0$ under the evaluation $\alpha\mapsto 0$, corresponding to the ideal $Q$, so $(\alpha)=Q^2$, and the order of $[Q]$ divides $2$. You've already shown that $Q$ can't be principal, so we're done.

$\endgroup$
  • $\begingroup$ Excuse me, how did you deduce that the class group is generated by $Q$? Even though $Q$ and $\overline{Q}$ are inverses, how does it have to do with the generating of the class group? Thanks in advance. $\endgroup$ – awllower May 22 '13 at 11:26
  • 1
    $\begingroup$ @awllower The class group is generated by ideals of norm at most $\mu_K$, so we need only look for primes above $2$ and $3$. Since $2$, is inert, only the primes above $3$ could generate the class group. So $Q$ and $\overline{Q}$ are the generators of the class group. Since one is the inverse of the other, we only need one of them to generate. The integers have two generators, $\pm 1$, but we only need one of them to generate the other. We've shown that $\text{Cl}(\mathcal{O}_K)=\langle [Q],[\overline{Q}]\rangle=\langle [Q], -[Q]\rangle=\langle [Q]\rangle$. $\endgroup$ – Warren Moore May 22 '13 at 13:18
  • $\begingroup$ Thanks for answering. Espacially so stupid a question like the above. Sorry for that. $\endgroup$ – awllower May 22 '13 at 15:06
  • $\begingroup$ @awllower Absolutely no problem :) $\endgroup$ – Warren Moore May 22 '13 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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