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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?

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  • $\begingroup$ 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, 2013 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, 2013 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, 2013 at 15:25
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    $\begingroup$ @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, 2013 at 16:56

2 Answers 2

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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 $( (1 \pm \sqrt{-35})/2).$

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  • $\begingroup$ 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, 2013 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, 2013 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, 2013 at 17:54
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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.

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  • $\begingroup$ 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, 2013 at 11:26
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    $\begingroup$ @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 $\text{Cl}(\mathcal{O}_K)=\langle [Q],[\overline{Q}]\rangle=\langle [Q], -[Q]\rangle=\langle [Q]\rangle$. $\endgroup$ May 22, 2013 at 13:18
  • $\begingroup$ Thanks for answering. Espacially so stupid a question like the above. Sorry for that. $\endgroup$
    – awllower
    May 22, 2013 at 15:06
  • $\begingroup$ @awllower Absolutely no problem :) $\endgroup$ May 22, 2013 at 15:13

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