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I'm trying to calculate the class group of $\mathbb{Q} (\sqrt{-21})$, and I've managed to confuse myself. I've used the standard Minkowski bound to show that we only need to consider principal ideals (p) for p $\leq$ 5, and have found that (2) = $\mathcal{P}^2$, (3) = $\mathcal{Q}^2$, (5) = $\mathcal{R}_1 \mathcal{R}_2$ for $\mathcal{R}_1 \neq \mathcal{R}_2$.

Then the norm is $N(a + b\sqrt{-21}) = a^2 + 21 b^2$ and we want to find out some more information about our ideals (none of which are principal): take a = 2 and b = 1: then $(2 + \sqrt{-21}) = \mathcal{R}_1 \mathcal{R}_1$ WLOG (since 5 doesn't divide $2^2 + 1^2 \sqrt{-21}$, it is divisible by just one of the $\mathcal{R}_i$: as in the handout by Keith Conrad - see on the first page) but then $\mathcal{R}_1$ has order 2 and is self inverse, so then $\mathcal{R}_2$ is the inverse so $= \mathcal{R}_1$, so they are not distinct: except they should be. So where have I gone wrong?

I don't know whether or not I've made a logical error, but if not I suspect the problem may be arising because you can factorise 25 in 2 distinct ways. Can anyone help me? I hope this makes sense, I think I have probably left off some square brackets but I hope the general approach I am taking (very similar to Conrad's handout) is comprehensible. Many thanks! -Pete

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The ideals aren't necessarily equal, but they do lie in the same class. (Did you forget to quotient out by principal ideals?) – Zhen Lin May 15 '11 at 17:25
Ah, I see, that must be what I forgot: so what can we deduce (if anything) by the case with norm 25? The only other case I could find anything useful from at all was a = 3, b = 1: I couldn't spot anything else to calculate the group... – Peter J. May 15 '11 at 17:30
I don't understand. If $(5) = R_1 R_2$ then $(25) = R_1^2 R_2^2$. Did you mean to say that $(2 + \sqrt{-21}) = R_1^2$? – Qiaochu Yuan May 15 '11 at 17:33
Peter: once you know P and Q are nonprincipal, so [P] and [Q] are nontrivial ideal classes, you know that the ideal class group is generated by two elements of order 2. But that does not imply the ideal class group is isom. to C_2 x C_2 unless you know [P] and [Q] are different. Since they have order 2, if [P] = [Q] then [1] = [P][Q]^(-1) = [P][Q] = [PQ], so the ordinary ideal PQ would be principal. Now you want to show PQ is a nonprincipal ideal to get a contradiction. (Hint: think about the ideal norm of PQ. Or use that [PQ] = [R_1] and show R_1 is nonprincipal.) – KCd May 15 '11 at 20:07
@KCd: I'm sorry, I certainly didn't mean to offend, I was merely saying it lightheartedly as I felt you and your 'handouts' have often proved invaluable when no-one else has been able to offer a clear and relevant explanation to something I'm studying; I felt this was self-evident in the clarity of the document itself, and indeed it was the only useful thing which came up upon searching. However, it seems I have managed to upset, so apologies for that - I think I'm fine with the rest of the question now, so thank-you all for the help, it is greatly appreciated. – Peter J. May 15 '11 at 20:37

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