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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?)2011-05-15
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    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...2011-05-15
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    I don't understand. If $(5) = R_1 R_2$ then $(25) = R_1^2 R_2^2$. Did you mean to say that $(2 + \sqrt{-21}) = R_1^2$?2011-05-15
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    Yes, I'm sorry, got a bit messed up between what I was thinking and what I was writing! So we can deduce that $R_1$ has order 2 in the class group then? And then what happens with $R_2$? If they aren't equal but they lie in the same class, do we still get $[R_1]^{-1} = [R_2]$, and then deduce $[R_1] = [P][Q] = [R_2]$ and that we have the group $C_2 \times C_2$?2011-05-15
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    @Peter: yes, you know that the group is generated by $[P], [Q]$ and satisfies $[P]^2 = [Q]^2 = [1]$. Have you shown that $[P] \neq [Q]$?2011-05-15
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    I hadn't thought of that as being an issue: does it suffice to say [P][Q] = [R$_1$] is non-principal, so they can't be equal?2011-05-15
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    First of all, you shouldn't write "so says [me]" since it that sounds like "proof by authority". It would be better to write "like the description on the first page of...". For a = 3 and b = 1 you get (3+sqrt(-21)) = PQR_i for one of the ideals R_i, and then in the class group that becomes [P][Q][R_i] = [1], which shows [R_i] is in the subgroup generated by [P] and [Q]. Since (5) = R_1R_2, in the class group [R_1][R_2] = [1], so [R_1] and [R_2] are inverses. Thus one of these ideal classes being in the subgp <[P],[Q]> implies both are, so [R_1] and [R_2] can be ignored.2011-05-15
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    Don't make the mistake of saying [R_i]^2 = [1] implies [R_i] has order 2. That only tells you it has order 1 or 2. Once you know it's nonprincipal then it has order 2. But in any case the whole matter is moot since [P] and [Q] generate the class group so you just need to focus on those.2011-05-15
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    quanta: the ideal P has been defined in the question. It is the unique prime ideal factor of (2). It needs no further description to be pinned down.2011-05-15
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    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.)2011-05-15
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    Last comment, Peter: what I wrote wasn't an article. It was a course handout.2011-05-15
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    @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.2011-05-15
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    @Peter: You might consider writing up your answer as an "answer" to the question, eventually even accepting it, since you have managed to work it out. Giving all due thanks to Keith Conrad and sundry, of course. That will keep down the number of unanswered questions on the site.2011-05-15
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    Peter: I wasn't upset or offended.2011-05-16

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