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My first step in computing the class group of $\mathbb Q(\sqrt{-55})$ was to compute the Minkowski bound. Initially, I said $\lambda(-55)=2\sqrt{-55}/\pi<2(8)/3<6$ and I went the normal way of finding an ideal of norm 5, $(5, \sqrt{-55})$ and this has a contribution to the class group since it is non-principal. But then I realised that I could get a better bound on $\lambda(-55)=2\sqrt{-55}/\pi<2(7.5)/3=5$. So in fact, I should have got no contribution from the ideal $(5, \sqrt{-55})$ and should have got that this ideal is in the same class as an ideal of norm $<5$.

I don't see how to get an ideal of norm $<5$ from $(5, \sqrt{-55})$ and I'm not sure why my argument above was wrong.

In more complicated situations, I'm worried that by making a mistake similar to the above, I would get much larger class groups than is correct. How do I avoid this mistake.

Is there a way of spotting if an ideal is in the same ideal class as an integral ideal of smaller norm, in general?

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  • $\begingroup$ for imaginary quadratic fields use the method of quadratic forms - it gives all you need $\endgroup$ – user8268 May 18 '15 at 11:06
  • $\begingroup$ @user8268 I don't know what quadratic forms are. I just looked up a definition on Wikipedia, but I don't see how they're helpful. $\endgroup$ – James May 18 '15 at 11:08
  • $\begingroup$ @user8268 and James math.blogoverflow.com/2014/08/23/… $\endgroup$ – Will Jagy May 18 '15 at 17:49
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The first thing to note is that ideal classes are different from ideals, we say $I$ is in the class of $J$ if $\exists (a),(b)$ with $(a)I=(b)J$. With this in mind, consider $(4\sqrt{-55})(5,\sqrt{-55})=(20\sqrt{-55},-220)=(\sqrt{-55})$. Now consider the ideal $(2,\frac{1+\sqrt{-55}}{2})$. I claim this is in the same class as $(5,\sqrt{-55})$.

Indeed, consider $$(10\sqrt{-55})(2,\frac{1+\sqrt{-55}}{2})=(20\sqrt{-55},5\sqrt{-55}+275)=(\sqrt{-55})$$

This answers your original question.

To get an ideal of norm $<5$, you want to consider ideals with norm $2$ or $3$. The way we do this is to determine whether $(2) , (3)$ are ramified in $K=\mathbb{Q}(\sqrt{-55})$.

Since $-55\equiv 1\pmod{4}$ we know that Disc($K)=-55$, and since neither $2$ nor $3$ divides 55, we see that neither prime is ramified. Thus, it remains to decide whether they are split completely or inert.

$(3)$ remains a prime in $K$, since $3$ is not a quadratic residue modulo $-55$, so we can discount ideals of norm $3$.

$(2)\equiv1\pmod{8}$, so we know $(2)$ splits completely in $K$. The ideal I=$(2,\frac{1+\sqrt{-55}}{2})$ is a prime ideal of norm $2$, since $(2,\frac{1+\sqrt{-55}}{2})(2,\frac{1-\sqrt{-55}}{2})=(2)$. A computation shows that $I^2$ is not principal, and indeed, that $I^2\neq \bar I$. Hence, $I$ is not of order $3$. We compute either $I^3$, to give us $\bar I $, or $(I^2)^2$, to give us a principal ideal. Either way, we see that $I$ has order 4, and the class group is henceforth $\mathbb{Z}/4\mathbb{Z}$.

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  • $\begingroup$ How do you get $(20\sqrt{-55},-220)=(\sqrt{-55})$? $\endgroup$ – James May 18 '15 at 14:08
  • $\begingroup$ I've made a mistake in my answer actually, it should be $(4\sqrt{-55})$. However, for an arbitrary ring, $R$, consider the ideal generated by $(f,g)$, where $f$ and $g$ are NOT coprime in $R$. What can this also be written as? $\endgroup$ – Ramified_Minds May 18 '15 at 14:49
  • $\begingroup$ In fact I think the first bit of my answer is slightly sketchy as a whole. However, you don't actually need to explicitly find the ideal that is in the class of $(5,\sqrt{-55})$ to compute the class group $\endgroup$ – Ramified_Minds May 18 '15 at 15:06
  • $\begingroup$ Regarding your first comment, I'm not sure really. If $d$ is a common divisor of $f, \, g$ then $(f,g)\subseteq (d)$ but I'm not sure what $(f,g)$ equals unless I somehow know that this is principal. $\endgroup$ – James May 18 '15 at 15:23

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