Class group of $\mathbb Q(\sqrt{-55})$ and finding representatives for ideal classes

Question

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?

Answer 1

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 $\mathbb{Z}/4\mathbb{Z}$.

Answer 2

Alternatively: since $(3\pm\sqrt{-55})/2$ has norm $2^4$, the smallest power of $2$ that can occur as a norm of an imaginary element in the ring, we have $(2)=P^n(\overline{P})^{4-n}$ for $n\in\{0,1,2,3,4\}$; but if both exponents were greater than zero then $2$ would have a proper divisor in the ring which is false. So $(2)=P^4$ or $(2)=(\overline{P})^4$, and either way a four-cycle $\mathbb{Z}/\mathbb{4Z}$ is generated. Since no other ideal below the Minkowski bound is available, this four-cycle is the entire class group.

How would ideals with norm $5$ fit into this class group? First off, $(5,\sqrt{-55})$ is actually not prime in this ring. It would be so if the ring consisted only of $\mathbb{Z}[\sqrt{-55}]$ integers, but that is false when the radicand in a quadratic field is $\equiv1\bmod4$. The actual prime ideals of norm $5$ take the form

$Q_k=(5,\frac12(5+k\sqrt{-55})),$

where $k$ is an odd integer not divisible by $5$. $Q_k$ is identical with $Q_{-k}$ and $Q_{10\pm k}$, so only $Q_1$ and $Q_3$ are actually distinct. We then find that

$Q_1^2=Q_3^2=(5,\sqrt{-55})$

$Q_1^3=Q_3, Q_3^3=Q_1$ (cubing the ideal multiplies $k$ by $3$, but then $Q_9$ reduces to $Q_1$ from the equalities given above)

$Q_k^4=(5,\sqrt{-55})^2=(5)$

These relationships give the same four-cycle as we found with the norm-$2$ ideals.