I am attempting to compute the ideal class group of the real quadratic field $K = \Bbb Q(\sqrt{65})$, which has ring of integers $\mathcal{O}_K = \Bbb Z\left[\frac{1 + \sqrt{65}}{2}\right]$.

The discriminant $d_K$ of $K$ is

\begin{align} d_K &= \begin{vmatrix} 1 &\frac{1 + \sqrt{65}}{2}\\ 1 &\frac{1 - \sqrt{65}}{2}\end{vmatrix}^2\\ &= 65 \end{align}

and, as a real quadratic field, there are no complex embeddings so the Minkowski bound $M_K$ is

\begin{align} M_K &= \frac{2!}{2^2}\left( \frac{2}{\pi}\right)^0 \sqrt{65}\\ &\approx 4.03 \end{align}

so the ideal class group is generated by primes lying above $2$ and $3$. Checking the factorisation of $X^2 - X - 16 \bmod p$ for $p = 2, 3$ we get

$$X^2 - X - 16 \equiv X(X+1) \bmod 2\\ X^2 - X - 16 \equiv X^2 + 2X + 2 \bmod 3$$

so $2$ splits and $3$ is inert, so that $(2) = \mathfrak{p}_2\bar{\mathfrak{p}}_2$ and $(3) = \mathfrak{p}_3$ with $N(\mathfrak{p}_2) = N(\bar{\mathfrak{p}}_2) = 2$ and $N(\mathfrak{p}_3) = 9$.

Since $\mathfrak{p}_3$ is principal, it does not generate the ideal class group so we need only think about $\mathfrak{p}_2$.

What follows might seem like a silly quesion so brace yourselves!

If the norm equation $a^2 + ab - 16b^2 = 2$ has solutions in the integers, should it have solutions modulo every prime?

I ask because the equation has no solutions modulo $5$, and this is perhaps a way of showing that there is no element of norm $2$ and so $\mathfrak{p}_2$ cannot be principal (the same computation shows there are no elements of norm $-2$, so this is also covered).

The confusion stems from the fact that the book I'm using quotes the ideal class $[(5, \sqrt{65})]$ as a generator for the class group, though this ideal has norm $5 > 4.03$, hence the confusion. In addition, following an argument by Keith Conrad in his expository papers, it seems that since $(5) = \mathfrak{p}_5^2$, if $\mathfrak{p}_5$ were principal we should be able to write

$$5 = (a + b\sqrt{65})^2 u$$

for some $u \in \mathcal{O}_K^\times$ and $a, b \in \frac{1}{2}\Bbb Z$. The units of $\mathcal{O}_K$ look like $\pm(8 + \sqrt{65})^n$ for $n \in \Bbb Z$, and since $N(8 + \sqrt{65}) = -1$, the units of norm $1$ have the form $(8 + \sqrt{65})^{2k}$ for some $k \in \Bbb Z$. In particular, this means that the $u$ term in the above equation can be absorbed into the $(a+b\sqrt{65})^2$ term, but this implies that $\sqrt{5} \in \mathcal{O}_K$, and so $\mathfrak{p}_5$ is not principal, which seems to contradict the Minkowski bound.

  • $\begingroup$ How do you get from $(5)=p_5^2$ to $5=(a+b\sqrt{65})^2u$? $\endgroup$ – Lord Shark the Unknown Mar 2 '18 at 13:01
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    $\begingroup$ I can't verify your calculation of the unit group -- where there must be a problem as you point out, since $\sqrt 5$ surely is not in the field - but on the other points: if an integer equation has a solution it has a solution mod n for all n. Second, the Minkowski bound does not say that all ideals of larger norm are principal, just where you can find representatives for each class. $(2, \frac{1+\sqrt{65}}{2})$ is non-principal but its square is. So the class group is of order $2$ and generated by any non-principal ideal. $\endgroup$ – John Brevik Mar 2 '18 at 14:01
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    $\begingroup$ @JohnBrevik Wonderful, so the fact that $a^2 + ab - 16b^2 = 2$ has no solutions mod $5$ (and the same for $-2$) shows that my $\mathfrak{p}_2$ is non-principal, right? Thanks a lot. $\endgroup$ – ÍgjøgnumMeg Mar 2 '18 at 14:04
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    $\begingroup$ Yes, that's right. It's a valid and useful line of reasoning! $\endgroup$ – John Brevik Mar 2 '18 at 14:20
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    $\begingroup$ @JohnBrevik Perfect, in that case I think the problem is sorted. I thought that $(5, \sqrt{65})$ had to be principal because it has norm greater than the Minkowski bound. The fundamental unit I quoted is correct, so this argument appears to work to show that $(5, \sqrt{65})$ is non-principal, and so from what you say, I can take it as a representative for the ideal class of order $2$ in the class group. Thanks! $\endgroup$ – ÍgjøgnumMeg Mar 2 '18 at 14:23

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