What does it mean that $\mathbb{Q}[x]/(x^2 - x + 15)$ has class number 3?

Question

I found on the internet that $\mathbb{Q}[x]/(x^2 - x + 15)$ has class number 3 (e.g. here). Could someone show me a prime number that factors in three different ways? Or how to generate the answer by computer, possibly with SAGE?

I am sure there are quick answers if I knew more number theory, it's at least implied that we can do this result more explicitly. Maybe we need to compute the ring of integers $\mathcal{O}$ first, the discriminant is $D = (-1)^2 - 4 \times 1 \times 15 = -59$ so the ring of integers should be $\mathbb{Z}[\sqrt{-59}]$ which looks possibly correct.

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The technical definition of class group is less intresting to me than numerical examples. Yet here's my guess:

The definition of the ideal class group is defined as follows (e.g. in these notes): $$ \frac{J_K}{P_K} = \frac{\text{fractional ideals}}{\text{principal fractional ideals}} $$ This is the quotient of two abelian groups, we're saying the quotient group is $\mathbb{Z}/3\mathbb{Z}$.

We seem to be measuring the text to which the multiplicative group $\big(\mathbb{Q}[x]/(x^2 - x + 15)\big)^\times$ can be written as the ratio of two elements n $\mathbb{Z}[\sqrt{-59}]$.

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The textbook example of class number is $\mathbb{Q}(\sqrt{-5})$ and the factorization $6 = 2 \times 3 = (1 + \sqrt{-5})\times (1 - \sqrt{-5}) $ so the ideal $(2, 1 + \sqrt{-5})$ is not principal. So the class number is two.

Also $9 = 3 \times 3 = (2 + \sqrt{-5}) \times (2 - \sqrt{-5})$ and that $(3, 2 + \sqrt{-5})$ is not principal.

$\mathbb{Z}[i] = \mathbb{Q}[x]/(x^2+1)$ is in fact a UFD.

Answer 1

Your field is $\Bbb Q[\sqrt{-59}]$ and the ring of integers is slightly larger than $\Bbb Z[\sqrt{-59}]$: it is $R = \Bbb Z[\frac{1 + \sqrt{-59}}2]$.

Let's write $a$ for $\frac{1 + \sqrt{-59}}2$, so that $a^2 - a + 15 = 0$.

Consider the ideal $(3, a)$ of the ring $R$. An easy calculation shows that

$(3, a)^2 = (9, 3a, a - 15) = (9, a + 3)$;

$(3, a)(9, a + 3) = (27, 3a + 9, 9a, 4a - 15) = (a + 3)$.

Thus we have $(3, a)^3 = (a + 3)$ is a principal ideal.

Now let's show that $(3, a)$ itself is not principal.

If $(3, a) = (c)$ for some $c\in R$, then we have $3\in cR$ and $a \in cR$. Taking norm gives $N(c)\mid N(3) = 9$ and $N(c) \mid N(a) = 15$, so we must have $N(c) = 1$ or $3$.

But $N(c) = 3$ is not possible: if we write $c = u + av$ with $u, v\in \Bbb Z$, then the norm $N(c)$ is equal to $u^2 - uv + 15v^2$. It is easy to show that this cannot be equal to $3$ for any pair of integers $u, v$.

Thus $N(c) = 1$, so $c$ is a unit and we get $(3, a) = (1)$. We then get $(a + 3) = (3, a)^3 = (1)$, contradicting the fact that $N(a + 3) \neq 1$.