I found many examples on computing ideal class numbers, but none gave an explicit statement on what we are examining when we are running through a list of elements with their norms written out.

The following came from Keith Conrad; compute the class number of $K = \mathbb Q(\sqrt {79})$. By Minkowski's bound, we need to examine the prime ideals generating $(2), (3), (5), (7)$. Then, he proceeded as follows:

Source:  Keith Conrad Blurbs

Since $(9 + \sqrt {79}) = (2, 1 + \sqrt {79})$, $\mathfrak p_2 \sim (1)$, so it won't help generating non principal ideals. Then, from $a \in \{ \, 8, 10 \, \}$, he concluded that $\mathfrak p_5 \sim \mathfrak p_7 \sim \mathfrak p_3^{-1}$, adding that the class group is generated by $[ \mathfrak p_3 ]$. I suppose that he meant that $\mathfrak p_5 = c_1 \mathfrak p_3$ and $\mathfrak p_7 = c_2 \mathfrak p_3$ for the appropriate $c_1, c_2 \in K^\times$. How do I verify this without looking for $c_1, c_2$?

With $a = 5$, we have $\mathfrak p_3^3 \sim (1)$. After confirming that $\mathfrak p_3$ is not principal, we conclude that the class number is $3$. Well, why did we have to go this route? I at first used $a \in \{ \, 3, 7 \, \}$ to obtain $\mathfrak p_3 \sim \mathfrak p_7$. Then, with $a = 10$, I wrote $\mathfrak p_3^2 \sim (1) \rightarrow$ the class number is $2$. Now, why is that wrong?


By Minkowski, every ideal class contains an ideal with small norm, where small is made precise e.g. by the Minkowski bound. The ideals with small norm generate the class group, and by multiplicativity, so do the prime ideals with small norm.

Thus for computing the class group, you a) write down all prime ideals of small norm b) determine all relations among them, i.e. find which products of powers of these prime ideals are principal. The class group is the free group on these prime ideals modulo the subgroup generated by these relations.

Your error, by the way, consists in not distinguishing prime ideals and their conjugates. If there is an element of order $pq$, then either ${\mathfrak p} {\mathfrak q}$ or ${\mathfrak p} {\mathfrak q}^{-1}$ is principal.


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