# What is the Strategy in Computing Ideal Class Number?

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:

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?

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.