The possible prime divisors of an ideal $J$ (whose norm is no more than 5) are necessarily among the primes lying over 2,3,5 To show that $R=\mathbb Z[\sqrt2]$ is a PID, one way is to show that it has class number $1$. By Minkowski bound estimate, we know every ideal class contains an ideal $J$ with norm no greater than $5$. I don't know to verify the following claim (my question comes from Marcus' Number Fields, page 132):

The possible prime divisors of $J$ are necessarily among the primes lying over $2,3,5$.

I guess this statement must be related to the definition of the norm of an ideal:
$$N(J)=|R/J|.$$
But I have no idea how to prove it. Thanks in advance!
 A: Write $K = \mathbb{Q}(\sqrt{2})$, $\mathcal{O}_K = \Bbb Z[\sqrt{2}]$. If $\frak{p}$ is a (nonzero) prime dividing $J$, write $\mathfrak{p}\mathfrak{a} = J$. Then
\begin{align*}
5&\geq N(J)\\
&= N(\mathfrak{p}\mathfrak{a})\\
&= N(\mathfrak{p}) N(\mathfrak{a}).
\end{align*}
$\mathfrak{p}$ is not the unit ideal, so $N(\mathfrak{p}) = 2, 3, 4,$ or $5$.
Now, a prime ideal of $\mathcal{O}_K$ lies over a prime ideal of $\Bbb Z$; i.e., $\mathfrak{p}\cap\Bbb Z = (p)$ for some prime $p$ in $\Bbb Z$. If $p$ is prime in $\Bbb Z$, then $N((p)) = \left|N(p)\right| = \left|(p + 0\sqrt{2})(p - 0\sqrt{2})\right| = p^2$ But if $\mathfrak{p}$ lies over $(p)$, then $\mathfrak{p}\mathfrak{b} = (p)$ for some ideal $\mathfrak{b}$ of $\mathcal{O}_K$. Then we have
\begin{align*}
p^2 &= N(\mathfrak{p}\mathfrak{b})\\
&= N(\mathfrak{p})N(\mathfrak{b}),
\end{align*}
which implies that $N(\mathfrak{p})$ divides $p^2$ if it lies over $p$.
Combining all these above, it follows that because $N(\mathfrak{p})$ can only be $2,3,4,$ or $5$, and $N(\mathfrak{p})$ divides $p^2$ if it lies over $p$, $\mathfrak{p}$ lies over $(2)$, $(3)$, or $(5)$.
