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I was working on some examples on how to compute the class number of a quadratic number field: I do understand that for some quadratic number field $\mathbb{Q}(\sqrt{d})$, with $d\in \mathbb{Z}, d \neq \{0,1\}$ and squarefree, I need to compute the Minkowski bound and then look at ideals with norm smaller than the Minkowski bound, which then gives me the class number.

However, I was wondering why I only look at ideals with a prime number as the norm? One example was the number field $K:=\mathbb{Q}(\sqrt{-163})$ (which has class number $1$) where I computed that in every class of the class group, there exists some ideal $\mathfrak{A}$ with $N(\mathfrak{A}) < 8.1$. I could easily compute that there are no ideals with norm $2,3,5$ or $7$ as those are inert in $\mathcal{O}_K$.

Now I'm having some trouble to understand why I only need to look at those ideals and ignore those with norm $4,6$ or $8$ (as was done in the example). I assume that there's maybe some argument working with the factorization of ideals into prime ideals (which works in $\mathcal{O}_K$ as a Dedekind domain). I also looked up other questions on this topic but did unfortunately not find a definite answer.

Thank you for your help and explanations!

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    $\begingroup$ When the degree is $\ge3$, you may need to go farther afield than ideals with prime norm. $\endgroup$ – Lubin Feb 6 '17 at 5:58
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You know that the ideal class group Cl(K) is generated by the classes of prime ideals, due to the existence of factorization of ideals; also, in your example, if an unramified prime has not prime norm, it must have norm the square of a prime, and be principal: if $p\in \mathbb{Z},p>0$ is a rational prime,$p\neq 163$, then p$\mathcal{O}_K=P$ is prime (and principal) or p$\mathcal{O}_K$=$P_1 P_2$, with $P_1,P_2$ distinct prime ideals; in the former case P has norm $p^2$, in the latter both ideals have norm p (due to multiplicativity of the norm). If you check that all ideals with prime norm are trivial in Cl(K), then Cl(K) has to be trivial, as well.

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There is a lot to say :

$\mathcal{O}_K$ is the integral closure of $\mathbb{Z}$ in the finite extension $K/\mathbb{Q}$, so it is a Dedekind domain :

  • every ideal is uniquely a product of prime ideals,

  • if $\mathfrak{p} $ is a prime ideal then $\mathfrak{p} \cap \mathbb{Z} = (p)$ for some prime number $p$, $\mathfrak{p}$ is maximal, $F = \mathcal{O}_K/\mathfrak{p}$ is a field extension of $\mathbb{F}_p = \mathbb{Z}/(p)$.

  • $p \mathcal{O}_K \subset \mathfrak{p}$ and $N(p\mathcal{O}_K)= p^n$ where $n= [K:\mathbb{Q}]$ so that $N(\mathfrak{p}) = p^f $ for some $f\le n$ (if $K/\mathbb{Q}$ is Galois then $f | n$) which is also $[F:\mathbb{F}_p]$, see this

  • In your case $n=2$, so it is enough to check the ideals whose norm is a prime or the square of a prime

  • Any ideal $I$ is generated by two elements : $I = (a,b)$

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  • $\begingroup$ This is just a note for explaining why $p\in \mathfrak{p}$. If $\mathfrak{p}$ is a nonzero prime ideal, then $\mathcal{O}_K/\mathfrak{p}$ is a finite field. Suppose that the order of this field is $p^r$. Then $p+\mathfrak{p}=0+\mathfrak{p}=\mathfrak{p}\in \mathcal{O}_K/\mathfrak{p}$ and $p\in \mathfrak{p}$. $\endgroup$ – bfhaha May 31 '18 at 9:32
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You do not, you only consider prime ideals, which may or may not have prime norm. The reason we only consider the prime ideals is because the ideal group is generated by the primes (that's what unique factorization means!) so if all the generators are principal, any product of them is principal. So it is necessary and sufficient that all prime ideals be principal in order to show the ring is a PID.

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