Let $K$ be a quadratic number field, assume that the related Minkowski bound is $10$. We know that the class group $Cl(\mathcal{O}_K)$ is generated by the classes of prime ideals $[\mathfrak{p}]$ where $N(\mathfrak{p}) \le 10$. We investigate the structure of $Cl(\mathcal{O}_K)$ by looking at the relations between factorizations of $p\mathcal{O}$'s.
For example, let us say that for some suitable prime $p$, we have the following factorization: $p\mathcal{O}=\mathfrak{p}_1\mathfrak{p}_2$. Then, we can say that $\mathfrak{p}_2$ is in the "inverse class" of $\mathfrak{p}_1$, let us write it as $[\mathfrak{p}_1]^{-1} = [\mathfrak{p}_2]$.
1. What is the relation between the inverse of the ideal $\mathfrak{p}_1$ -which is a fractional ideal - and $[\mathfrak{p}_1]^-1$?
2. Assume that the prime ideals with norm less than $10$ are $\mathfrak{p}_2,\mathfrak{p}_3,\mathfrak{p}_5$ and let us say that they are non-principal. Also, say that $N(\alpha\mathcal{O})=30$. Can we say that the factorization of $\alpha\mathcal{O}$ is exactly $\mathfrak{p}_2\mathfrak{p}_3\mathfrak{p}_5$, even if we do not know whether there exist some other prime ideals with norms $2,3$ or $5$?
• If $p$ splits in $K$, then $\mathfrak{p_1}^{-1} = \frac{1}{p} \cdot \mathfrak{p_2}$. In particular the ideal class of (the inverse ideal) $\mathfrak{p_1}^{-1}$ agrees with the inverse of the ideal class of $\mathfrak{p_1}$. – Brandon Carter Apr 16 '18 at 21:20
• For the second, are you assuming that $\mathfrak{p_2}, \mathfrak{p_3}, \mathfrak{p_5}$ are the only prime ideals of norm less than 10 (i.e. 2 and 3 are inert or ramified and 5 is ramified)? If so, the answer to your question is yes. If not, the answer is no. – Brandon Carter Apr 16 '18 at 21:22