If I have two prime ideals $\mathfrak{p}$ and $\mathfrak{q}$ with $\mathfrak{p} = (a)\mathfrak{q}$ where $(a)$ is a principal fractional ideal (that is, we can have $a$ not necessarily in our ring of integers), is the ideal class of $\mathfrak{p}$ the same as that of $\mathfrak{q}$?


Yes. Let $K$ be a number field with ring of integers $\mathcal{O}_K.$ Then the class group of $K,$ denoted $Cl(K),$ sits in exact sequence

$$0 \rightarrow \mathcal{O}_K^{\times} \rightarrow K^{\times} \rightarrow I_K \rightarrow Cl(K) \rightarrow 0,$$

where $I_K$ is the group of fractional ideals of $K$ and the map $K^{\times} \rightarrow I_K$ is given by sending $\alpha$ to the fractional ideal $\alpha\mathcal{O}_K.$ So two fractional ideals $\mathfrak{p}$ and $\mathfrak{q}$ are equivalent if and only if they satisfy $\mathfrak{p} = (a)\mathfrak{q}$ for $a \in K^{\times}.$

  • 1
    $\begingroup$ This answer is rather contentless. $\endgroup$ – jspecter Nov 16 '11 at 5:20

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