Let $K$ be a number field and $\mathcal{O}_K$ its ring of integers. Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$, and $I$ be an ideal of $\mathcal{O}_K$ such that $N(I) = N(\mathfrak{p})$, where $N(\cdot)$ is the ideal norm (i.e., $N(I) = |\mathcal{O}_K/I|$). Is it true that:
(a) $I$ is a prime ideal;
(b) $[I]$ and $[\mathfrak{p}]$ have the same order in the ideal class group?
Any help appreciated.
Best Answer
The following is perhaps too long for a comment, so I'll put it in the answer. By looking at LMFDB I found that the cubic field $\mathbb{Q}(\sqrt[3]{11})$ has class number $2$. The Minkowski bound is $M = \dfrac{4}{\pi}\cdot \dfrac{6}{27} \sqrt{3267} = \dfrac{88}{\sqrt{3}\pi}<17$, so we need to consider the factorizations of $(p)$ for $p=2,3,5,7,11,13$ in $\mathcal{O}_{\mathbb{Q}(\sqrt[3]{11})}$. The LMFDB page shows that $7$ and $13$ remains inert in $\mathbb{Q}(\sqrt[3]{11})$. $3$ and $11$ factor into prime elements in $\mathcal{O}_{\mathbb{Q}(\sqrt[3]{11})}$: $3 = (-2+\sqrt[3]{11})^2(20+9\sqrt[3]{11}+4\sqrt[3]{121}), 11=(\sqrt[3]{11})^3$. The factorizations of $(2)$ and $(5)$ are $$(2) = (2,5-\sqrt[3]{121})(2,25+11\sqrt[3]{11}+5\sqrt[3]{121})$$ $$(5) = (5,9-4\sqrt[3]{11})(5,81+36\sqrt[3]{11}+16\sqrt[3]{121})$$
The ideal $(2,5-\sqrt[3]{121})$ has norm $2$ and $(5,9-4\sqrt[3]{11})$ has norm $5$, so they are prime ideals. Since $p=2,5$ have Frobenius cycle type $2,1$, the other factors are also prime ideals. $(2,5-\sqrt[3]{121})$ is not principal: $(2,5-\sqrt[3]{121})(5,9-4\sqrt[3]{11}) = (-1+\sqrt[3]{11})$ is principal, if $(2,5-\sqrt[3]{121})$ is principal then $(2)$ and $(5)$ both factor into principal ideals, which would imply that $\mathbb{Q}(\sqrt[3]{11})$ has class number $1$, a contradiction.
As a result, if we pick $I = (5-\sqrt[3]{121})$ and $\mathfrak{p} = (2,25+11\sqrt[3]{11}+5\sqrt[3]{121})$, then $N(I)=N(\mathfrak{p}) = 4$, and $I$ is not a prime ideal; what's more, $I$ is principal while $\mathfrak{p}$ is not. (The same happens for $I = (9-4\sqrt[3]{11})$ and $\mathfrak{p} = (5,81+36\sqrt[3]{11}+16\sqrt[3]{121})$.)