Integral ideals intersected with $\mathbb Z$ Let $I \subseteq \mathcal O_K$ be an integral ideal of a number field $K$ such that $p \nmid I$ for all primes $p \in \mathbb N$. I want to show that $I \cap \mathbb Z = N(I) \mathbb Z$.
It's clear that we have $N(I) \mathbb Z \subseteq I \cap \mathbb Z$ because $N(I) \in I$. But I don't know how to show $N(I) \mid m$ for all $m \in \mathbb Z \cap I$.
Edit: Since Stefan4024 showed that the statement isn't true in general let's make it a little less general: Does the statement hold in quadratic number fields?
 A: I'm not sure the other inclusion holds.
For example let $K=\mathbb{Q}\left(\sqrt[3]{2}\right)$. Then we have that $\mathcal O_K = \mathbb Z\left[\sqrt[3]{2}\right]$. Now consider how $5\mathcal O_K$ factors into ideals. This can be done by factorzing $x^3-2 \in \mathbb{F}_5[x]$, which factors as $(x+2)(x^2 + 3x + 4)$. Thus:
$$5\mathcal O_K = \left(5,\sqrt[3]{2}+2\right)\left(5,\sqrt[3]{4}+3\sqrt[3]{2} + 4\right) = \mathfrak p_1 \mathfrak p_2$$
Take $I = \mathfrak p_2$. It's not hard to conclude that $N(I) = 25$. Also no prime $p$ divides the ideal $I$, as all $p\mathcal O_K$ have cube norm. On the other side we have that $p_2 \cap \mathbb{Z} = 5\mathbb{Z}  \not = 25\mathbb{Z}$
Hence the proof.

[UPDATE]: However the claim is true in the quadratic case. Let $I = \mathfrak q_1^{e_1} \cdots \mathfrak q_s^{e_s}$ be the prime ideal factorization of $I$. Obviously $\mathfrak q_i$ can't remain inert over $q_i = \mathfrak q_i \cap \mathbb Z$, as then $q_i \mid I$, which can't be the case. Therefore all prime ideals have prime norm. Also if it ramifies for the same reasons the exponent $e_i$ must be $1$. Finally if it splits, then the other factor can't be included in the factorization. Thus no prime ideal factors have same norm and $N(I) = q_1^{e_1}\cdots q_s^{e_s}$.
Now let $m \in I \cap \mathbb{Z}$. Then we have that $I \mid m\mathcal O_K$ and so $m \mathcal O_K = \mathfrak q_1^{e_1} \cdots \mathfrak q_s^{e_s}\mathfrak p_1^{k_1} \cdots \mathfrak p_r^{k_r}$ for some prime ideals $\mathfrak p_i$. Now $\mathfrak q_1 \mid m\mathcal O_K$ and so $q_1 = \mathfrak q_1 \cap \mathbb{Z} \mid m$. If $q_1$ ramifies in $\mathcal O_K$ we get that $e_1 = 1$ and so $N(I)$ can't have greater $q_1$ exponent than $m$. If $q_1$ splits, WLOG let the "pair ideal" of $\mathfrak q_1$ be $\mathfrak p_1$ (Note that by above assertation it has to be one of the $\mathfrak p_i$'s). Then $\left( \frac{m}{q_1} \right) \mathcal O_K = \mathfrak q_1^{e_1-1} \cdots \mathfrak q_s^{e_s}\mathfrak p_1^{k_1-1} \cdots \mathfrak p_r^{k_r}$. If $e_1 \not = 1$, then repeat the same argument till you get rid of $\mathfrak q_1$ factor and conclude that $N(I)$ can't have greater $q_i$ exponent than $m$. Finally do this for all other factors of $N(I)$ to get that $N(I) \mid m$ and $I \cap \mathbb{Z} = N(I)\mathbb{Z}$
