let \(\displaystyle 0 \neq a \in P\) and suppose that \(\displaystyle f(x)=x^n + c_1x^{n-1} + \cdots + c_n\) is the minimal polynomial of \(\displaystyle f(x)\) over \(\displaystyle \mathbb{Q}.\) it's a known fact that, since \(\displaystyle a \in \mathcal{O}_K,\) we have \(\displaystyle c_j \in \mathbb{Z},\) for all \(\displaystyle j.\)

now if \(\displaystyle n=1,\) then since \(\displaystyle 0=f(a)=a+c_1,\) we have \(\displaystyle a=-c_1 \in \mathbb{Z}\) and so \(\displaystyle 0 \neq a \in P \cap \mathbb{Z}.\) if \(\displaystyle n \geq 2,\) then \(\displaystyle c_n \neq 0\) because of minimality of \(\displaystyle f(x).\) thus \(\displaystyle f(a)=0\) gives us

\(\displaystyle c_n=-(a^n + c_1a^{n-1} + \cdots + c_{n-1}a) \in P\) and so \(\displaystyle 0 \neq c_n \in P \cap \mathbb{Z}.\)

**Note**: in the proof we didn't need \(\displaystyle P\) to be a prime ideal. so the above result holds for any non-zero ideal \(\displaystyle P\) of \(\displaystyle \mathcal{O}_K.\)