Prime ideal contains an integer

Question

If $P$ is a non-zero prime ideal inside the ring of integers for some number field $L$, I want to show that $P$ contains a rational integer. My idea was to take some $x$ in $P$ and look at its norm (which is an integer). But I don't know how to show that this norm is in $P$.

Answer 1

Consider the characteristic polynomial $Q_a$ of an element $a$, that is $Q_a(X)= \prod_{i=1}^n(X - \sigma_i(a)) = \sum_{i=0}^n b_i X^i$ where $\sigma_i$ are all embeddings of $L$.

Recall that this is an integral polynomial and its constant coefficient $b_0$ is $(-1)^n$ times the norm of $a$.

On the other hand as $a$ is a root of $Q_a$ one has $-b_0 = \sum_{i=1}^n b_i a^i = a( \sum_{i=0}^{n-1} b_i a^{i-1}) $.

In particular in the ring of algebraic integers the norm of $a$ is a multiple of $a$; recall that the $b_i$ are all rational integers.

Now if $a$ is in an ideal $P$, then every multiple of $a$ is also in $P$; thus $b_0$ and hence $(-1)^n b_0$, that is the norm of $a$, are also in $P$.

I believe you do not need that the ideal is prime.

The other thing is that it always contains $0$, so literally as asked the question has an easier answer. In fact for prime ideals one can show that the contain exactly one prime number. Maybe this is what you are ultimately after.