Every ideal $I\not=0$ of $\mathcal{O}_K$ contains an integer $m\in\mathbb{Z}$ distinct from zero.

Question

Suppose $K$ is a number field of degree $n$. How could we prove that every ideal $I\not=0$ in the ring of integers $\mathcal{O}_K$ contains an integer different from zero?

We know that $I$ contains some element different from zero, say $x\in I$. I think we should consider, for instance, the norm of this element in $K$, for it is an integer number. Unfortunately, it is not clear at all that it could be obtained via a finite number of operations of elements of $I$ and $K$, hence lie in $I$; indeed, the norm of $x$ is the product of the roots of the characteristic polinomial, but those possibly could lie outside of $K$.

We could also consider the discriminant $\Delta(x,x^2,\dots,x^n)$, or other variants, which also lie in $\mathbb{Z}$, however, I am concerned that it could be equal to zero, and if is not clear that the trace is an element of the ideal, it should be way more difficult to check that the discriminant is.

I have found this post in which an accepted answer is given. However, it has a flaw, as the first comment suggests.

I don't know which element should be picked, and how could we prove it is an element of $I$.

Thanks in advance for your answers.

Answer 1

Let me clarify the linked answer:

Take $x \in I$ integral and nonzero. We claim that $N(x)/x \in \mathcal O_K$. Take an embedding of $K$ into a field $L$ that contains all conjugates of $x$. On the one hand, $N(x) / x \in \mathcal O_L$ because it is a product of conjugates of $x$. On the other hand, $N(x) / x \in K$. Because $\mathcal O_K$ is integrally closed, $\mathcal O_L \cap K = \mathcal O_K$, so that $N(x) / x \in \mathcal O_K$.

Thus the nonzero integer $N(x) = x \cdot N(x)/x$ lies in $I$.

Answer 2

If $x\ne0$, then its minimal polynomial $X^n+a_{n-1}X^{n-1}+\cdots+a_1 X + a_0$ has integer coefficients and $a_0\ne 0$. Therefore, $$ a_0 = -(x^n+a_{n-1}x^{n-1}+\cdots+a_1 x) \in I \cap \mathbb Z^ * $$