Ideal $I\cap \Bbb Z\ne(0)$

Question

What does this mean: $I$ is an ideal of $\Bbb Z[x]$. Let $I\cap \Bbb Z\ne(0)$

What does it mean to take the intersection of an ideal and the integers? All elements of the ideal that are integers? We want the intersection of some ideal(prime here) with the integers to not be the trivial ideal?

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Essentially we are taking some ideal $I\cap \Bbb Z \ne (0)$ and then we are deducing what $I$ looks like.

Answer 1

The ring $\mathbb{Z}$ can be considered as a subset of $\mathbb{Z}[x]$ (the constant polynomials). Of course any ideal of $\mathbb{Z}[x]$ is also a subset of $\mathbb{Z}[x]$. Thus we can form their intersection, and in fact it will be an ideal of $\mathbb{Z}$.

Answer 2

It's a roundabout way of saying that $p$ must be a nonzero integer.

If $p=0$ then of course $(p)\cap\mathbb Z=\{0\}\cap\mathbb Z=\{0\}=(0)$.

And if $p\notin\mathbb Z$, then $p$ has positive degree, and thus every nonzero multiple of $p$ has positive degree too. So the intersection of the principal ideal $(p)$ with $\mathbb Z$ contains only $0$.