 Solution

Suppose, for a contradiction, that $P_2\cap \mathbb{Z}=(0)$. Then $P_1\cap \mathbb{Z}=(0)$ too. Both prime ideals are disjoint from the multiplicatively closed set $Y=\mathbb{Z}\setminus \{0\}$. The localization iseach fraction on the left has rational coefficients, and every polynomial with rational coefficients has a common nonzero integer denominator.

By b, the extended prime ideals form a strict inclusionThe first is nonzero, because $P_1$ contains a nonzero polynomial and the map $\mathbb{Z}[t]\to \mathbb{Q}[t]$ is injective.

The polynomial ring $\mathbb{Q}[t]$ is a principal ideal domain: in any nonzero ideal, a nonzero polynomial of least polynomial degree generates the ideal, since the polynomial division algorithm gives a remainder of smaller polynomial degree, which must be zero. By a, every nonzero prime ideal of this principal ideal domain is maximal. The strict inclusion of extended prime ideals is therefore impossible. We obtain