 Solution

The set $\tilde{P}$ contains zero. It is closed under negatives because $-f\in P$ whenever $f\in P$. If $f,g\in P$ and $n,m>0$, thensince $P$ is an ideal of $\mathbb{Z}[t]$. Also, any $q\in \mathbb{Q}[t]$ can be written $q=h/d$ with $h\in \mathbb{Z}[t]$ and $d>0$, givingThis proves that $\tilde{P}$ is an ideal of $\mathbb{Q}[t]$.

Clearly $P\subseteq \tilde{P}\cap \mathbb{Z}[t]$, by taking denominator one. Conversely, if $h\in \tilde{P}\cap \mathbb{Z}[t]$, then $nh\in P$ for some positive integer $n$. The condition $P\cap \mathbb{Z}=\{0\}$ implies $n\notin P$. Since $P$ is a prime ideal, $nh\in P$ forces $h\in P$. HenceIn particular, $\tilde{P}$ is nonzero because it contains the nonzero ideal $P$, and it is proper: otherwise its intersection with $\mathbb{Z}[t]$ would contain $1$. This is the concrete extension of an ideal and contraction of an ideal calculation for the localization that inverts nonzero integers.