 Prime ideals of the integer polynomial ring above zero

The prime ideals $P$ of $\mathbb{Z}[t]$ with $P\cap \mathbb{Z}=\{0\}$ are the zero ideal and the principal ideals $(f)$ generated by nonconstant primitive polynomials that are irreducible in $\mathbb{Q}[t]$.

To prove the description, invert the nonzero integers to obtain the localization $\mathbb{Q}[t]$. The extension of an ideal $P$ consists of the polynomials $f/n$ with $f\in P$ and $n>0$. Its contraction is $P$: if $nh\in P$ and $n\ne 0$, the defining condition puts $n\notin P$, so primeness gives $h\in P$. The prime ideal correspondence for localization makes this extension a proper prime ideal of $\mathbb{Q}[t]$.

If $P\ne 0$, its extension has a nonzero generator because $\mathbb{Q}[t]$ is a principal ideal domain. Dividing this generator by its rational polynomial content gives an integer-coefficient primitive polynomial $f$ generating the same extended ideal. Divisibility descent by Gauss lemma for polynomials then gives $P=(f)$ in $\mathbb{Z}[t]$. The proper prime ideal $(f)$ in $\mathbb{Q}[t]$ makes $f$ irreducible there. Conversely, any such polynomial generates a prime ideal in $\mathbb{Q}[t]$ whose contraction is $(f)\subset \mathbb{Z}[t]$ and whose intersection with $\mathbb{Z}$ is zero.

The intersection condition is essential: $(2,t)$ is a maximal ideal of $\mathbb{Z}[t]$, since its quotient ring is ${\mathbb{F}}_2$, but it is not principal. A common divisor of $2$ and $t$ must be a unit, which would generate the whole ring.