= Prime ideals of the integer polynomial ring above zero
{title2=$P\cap\mathbb Z=0\Longrightarrow P=0\text{ or }P=(f)\text{ with primitive }f$}
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 $n\ne0$, 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 $P\ne0$, 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>.
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