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For a nonzero integer-coefficient polynomial $F$, its polynomial content $c(F)$ is the positive greatest common divisor of all its coefficients; one may set $c(0)=0$. A primitive polynomial has content one.

For $f\in \mathbb{Q}[t]\setminus \{0\}$, choose a positive integer $m$ clearing its denominators, so $F=mf\in \mathbb{Z}[t]$. Put $d=c(F)>0$. Thenwhere $f_1=F/d$ is an integer-coefficient primitive polynomial. This proves existence with $\alpha =d/m>0$.

For uniqueness, suppose $\alpha f_1=\beta g_1$, where both integer-coefficient polynomials are primitive and $\alpha ,\beta >0$. Write $\alpha /\beta =a/b$ in lowest terms with $a,b$ positive. Then $g_1=(a/b)f_1$ has integer coefficients. Coprimality forces $b$ to divide every coefficient of $f_1$, hence $b=1$. Now $g_1=a{f}_1$ has content $a$, so $a=1$. Therefore $\alpha =\beta$ and $f_1=g_1$.

Define the rational polynomial content by this unique positive multiplier. If $f=\alpha f_1$ and $g=\beta g_1$, Gauss lemma for polynomials says that $f_1{g}_1$ is primitive. ThusThe uniqueness argument also shows that the construction is independent of the chosen denominator-clearing integer $m$.

For a nonzero polynomial $f={\sum }_i{a}_i{t}^i\in \mathbb{Z}[t]$, its polynomial content is the positive greatest common divisor of its coefficients:For nonzero $g\in \mathbb{Q}[t]$, choose a positive integer $n$ with $ng\in \mathbb{Z}[t]$ and define its rational polynomial content byTo check independence of the denominator, write $g=f/n=h/m$ with $f,h\in \mathbb{Z}[t]$ and $n,m>0$. Then $mf=nh$, while multiplication of all coefficients by a positive integer givesDividing by $mn$ proves $c(f)/n=c(h)/m$, as required.

For multiplication, write $g_i=f_i/n_i$. The integer-coefficient content identity permitted in the source givesThusWe will also use the normalization $g/c(g)\in \mathbb{Z}[t]$, with content one: if $g=f/n$, then $g/c(g)=f/c(f)$. It is a primitive polynomial. For a nonzero rational constant $r$, the same definition gives $c(r)=|r|$.

The nonzero ideal $\tilde{P}$ is principal because $\mathbb{Q}[t]$ is a principal ideal domain. Choose a nonzero generator $h\in \mathbb{Q}[t]$. Part (a) givesThe nonzero rational scalar $c(h)$ is a unit in $\mathbb{Q}[t]$, so this rescaling does not change the generated principal ideal:Part (b) now places $f$ in $P$, since $f\in \tilde{P}\cap \mathbb{Z}[t]$. Thus $(f{)}_{\mathbb{Z}[t]}\subseteq P$.

For the reverse inclusion, let $g\in P$ be nonzero. The displayed equality supplies $q\in \mathbb{Q}[t]$ with $g=fq$. Multiplicativity of rational polynomial content givesSince $g$ has integer coefficients, $c(g)$ is a positive integer. The normalization from part (a) gives $q/c(q)\in \mathbb{Z}[t]$, so multiplying it by the integer $c(q)=c(g)$ proves $q\in \mathbb{Z}[t]$. Consequently $g\in (f{)}_{\mathbb{Z}[t]}$. The zero polynomial belongs as well, provingThis is the divisibility descent in Gauss lemma for polynomials: dividing an integer-coefficient polynomial by a primitive polynomial cannot introduce a denominator into an exact quotient. Here it yields the description of prime ideals of the integer polynomial ring above zero. The generator has positive polynomial degree, since a primitive constant would be a unit and would put $1$ in $P$.

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.