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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 $c(r)=|r|$.

First extract the polynomial content from $f$:The primitive quartic is Eisenstein at the prime $2$: every nonleading coefficient is divisible by $2$, while the constant term is not divisible by $4$. Eisenstein's criterion makes it irreducible over $\mathbb{Q}$, and Gauss's lemma makes it irreducible in $\mathbb{Z}[X]$. The constant factor $2$ is itself irreducible in $\mathbb{Z}[X]$.

The polynomial $g$ is already irreducible over $\mathbb{Q}$: it is Eisenstein at $3$. Its nonleading coefficients, including the missing zero coefficients, are divisible by $3$, and $-21$ is not divisible by $9$. Thus $g$ is its own sole nonconstant irreducible factor over $\mathbb{Q}$.

Over the field ${\mathbb{Z}}_3$, grouping terms givesThe first two factors are linear and therefore irreducible. The quadratic takes the values $1,2,2$ at $0,1,2$ respectively, so it has no root in ${\mathbb{Z}}_3$. A degree-two polynomial over a field is reducible exactly when it has a root, proving its irreducibility.

All three factorizations are unique up to order and multiplication by units. The ring $\mathbb{Z}$ is a unique factorization domain, hence so is $\mathbb{Z}[X]$ by Gauss's lemma. A polynomial ring over a field is a Euclidean domain, so both $\mathbb{Q}[X]$ and ${\mathbb{Z}}_3[X]$ have unique factorization. The units are $\pm 1$ in the first ring, nonzero rational constants in the second, and $1,2$ in the third.

A nonzero primitive polynomial in $\mathbb{Z}[x]$ has greatest common divisor of its coefficients equal to one. Suppose primitive polynomials $f,g$ had a product whose coefficients were all divisible by some prime $p$. Their reductions in ${\mathbb{F}}_p[x]$ are both nonzero, because neither polynomial has every coefficient divisible by $p$. A polynomial ring over a field is an integral domain, so their product is nonzero modulo $p$, a contradiction. Hence the product of primitive polynomials is primitive.

To deduce unique factorization, write any nonzero integer polynomial as its positive polynomial content times a primitive polynomial, with its sign treated as a unit. The preceding result gives multiplicativity of content. Factor the primitive part over the Euclidean polynomial ring $\mathbb{Q}[x]$, and rescale every irreducible rational factor to a primitive integer polynomial. If $f=q{\prod }_i{f}_i$ with all polynomials primitive and $q=a/b$ in lowest terms, the equality $bf=a{\prod }_i{f}_i$ has contents $b$ and $|a|$. Thus $b=|a|$, forcing $q=\pm 1$. The rational factorization therefore lifts to an integer factorization.

A primitive integer polynomial is irreducible over $\mathbb{Z}$ exactly when it is irreducible over $\mathbb{Q}$: an integer factorization gives a rational one, and a rational factorization lifts by the same primitive-content argument. Factor the content into integer primes. Existence and uniqueness over the integers and over $\mathbb{Q}[x]$ now give existence and uniqueness of the resulting factorization into constant prime factors and primitive irreducible polynomial factors, up to order and signs. Therefore $\boxed{\mathbb{Z}[x]\text{ is a unique factorization domain}}$. This is the full lifting step in Gauss lemma for polynomials.

A unique factorization domain is an integral domain in which every nonzero nonunit is a finite product of irreducible elements, uniquely up to reordering and multiplication of factors by a unit in a ring. A principal ideal domain is an integral domain in which every ideal is generated by one element.

Gauss lemma for polynomials says that over a unique factorization domain $D$, the product of two primitive polynomials is primitive, and the polynomial content satisfies $c(fg)=c(f)c(g)$ up to a unit. In particular a primitive polynomial of positive degree is irreducible over $D$ exactly when it is irreducible over the fraction field of $D$. Consequently a polynomial ring over a unique factorization domain is itself a unique factorization domain.

The Eisenstein criterion says that if $f(X)=a_n{X}^n+\cdots +a_0\in D[X]$ has $n\ge 1$ and there is a prime element $p$ with $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$, then $f$ is irreducible over the fraction field. If $f$ is primitive, it is also irreducible in $D[X]$. Over $\mathbb{Z}$ one may take $p$ to be an ordinary prime number. The primitivity qualification prevents a nonunit constant content from being overlooked.

Eisenstein criterion states that ifhas a prime number $p$ such that $p\nmid a_n$, $p\mid a_i$ for every $i<n$, and $p^2\nmid a_0$, then $f$ is irreducible over $\mathbb{Q}$.

After dividing out the polynomial content, suppose for contradiction that the resulting primitive polynomial factors as $f=gh$ with positive-degree $g,h\in \mathbb{Z}[X]$, using Gauss lemma for polynomials. Reduction modulo $p$ gives $\bar{f}=a_n{X}^n$, so in the polynomial ring ${\mathbb{F}}_p[X]$ both $\bar{g}$ and $\bar{h}$ are monomials. Because both factors have positive degree, their constant coefficients are divisible by $p$. Their product $a_0$ is then divisible by $p^2$, a contradiction.

For a nonzero polynomial $g\in \mathbb{Q}[t]$, choose a positive integer $n$ clearing its denominators and set $c(g)=c(ng)/n$, where the polynomial content of an integer-coefficient polynomial is the positive greatest common divisor of its coefficients. This is independent of $n$: if $f/n=h/m$, then $mf=nh$, hence $mc(f)=nc(h)$.

The integer-coefficient product formula from Gauss lemma for polynomials givesMoreover $g/c(g)$ is an integer-coefficient primitive polynomial: if $g=f/n$, then $g/c(g)=f/c(f)$. For example,This normalization proves a useful divisibility descent. If $f\in \mathbb{Z}[t]$ is primitive, $g\in \mathbb{Z}[t]$ and $g=fq$ for $q\in \mathbb{Q}[t]$, then $c(q)=c(g)$ is an integer. Since $q/c(q)\in \mathbb{Z}[t]$, it follows that $q\in \mathbb{Z}[t]$.