A polynomial ring over a UFD is a UFD.

Gauss's lemma holds more generally over arbitrary unique factorization domains. There the content $c (P)$ of a polynomial $P$ can be defined as the greatest common divisor of the coefficients of $P$ (like the gcd, the content is actually a set of associate elements). A polynomial $P$ with coefficients in a UFD $R$ is then said to be primitive if the only elements of $R$ that divide all coefficients of $P$ at once are the invertible elements of $R$ ; i.e., the gcd of the coefficients is one.

Primitivity statement: If $R$ is a UFD, then the set of primitive polynomials in $R [X]$ is closed under multiplication. More generally, the content of a product $fg$ of polynomials is the product $c(f)c(g)$ of their individual contents.

Irreducibility statement: Let $R$ be a unique factorization domain and $F$ its field of fractions. A non-constant polynomial $f$ in $R[x]$ is irreducible in $R[x]$ if and only if it is both irreducible in $F[x]$ and primitive in $R[x]$ .

(For the proofs, see #General version below.)

Let $R$ be a unique factorization domain with field of fractions $F$ . If $f\in F[x]$ is a polynomial over $F$ then for some $d$ in $R$ , $df$ has coefficients in $R$ , and so – factoring out the gcd $q$ of the coefficients – we can write $df=qf'$ for some primitive polynomial $f'\in R[x]$ . As one can check, this polynomial $f'$ is unique up to the multiplication by a unit and is called the primitive part (or primitive representative) of $f$ and is denoted by $\operatorname {pp} (f)$ . The procedure is compatible with product: $\operatorname {pp} (fg)=\operatorname {pp} (f)\operatorname {pp} (g)$ .

The construct can be used to show the statement:

Indeed, by induction, it is enough to show $R[x]$ is a UFD when $R$ is a UFD. Let $f\in R[x]$ be a non-zero polynomial. Now, $F[x]$ is a unique factorization domain (since it is a principal ideal domain) and so, as a polynomial in $F[x]$ , $f$ can be factorized as:

where $g_{i}$ are irreducible polynomials of $F[x]$ . Now, we write $f=cf'$ for the gcd $c$ of the coefficients of $f$ (and $f'$ is the primitive part) and then:

Now, $c$ is a product of prime elements of $R$ (since $R$ is a UFD) and a prime element of $R$ is a prime element of $R[x]$ , as $R[x]/(p)\cong R/(p)[x]$ is an integral domain. Hence, $c$ admits a prime factorization (or a unique factorization into irreducibles). Next, observe that $f'=\operatorname {pp} (g_{1})\cdots \operatorname {pp} (g_{r})$ is a unique factorization into irreducible elements of $R[x]$ , as (1) each $\operatorname {pp} (g_{i})$ is irreducible by the irreducibility statement and (2) it is unique since the factorization of $f'$ can also be viewed as a factorization in $F[x]$ and factorization there is unique. Since $c$ and $f'$ are uniquely determined by $f$ up to unit elements, the above factorization of $f$ is a unique factorization into irreducible elements. $\square$

The condition that "R is a unique factorization domain" is not superfluous because it implies that every irreducible element of this ring is also a prime element, which in turn implies that every non-zero element of R has at most one factorization into a product of irreducible elements and a unit up to order and associate relationship. In a ring where factorization is not unique, say $pa = qb$ with p and q irreducible elements that do not divide any of the factors on the other side, the product $(p + qX)(a + qX) = pa + (p + a) qX + q 2 X 2 = q (b + (p + a) X + qX 2)$ shows the failure of the primitivity statement. For a concrete example one can take $R = Z [i \sqrt5]$ , $p = 1 + i \sqrt5$ , $a = 1 - i \sqrt5$ , $q = 2$ , $b = 3$ . In this example the polynomial $3 + 2 X + 2 X 2$ (obtained by dividing the right hand side by $q = 2$ ) provides an example of the failure of the irreducibility statement (it is irreducible over R, but reducible over its field of fractions $Q [i \sqrt5]$ ). Another well-known example is the polynomial $X 2 - X - 1$ , whose roots are the golden ratio $φ = (1 + \sqrt5)/2$ and its conjugate $(1 - \sqrt5)/2$ showing that it is reducible over the field $Q [\sqrt5]$ , although it is irreducible over the non-UFD $Z [\sqrt5]$ which has $Q [\sqrt5]$ as field of fractions. In the latter example the ring can be made into an UFD by taking its integral closure $Z [φ]$ in $Q [\sqrt5]$ (the ring of Dirichlet integers), over which $X 2 - X - 1$ becomes reducible, but in the former example R is already integrally closed.

Remark: Over a GCD domain (e.g., a unique factorization domain), the gcd of all the coefficients of a polynomial $f$ , unique up to unit elements, is also called the content of $f$ .

Let $R$ be a commutative ring. If $f$ is a polynomial in $R[x_{1},\dots ,x_{n}]$ , then we write $\operatorname {cont} (f)$ for the ideal of $R$ generated by all the coefficients of $f$ ; it is called the content of $f$ . Note that $\operatorname {cont} (af)=a\operatorname {cont} (f)$ for each $a$ in $R$ . The next proposition states a more substantial property.

A polynomial $f$ is said to be primitive if $\operatorname {cont} (f)$ is the unit ideal $(1)$ .^[4] When $R=\mathbb {Z}$ (or more generally when $R$ is a Bézout domain), this agrees with the usual definition of a primitive polynomial. (But if $R$ is only a UFD, this definition is inconsistent with the definition of primitivity in #Statements for unique factorization domains.)

Proof: This is easy using the fact^[5] that ${\sqrt {I}}=(1)$ implies $I=(1).$ $\square$

Proof: ( $\Rightarrow$ ) First note that the gcd of the coefficients of $f$ is 1 since, otherwise, we can factor out some element $c\in R$ from the coefficients of $f$ to write $f=cf'$ , contradicting the irreducibility of $f$ . Next, suppose $f=gh$ for some non-constant polynomials $g,h$ in $F[x]$ . Then, for some $d\in R$ , the polynomial $dg$ has coefficients in $R$ and so, by factoring out the gcd $q$ of the coefficients, we write $dg=qg'$ . Do the same for $h$ and we can write $f=cg'h'$ for some $c\in F$ . Now, let $c=a/b$ for some $a,b\in R$ . Then $bf=ag'h'$ . From this, using the proposition, we get:

That is, $b$ divides $a$ . Thus, $c\in R$ and then the factorization $f=cg'h'$ constitutes a contradiction to the irreducibility of $f$ .

( $\Leftarrow$ ) If $f$ is irreducible over $F$ , then either it is irreducible over $R$ or it contains a constant polynomial as a factor, the second possibility is ruled out by the assumption. $\square$

Proof of the proposition: Clearly, $\operatorname {cont} (fg)\subset \operatorname {cont} (f)\operatorname {cont} (g)$ . If ${\mathfrak {p}}$ is a prime ideal containing $\operatorname {cont} (fg)$ , then $fg\equiv 0$ modulo ${\mathfrak {p}}$ . Since $R/{\mathfrak {p}}[x_{1},\dots ,x_{n}]$ is a polynomial ring over an integral domain and thus is an integral domain, this implies either $f\equiv 0$ or $g\equiv 0$ modulo ${\mathfrak {p}}$ . Hence, either $\operatorname {cont} (f)$ or $\operatorname {cont} (g)$ is contained in ${\mathfrak {p}}$ . Since ${\sqrt {\operatorname {cont} (fg)}}$ is the intersection of all prime ideals that contain $\operatorname {cont} (fg)$ and the choice of ${\mathfrak {p}}$ was arbitrary, $\operatorname {cont} (f)\operatorname {cont} (g)\subset {\sqrt {\operatorname {cont} (fg)}}$ .

We now prove the "moreover" part. Factoring out the gcd's from the coefficients, we can write $f=af'$ and $g=bg'$ where the gcds of the coefficients of $f',g'$ are both 1. Clearly, it is enough to prove the assertion when $f,g$ are replaced by $f',g'$ ; thus, we assume the gcd's of the coefficients of $f,g$ are both 1. The rest of the proof is easy and transparent if $R$ is a unique factorization domain; thus we give the proof in that case here (and see ^{[note 4]} for the proof for the GCD case). If $\gcd(\operatorname {cont} (fg))=(1)$ , then there is nothing to prove. So, assume otherwise; then there is a non-unit element dividing the coefficients of $fg$ . Factorizing that element into a product of prime elements, we can take that element to be a prime element $\pi$ . Now, we have:

Thus, either $(\pi )$ contains $\operatorname {cont} (f)$ or $\operatorname {cont} (g)$ ; contradicting the gcd's of the coefficients of $f,g$ are both 1. $\square$

If $f(x)$ is a in one variable with coefficients in a unique factorization domain $R$ (or more generally a GCD domain), then a root of $f$ that is in the field of fractions $F$ of $R$ is in $R$ .^{[note 5]}

monic polynomial

It follows from Gauss's lemma that for each unique factorization domain $R$ , the polynomial ring $R[X_{1},X_{2},...,X_{n}]$ is also a unique factorization domain (see #Statements for unique factorization domains). Gauss's lemma can also be used to show Eisenstein's irreducibility criterion. Finally, it can be used to show that cyclotomic polynomials (unitary units with integer coefficients) are irreducible.

Gauss's lemma implies the following statement:

If $R=\mathbb {Z}$ , then it says a rational root of a monic polynomial over integers is an integer (cf. the rational root theorem). To see the statement, let $a/b$ be a root of $f$ in $F$ and assume $a,b$ are relatively prime. In $F[x]$ we can write $f=(x-a/b)g$ with $cg\in R[x]$ for some $c\in R$ . Then

is a factorization in $R[x]$ . But $bx-a$ is primitive (in the UFD sense) and thus $cb$ divides the coefficients of $cg$ by Gauss's lemma, and so

with $h$ in $R[x]$ . Since $f$ is monic, this is possible only when $b$ is a unit.

A similar argument shows:

The irreducibility statement also implies that the minimal polynomial over the rational numbers of an algebraic integer has integer coefficients.

; Macdonald, I.G. (1969), Introduction to Commutative Algebra, Westview Press, ISBN 978-0-201-40751-8

Atiyah, Michael Francis

(1995), Commutative algebra, Graduate Texts in Mathematics, vol. 150, Berlin, New York: Springer-Verlag, doi:10.1007/978-1-4612-5350-1, ISBN 978-0-387-94268-1, MR 1322960, ISBN 978-0-387-94269-8

monic polynomial

Atiyah, Michael Francis

Eisenbud, David