# Gauss's lemma (polynomial)

In algebra, **Gauss's lemma**,[1] named after Carl Friedrich Gauss, is a statement about polynomials over the integers, or, more generally, over a unique factorization domain (that is, a ring that has a unique factorization property similar to the fundamental theorem of arithmetic). Gauss's lemma underlies all the theory of factorization and greatest common divisors of such polynomials.

Gauss's lemma asserts that the product of two primitive polynomials is primitive (a polynomial with integer coefficients is *primitive* if it has 1 as a greatest common divisor of its coefficients).

A corollary of Gauss's lemma, sometimes also called *Gauss's lemma*, is that a primitive polynomial is irreducible over the integers if and only if it is irreducible over the rational numbers. More generally, a primitive polynomial has the same complete factorization over the integers and over the rational numbers. In the case of coefficients in a unique factorization domain R, "rational numbers" must be replaced by "field of fractions of R". This implies that, if R is either a field, the ring of integers, or a unique factorization domain, then every polynomial ring (in one or several indeterminates) over R is a unique factorization domain. Another consequence is that factorization and greatest common divisor computation of polynomials with integers or rational coefficients may be reduced to similar computations on integers and primitive polynomials. This is systematically used (explicitly or implicitly) in all implemented algorithms (see Polynomial greatest common divisor and Factorization of polynomials).

Gauss's lemma, and all its consequences that do not involve the existence of a complete factorization remain true over any GCD domain (an integral domain over which greatest common divisors exist). In particular, a polynomial ring over a GCD domain is also a GCD domain. If one calls *primitive* a polynomial such that the coefficients generate the unit ideal, Gauss's lemma is true over every commutative ring.[2] However, some care must be taken, when using this definition of *primitive*, as, over a unique factorization domain that is not a principal ideal domain, there are polynomials that are primitive in the above sense and not primitive in this new sense.

## The lemma over the integers

If is a polynomial with integer coefficients, then is called *primitive* if the greatest common divisor of all the coefficients is 1; in other words, no prime number divides all the coefficients.

**Gauss's lemma (primitivity)** — If *P* and *Q* are primitive polynomials over the integers, then product *PQ* is also primitive.

**Proof:** Clearly the product *f*(*x*).*g*(*x*) of two primitive polynomials has integer coefficients. Therefore, if it is not primitive, there must be a prime *p* which is a common divisor of all its coefficients. But *p* can not divide all the coefficients of either *f*(*x*) or *g*(*x*) (otherwise they would not be primitive). Let *a _{r}x^{r}* be the first coefficient of

*f*(

*x*) not divisible by

*p*and let

*b*be the first coefficient of

_{s}x^{s}*g*(

*x*) not divisible by

*p*. Now consider the term

*x*in the product. Its coefficient must take the form:

^{r + s}The first term is not divisible by *p* (because *p* is prime), yet all the remaining ones are, so the entire sum cannot be divisible by *p*. By assumption all coefficients in the product are divisible by *p*, leading to a contradiction. Therefore, the coefficients of the product can have no common divisor and are thus primitive.

**Gauss's lemma (irreducibility)** — A non-constant polynomial in **Z**[*X*] is irreducible in **Z**[*X*] if and only if it is both irreducible in **Q**[*X*] and primitive in **Z**[*X*].

The proof is given below for the more general case. Note that an irreducible element of **Z** (a prime number) is still irreducible when viewed as constant polynomial in **Z**[*X*]; this explains the need for "non-constant" in the statement.

## Statements for unique factorization domains

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 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 of polynomials is the product of their contents.

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

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

Let be a unique factorization domain with field of fractions . If is a polynomial over , then, for some , has coefficients in and so, factoring-out the gcd of the coefficients, we can write: for some primitive polynomial . As one can check, this polynomial is unique up to the multiplication by a unit element and is called the **primitive part** (or **primitive representative**) of and is denoted by . The procedure is compatible with product: .

The construct can be used to show the statement:

- A polynomial ring over a UFD is a UFD.

Indeed, by induction, it is enough to show is a UFD when is a UFD. Let be a nonzero polynomial. Now, is a unique factorization domain (since it is a principal ideal domain) and so, as a polynomial in , can be factorized as:

where are irreducible polynomials of . Now, we write for the gcd of the coefficients of (and is the primitive part) and then:

Now, is a product of prime elements of (since is a UFD) and a prime element of is a prime element of , as is an integral domain. Hence, admits a prime factorization (or a unique factorization into irreducibles). Next, observe that is a unique factorization into irreducible elements of , as (1) each is irreducible by the irreducibility statement and (2) it is unique since the factorization of can also be viewed as a factorization in and factorization there is unique. Since are uniquely determined by , up to unit elements, the above factorization of is a unique factorization into irreducible elements.

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 nonzero 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*) shows the failure of the primitivity statement. For a concrete example one can take

^{2}*R*=

**Z**[

*i*√

*5*],

*p*=

*1*+

*i*√

*5*,

*a*=

*1*-

*i*√

*5*,

*q*=

*2*,

*b*=

*3*. In this example the polynomial

*3*+

*2X*+

*2X*(obtained by dividing the right hand side by

^{2}*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*√

*5*]). Another well known example is the polynomial

*X*−

^{2}*X*−

*1*, whose roots are the golden ratio φ = (

*1*+ √

*5*)/

*2*and its conjugate (

*1*− √

*5*)/

*2*showing that it is reducible over the field

**Q**[√

*5*], although it is irreducible over the non-UFD

**Z**[√

*5*] which has

**Q**[√

*5*] 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**[√

*5*] (the ring of Dirichlet integers), over which

*X*−

^{2}*X*−

*1*becomes reducible, but in the former example

*R*is already integrally closed.

## General version

Let be a commutative ring. If is a polynomial in , then we write for the ideal of generated by all the coefficients of ; it is called the content of . Note that for each . The next proposition states a more substantial property.

**Proposition[3]** — For each pair of polynomials in ,

where denotes the radical of an ideal. Moreover, if is a GCD domain (e.g., a unique factorization domain), then

where denotes the unique minimal principal ideal containing a finitely generated ideal .[note 1]

A polynomial is said to be *primitive* if is the unit ideal.[4] When (or more generally a Bézout domain), this agrees with the usual definition of a primitive polynomial. (But if is only a UFD, this definition is inconsistent with the definition of primitivity in #Statements for unique factorization domains.)

**Corollary[2]** — Two polynomials are primitive if and only if the product is primitive.

*Proof:* This is easy using the fact[5] that implies

**Corollary[6]** — Suppose is a GCD domain (e.g., a unique factorization domain) with the field of fractions . Then a non-constant polynomial in is irreducible if and only if it is irreducible in and the gcd of the coefficients of is 1.

*Proof:* () First note that the gcd of the coefficients of is 1 since, otherwise, we can factor out some element from the coefficients of to write , contradicting the irreducibility of . Next, suppose for some non-constant polynomials in . Then, for some , the polynomial has coefficients in and so, by factoring out the gcd of the coefficients, we write . Do the same for and we can write for some . Now, let for some . Then . From this, using the proposition, we get:

- .

That is, divides . Thus, and then the factorization constitutes a contradiction to the irreducibility of .

() If is irreducible over , then either it is irreducible over or it contains a constant polynomial as a factor, the second possibility is ruled out by the assumption.

*Proof of the proposition:* Clearly, . If is a prime ideal containing , then modulo . Since is a polynomial ring over an integral domain and thus is an integral domain, this implies either or modulo . Hence, either or is contained in . Since is the intersection of all prime ideals that contain and the choice of was arbitrary, .

We now prove the "moreover" part. Factoring out the gcd’s from the coefficients, we can write and where the gcds of the coefficients of are both 1. Clearly, it is enough to prove the assertion when are replaced by ; thus, we assume the gcd's of the coefficients of are both 1. The rest of the proof is easy and transparent if is a unique factorization domain; thus we give the proof in that case here (and see [note 2] for the proof for the GCD case). If , then there is nothing to prove. So, assume otherwise; then there is a non-unit element dividing the coefficients of . Factorizing that element into a product of prime elements, we can take that element to be a prime element . Now, we have:

- .

Thus, either contains or ; contradicting the gcd's of the coefficients of are both 1.

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

## Applications

It follows from Gauss's lemma that for each unique factorization domain , the polynomial ring 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 is a monic polynomial in one variable with coefficients in a unique factorization domain (or more generally a GCD domain), then a root of that is in the field of fractions of is in .[note 3]

If , 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 be a root of in and assume are relatively prime. In , we can write with for some . Then

- ,

is a factorization in . But is primitive (in the UFD sense) and thus divides the coefficients of by Gauss's lemma and so

- ,

with in . Since is monic, this is possible only when is a unit.

A similar argument shows:

- Let be a GCD domain with the field of fractions and . If for some polynomial that is primitive in the UFD sense and , then .

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

## Notes and references

- A generator of the principal ideal is a gcd of some generators of
*I*(and it exists because is a GCD domain). -
*Proof for the GCD case*: The proof here is adopted from Mines, R.; Richman, F.; Ruitenburg, W. (1988).*A Course in Constructive Algebra*. Universitext. Springer-Verlag. ISBN 0-387-96640-4. We need the following simple lemma about gcd:- If , then .

- .

- In other words, it says that a unique factorization domain is integrally closed.

- Article 42 of Carl Friedrich Gauss's
*Disquisitiones Arithmeticae*(1801) - Atiyah & MacDonald, Ch. 1., Exercise 2. (iv) and Exercise 3.
- Eisenbud, Exercise 3.4. (a)
- Atiyah & MacDonald, Ch. 1., Exercise 2. (iv)
- Atiyah & MacDonald, Ch. 1., Exercise 1.13.
- Eisenbud, Exercise 3.4.c; The case when
*R*is a UFD.

- Atiyah, Michael Francis; Macdonald, I.G. (1969),
*Introduction to Commutative Algebra*, Westview Press, ISBN 978-0-201-40751-8 - Eisenbud, David (1995),
*Commutative algebra*, Graduate Texts in Mathematics,**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