Gauss’s lemma I

There are a few different things that are sometimes called “Gauss’s Lemma”. See also Gauss’s Lemma II.

Gauss’s Lemma I: If R is a UFD and f(x) and g(x) are both primitive polynomials in R[x], so is f(x)g(x).

Proof: Suppose f(x)g(x) not primitive. We will show either f(x) or g(x) isn’t as well. f(x)g(x) not primitive means that there exists some non-unit d in R that divides all the coefficients of f(x)g(x). Let p be an irreduciblePlanetmathPlanetmathPlanetmathPlanetmath factor of d, which exists and is a prime elementMathworldPlanetmath because R is a UFD. We consider the quotient ringMathworldPlanetmath of R by the principal idealMathworldPlanetmathPlanetmathPlanetmathPlanetmath pR generated by p, which is a prime idealMathworldPlanetmathPlanetmath since p is a prime element. The canonical projection RR/pR induces a surjective ring homomorphismMathworldPlanetmath θ:R[X](R/pR)[X], whose kernel consists of all polynomialsMathworldPlanetmathPlanetmath all of whose coefficients are divisible by p; these polynomials are therefore not primitive.

Since pR is a prime ideal, R/pR is an integral domainMathworldPlanetmath, so (R/pR)[x] is also an integral domain. By hypothesis θ sends the product f(x)g(x) to 0(R/pR)[X], which is therefore the product of θ(f(x)) and θ(g(x)), and one of these two factors in (R/pR)[x] must be zero. But that means that f(x) or g(x) is in the kernel of θ, and therefore not primitive.

Title Gauss’s lemma I
Canonical name GausssLemmaI
Date of creation 2013-03-22 13:07:49
Last modified on 2013-03-22 13:07:49
Owner bshanks (153)
Last modified by bshanks (153)
Numerical id 17
Author bshanks (153)
Entry type Theorem
Classification msc 12E05
Related topic GausssLemmaII