# 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 irreducible factor of $d$, which exists and is a prime element because $R$ is a UFD. We consider the quotient ring of $R$ by the principal ideal $pR$ generated by $p$, which is a prime ideal since $p$ is a prime element. The canonical projection $R\to R/pR$ induces a surjective ring homomorphism $\theta:R[X]\to(R/pR)[X]$, whose kernel consists of all polynomials 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 domain, so $(R/pR)[x]$ is also an integral domain. By hypothesis $\theta$ sends the product $f(x)g(x)$ to $0\in(R/pR)[X]$, which is therefore the product of $\theta(f(x))$ and $\theta(g(x))$, 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 $\theta$, and therefore not primitive.

Title Gauss’s lemma I GausssLemmaI 2013-03-22 13:07:49 2013-03-22 13:07:49 bshanks (153) bshanks (153) 17 bshanks (153) Theorem msc 12E05 GausssLemmaII