# 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 |
---|---|

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 |