Gauss's lemma IGausssLemmaI2002-11-04 05:39:332008-05-01 01:10:32Theorem% this is the default PlanetMath preamble. as your knowledge
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There are a few different things that are sometimes called ``Gauss's Lemma''. See also Gauss's Lemma II.
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\\\emph{Gauss's Lemma I:} If $R$ is a UFD and $f(x)$ and $g(x)$ are both primitive polynomials in $R[x]$, so is $f(x) g(x)$.
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\\ \emph{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 the gcd of the coefficients of $f(x) g(x)$ is not a unit. Let $p$ be a prime factor of that gcd. We consider the image of $R$ mod $p$ - i.e. under the natural ring homomorphism $\theta: R \to R/pR$ - and extend to the polynomial ring.
Since $R$ is an integral domain, $R/pR$ is an integral domain, so $(R/pR)[x]$ is an integral domain. And we have
\begin{align*}
\overline{f(x)} \ \overline{g(x)} = 0
\end{align*}
where $\overline{f(x)}$ is the image of $f(x)$ in $(R/pR)[x]$, similarly $\overline{g(x)}$. So $\overline{f(x)} = 0$ or $\overline{g(x)} = 0$. So $f(x)$ or $g(x)$ is divisible by $p$, so one of them is not primitive.