Content of Rational Polynomial is Multiplicative/Proof 1

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Let $h \in \Q \sqbrk X$ be a polynomial with rational coefficients.

Let $\cont h$ denote the content of $h$.

Then for any polynomials $f, g \in \Q \sqbrk X$ with rational coefficients:

$\cont {f g} = \cont f \cont g$


Let $f^* = \dfrac 1 {\cont f} f$, $g^* = \dfrac 1 {\cont g} g$

By Content of Scalar Multiple:

$\cont {f^*} = \cont {g^*} = 1$

That is, $f^*$ and $g^*$ are primitive.

By Gauss's Lemma on Primitive Rational Polynomials, it follows that $f^* g^*$ is primitive.


\(\ds \cont {f g}\) \(=\) \(\ds \cont f \cont g \cont {f^* g^*}\) Content of Scalar Multiple
\(\ds \) \(=\) \(\ds \cont f \cont g\) because $\cont {f^* g^*} = 1$