Content of Rational Polynomial is Multiplicative
Theorem
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:
- $\cont {f g} = \cont f \cont g$
Proof 1
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.
Then:
| \(\ds \cont {f g}\) | \(=\) | \(\ds \cont f \cont g \cont {f^* g^*}\) | Content of Scalar Multiple | |||||||||||
| \(\ds \) | \(=\) | \(\ds \cont f \cont g\) | because $\cont {f^* g^*} = 1$ |
$\blacksquare$
Proof 2
From Rational Polynomial is Content Times Primitive Polynomial, let $\map f X$ and $\map g X$ be expressed as:
- $\map f X = \cont f \cdot \map {f^*} X$
- $\map g X = \cont g \cdot \map {g^*} X$
where:
We have, by applications of Rational Polynomial is Content Times Primitive Polynomial:
- $\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \, \map {g^*} X$
By Gauss's Lemma on Primitive Rational Polynomials we have that $\map {f^*} X \, \map {g^*} X$ is primitive.
As $\cont f > 0$ and $\cont g > 0$, then so is $\cont f \cont g > 0$.
By the uniqueness clause in Rational Polynomial is Content Times Primitive Polynomial, the result follows.
$\blacksquare$