Content of Rational Polynomial is Multiplicative

From ProofWiki
Jump to navigation Jump to search

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:

\(\displaystyle \cont {f g}\) \(=\) \(\displaystyle \cont f \cont g \cont {f^* g^*}\) Content of Scalar Multiple
\(\displaystyle \) \(=\) \(\displaystyle \cont f \cont g\) 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:

$\cont f, \cont g$ are the content of $f$ and $g$ respectively
$f^*, g^*$ are primitive.


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.

$\blacksquare$