# Content of Rational Polynomial is Multiplicative/Proof 2

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## 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

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.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Theorem $62$