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 $62$