Content of Rational Polynomial is Multiplicative/Proof 1
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
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 $\cont {f^* g^*} = 1$ |
$\blacksquare$