Content of Polynomials is Multiplicative

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Theorem

Rational polynomials

Let $h \in \Q \left[{X}\right]$ be a polynomial with rational coefficients.

Let $c_h = \operatorname{cont} \left({h}\right)$ denote the content of $h$.


Then for any polynomials $f, g \in \Q \left[{X}\right]$ with rational coefficients:

$\operatorname{cont} \left({f g}\right) = \operatorname{cont} \left({f}\right) \operatorname{cont} \left({g}\right)$

Dedekind domain

Let $R$ be a Dedekind domain.

Let $f, g \in R[X]$ be polynomials.

Let $\operatorname{cont} (f)$ denote the content of $f$.


Then $\operatorname{cont} (fg) = \operatorname{cont} (f) \operatorname{cont} (g)$ is the product of $\operatorname{cont}(f)$ and $\operatorname{cont}(g)$.


Proof

Let $\tilde f = c_f^{-1} f$, $\tilde g = c_g^{-1} g$

By Content of Scalar Multiple:

$c_{\tilde f} = c_{\tilde g} = 1$

That is, $\tilde f$ and $\tilde g$ are primitive.

By Gauss's Lemma on Primitive Polynomials, it follows that $\tilde f \tilde g$ is primitive.

Now,

\(\displaystyle c_{f g}\) \(=\) \(\displaystyle c_f c_g c_{\tilde f \tilde g}\) $\quad$ Content of Scalar Multiple $\quad$
\(\displaystyle \) \(=\) \(\displaystyle c_f c_g\) $\quad$ because $c_{\tilde f \tilde g} = 1$ $\quad$

$\blacksquare$