Gauss's Lemma on Primitive Rational Polynomials

Theorem
Let $f,g \in \Q\left[{X}\right]$ be primitive polynomials with rational coefficients.

Then $fg \in \Q\left[{X}\right]$ is also primitive.

Corollary 1
If $h \in \Q\left[X\right]$ is a polynomial with rational coefficients, let $\operatorname{cont}\left({h}\right)$ denote the content of $h$.

Then for any polynomials $f,g \in \Q\left[{X}\right]$ with rational coefficients we have:
 * $\displaystyle \operatorname{cont}\left({fg}\right) = \operatorname{cont}\left({f}\right)\operatorname{cont}\left({g}\right)$

Corollary 2
Let $\Z \left[{X}\right]$ be the Ring of Polynomial Forms over the integers.

Let $h \in \Z \left[{X}\right]$ have degree at least $1$.

Then $h$ is irreducible in $\Q \left[{X}\right]$ iff $h$ is irreducible in $\Z \left[{X}\right]$.

Necessary Condition
If $h$ is not irreducible in $\Z \left[{X}\right]$ then $h$ is obviously not irreducible in $\Q \left[{X}\right]$.

Sufficient Condition
Suppose now that $h$ is not irreducible in $\Q \left[{X}\right]$.

Note that $h$ is reducible iff $c h$ is reducible for any non-zero constant $c \in \Q$.

From Properties of Content, it follows that $\operatorname{cont} \left({h}\right) \in \Z$.

Let $\tilde h = \dfrac 1 {\operatorname{cont} \left({h}\right) } h$, which is an element of $\Z \left[{X}\right]$ with $\operatorname{cont} \left({\tilde h}\right) = 1$ by definition of content.

By assumption, $\tilde h$ has factors in $\Q \left[{X}\right]$.

Suppose $\tilde h = \tilde f \tilde g$, with $\tilde f$ and $\tilde g$ both non-constant.

Let $k_f$ and $k_g$ be the least common multiples of the denominators of the coefficients of $\tilde f$ and $\tilde g$, respectively.

Define $f = k_f \tilde f$ and $g = k_g \tilde g$.

From the definition of the least common multiple, it follows that $\operatorname{cont} \left({f}\right) = \operatorname{cont} \left({g}\right) = 1$.

Thus, by Properties of Content, we have:
 * $f, g \in \Z \left[{X}\right]$

Thus we have:
 * $f g = \dfrac {\tilde h} {k_f k_g} \in \Z \left[{X}\right]$

Computing the content of both sides of the equation, we conclude $k_f k_g = 1$.

This means that already $\tilde f, \tilde g \in \Z \left[{X}\right]$.

Now multiply both sides of the equation by $\operatorname{cont} \left({h}\right) \in \Z$:
 * $\operatorname{cont} \left({h}\right) \tilde f \tilde g = \operatorname{cont} \left({h}\right) \tilde h = h$

Thus $h$ is not irreducible in $\Z \left[{X}\right]$.