Gauss's Lemma on Primitive Rational Polynomials

Theorem
If $h\in\Z[X]$, then $h$ is irreducible in $\Q[X]$ iff $h$ is irreducible in $\Z[X]$.

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

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

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 $cont \left({h}\right) \in \Z$.

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

By assumption, $\tilde h$ factors in $\Q[X]$; 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 $cont \left({f}\right) = cont \left({g}\right) = 1$ and thus, by Properties of Content, we have $f, g \in \Z[X]$.

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

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

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

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

Thus $h$ is not irreducible in $\Z[X]$.