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.

Proof of Theorem
First we note that a polynomial has integer coefficients iff its content is an integer.

By hypothesis $f$ and $g$ have content $1 \in \Z$, so $f,g \in \Z\left[{X}\right]$.

Suppose that $fg$ is not primitive, say
 * $\displaystyle \operatorname{cont}\left({fg}\right) = d \neq 1$

By the fundamental theorem of arithmetic we can choose a prime $p$ dividing $d$.

Let $\pi : \Z \to \Z/p\Z$ be the canonical epimorphism to the ring of integers modulo $p$

We have that the ring of integers modulo a prime is a field, so $\Z/p\Z$ is a field.

Let $\Pi : \Z\left[{X}\right] \to \left({\Z/p\Z}\right)\left[{X}\right]$ be the induced homomorphism of the polynomial rings.

By construction, $p$ divides each coefficient of $fg$, so
 * $\Pi\left({fg}\right) = \Pi\left({f}\right)\Pi\left({g}\right) = 0$.

Since a polynomial ring over a field is an integral domain we know that $\left({\Z/p\Z}\right)\left[{X}\right]$ is an integral domain.

Thus we must have $\Pi\left({f}\right) = 0$ or $\Pi\left({g}\right) = 0$.

After possibly exchanging $f$ and $g$, we may assume that $\Pi\left({f}\right) = 0$.

Now by the characterization of the kernel of the induced homomorphism, if
 * $f = a_0 + a_1X + \cdots + a_nX^n$

we must have $\pi\left({a_i}\right) = 0$ for $i = 0,\ldots,n$.

That is, $a_i \in p\Z$ for $i = 0,\ldots,n$.

But this says precisely that $p$ divides each $a_i$, $i = 0,\ldots,n$.

Therefore $p$ divides the content of $f$, a contradiction.