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

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

Proof of Theorem
First we note that a polynomial has integer coefficients iff it's 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.

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

By Content of Scalar Multiple, we have $c_{\tilde f} = c_{\tilde g} = 1$.

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

By the theorem, it follows that $\tilde f \tilde g$ is primitive.

Now,

Proof of Corollary 2
If $h$ is not irreducible in $\Z \left[{X}\right]$ then $h$ is obviously not irreducible in $\Q \left[{X}\right]$.

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 $c \in \Q$.

Let $\tilde h = \dfrac 1 {\operatorname{cont} \left({h}\right) } h$.

By Content of Scalar Multiple, we have $\operatorname{cont} \left({\tilde h}\right) = 1$.

Therefore by Polynomial has Integer Coefficients iff Content is Integer, $\tilde h \in \Z\left[{X}\right]$.

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

Suppose $\tilde h = \tilde f \tilde g$, with $\tilde f$ and $\tilde g$ both of positive degree.

Let $c_f$ and $c_g$ be the contents of $\tilde f$ and $\tilde g$ respectively.

Define $f = c_f^{-1} \tilde f$ and $g = c_g^{-1} \tilde g$.

By Content of Scalar Multiple, it follows that $\operatorname{cont} \left({f}\right) = \operatorname{cont} \left({g}\right) = 1$.

Now we have:
 * $f g = \dfrac {\tilde f \tilde g} {c_f c_g} = \dfrac {\tilde h} {c_f c_g}$

Taking the content, by Content of Scalar Multiple we have:
 * $\operatorname{cont} \left({fg}\right) = \dfrac{1}{c_f c_g}\operatorname{cont} \left({\tilde h}\right)$

By the theorem we know that $\operatorname{cont} \left({fg}\right) = 1$.

Moreover we already have that $\operatorname{cont} \left({\tilde h}\right) = 1$.

Therefore we must have $c_f c_g = 1$.

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

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

This is a non-trivial factorization of $h$, thus $h$ is not irreducible in $\Z \left[{X}\right]$.