Gauss's Lemma on Primitive Rational Polynomials

Lemma
If $f,g\in \Z[x]$ are primitive, then so is $h = fg.$

Proof
First, we write

$f = b_ex^e + b_{e-1}x^{e-1} + \ldots + b_0$,

$g = c_fx^f + c_{f-1}x^{f-1} + \ldots + c_0$, and

$h = a_dx^d + a_{d-1}x^{d-1} + \ldots + a_0$.

Suppose $p$ is a prime dividing the content of $h$.

Let $i$ and $j$ be the smallest positive integers such that $p\nmid b_i$ and $p\nmid c_j$. Such $i$ and $j$ must exist since $f$ and $g$ are primitive.

Consider the coefficient $a_{i+j} = b_0c_{i+j} + b_1c_{i+j-1} + \ldots + b_ic_j + \ldots + b_{i+j}c_0$.

By the minimality of $i$ and $j$, every term in this sum is divisible by $p$ with the sole exception of $b_ic_j$.

Thus, $a_{i+j}$ is not divisible by $p$---a contradiction.

Corollary
If $f,g,h\in\Z[x]$ such that $h = fg$, then $cont(h) = cont(f)cont(g)$.

Proof
Note that we can write

$q = cont(q)\tilde{q}$,

such that the content of $\tilde{q}$ is $1$ for each polynomial $q$.

By the above results, we see that $cont(\tilde{f}\tilde{g}) = 1 = cont(\tilde{h})$. Thus, we must have $\tilde{h} = \tilde{f}\tilde{g}$, from which the desired result follows.

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

Proof
If $h$ is not irreducible in $\Z[x]$ then $h$ is obviously not irreducible in $\Q[x]$.

Suppose, for contradiction, that $h$ is irreducible in $\Z[x]$ but not in $\Q[x]$.

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

Let $\tilde{h} = \frac{1}{cont(h)}h$, which is an element of $\Z[x]$ with $cont(\tilde{h})=1$.

By our assumption, $\tilde{h}$ factors in $\Q[x]$; suppose $\tilde{h} = \tilde{f}\tilde{g}$.

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

Define $f = k_f\tilde{f}$ and $g = k_g\tilde{g}$. By the definition of the least common multiple, we see that $cont(f) = cont(g) = 1$ and $f,g\in \Z[x]$.

Clearly $fg \nmid \tilde{h}$ and comparing degrees we deduce the existence of a non-zero constant $c$ such that,

$\tilde{h} = cfg$.

Computing the content of both sides of the equation, it follows from the above result that $c=1$.

Multiplying both sides of the equation by $cont(h)\in\Z$ we arrive at a contradiction.