Gauss's Lemma on Primitive Rational Polynomials/Proof 2

Proof
Recall Polynomial has Integer Coefficients iff Content is Integer.

By hypothesis $f$ and $g$ have content $1 \in \Z$.

Therefore:
 * $f, g \in \Z \sqbrk X$

that $f g$ is not primitive, say:


 * $\cont {f g} = d \ne 1$

By the Fundamental Theorem of Arithmetic we can choose a prime $p$ dividing $d$.

From Quotient Epimorphism from Integers by Principal Ideal, let $\pi : \Z \to \Z / p \Z$ be the quotient epimorphism to the ring of integers modulo $p$.

From Ring of Integers Modulo Prime is Field, $\Z / p \Z$ is a field.

Let $\Pi : \Z \sqbrk X \to \paren {\Z / p \Z} \sqbrk X$ be the induced homomorphism of the polynomial rings.

By construction, $p$ divides each coefficient of $f g$, so:


 * $\map \Pi {f g} = \map \Pi f \, \map \Pi g = 0$

From Polynomial Forms over Field form Integral Domain, $\paren {\Z / p \Z} \sqbrk X$ is an integral domain.

Thus:
 * $\map \Pi f = 0$ or $\map \Pi g = 0$

After possibly exchanging $f$ and $g$, we may assume that $\map \Pi f = 0$.

Now by Kernel of Induced Homomorphism of Polynomial Forms, if:


 * $f = a_0 + a_1 X + \cdots + a_n X^n$

we must have:


 * $\map \pi {a_i} = 0$

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.

Hence our assumption that $f g$ is not primitive was invalid.

The result follows by Proof by Contradiction.