Euclid's Lemma for polynomials

Proof $\ \ \ p\mid fg,pg\ \Rightarrow\ p\mid (fg,pg) = (f,p)\ g\ =\ g,\ $ by $\ (f,p) = 1.\quad\ $ QED

Thus if $\,p\,$ is irreducible and $\,p\nmid f\,$ then $\, p\mid fg\Rightarrow p\mid g,\,$ i.e. irreducibles are prime.

This proof of Euclid's Lemma works in any GCD domain, e.g. any domain like $\rm\:F[x]\:$ enjoying a Euclidean algorithm to compute the GCD. See also this answer where I present this version of Euclid's lemma in Bezout, gcd, and ideal form.

Polynomials over a field are an Euclidean domain relative to the degree function. In particular, for every polynomials $a(x)$ and $b(x)$ in $F[x]$, with $b(x)\neq 0$, there exist unique polynomials $q(x),r(x)\in F[x]$ such that $$ a(x) = q(x)b(x) + r(x),\quad\mbox{$r(x)=0$ or $\mathrm{deg}(r)\lt \mathrm{deg}(b)$}$$

So the Euclidean algorithm also applies to polynomials. In particular, if $q(x) = \gcd(m(x),n(x))$, then there exist polynomial $\alpha(x),\beta(x)$ such that $q(x) = \alpha(x)m(x) + \beta(x)n(x)$.

Now you can use the same proof as the proof of Euclid's Lemma for integers: if $a|bc$ and $\gcd(a,b)=1$, then $a|c$.