Bézout's Identity for Polynomial Ring

Question

We know that for two non-zero polynomial forms $f,g\in F[X]$ over the field $F$, their greatest common divisor $d$ exists and is unique (when ignoring those multiplied by constants), and it is the last non-zero remainder in the Euclidean algorithm; and there exists $u,v\in F[x]$ such that $$ uf+vg=d; $$ this is called the Bézout's Identity for Polynomial Ring.

On my textbook, there is a corollary of this that states $f,g\in F[x]$ are coprime polynomials (have only the invertible elements in $F[x]$, i.e. elements of $F*$, as common divisors) if and only if there exists $u,v\in F[X]$ such that $$ uf+vg=1. $$ The $(\implies)$ direction of this corollary is trivial to me, but I think I need a little help on the $(\impliedby)$. Any kind of help is appreciated. Thanks.

Answer 1

Suppose $f,g$ have a common divisor $d$, that is $f=dr$ and $g=ds$.

Then $d\mid uf+vg=udr+vds=d(ur+vs)$. Therefore $d\mid 1$. That means they only share units as common divisors.

Answer 2

It's special case $d=1$ of the following [i.e. linear (Bezout) common divisors are always greatest]

Theorem $\ \gcd(f,g) =d\,\iff\, d\mid f,g\ \ \,\&\ \ d = uf+vg$ for some $\,u,v$

Proof $(\Rightarrow)$ Bezout. $(\Leftarrow)$ $d$ is common divisor, necessarily greatest by $\,c\mid f,g\,\Rightarrow\, c\mid d=uf+vg$