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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.

Hypatia du Bois-Marie
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2 Answers2

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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.

rschwieb
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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 $\,c\mid f,g\,\Rightarrow\, c\mid d=uf+vg$

Bill Dubuque
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  • Above I presume that $d$ is normalized to be monic (else replace $,\gcd(f,g) = d,$ by $,\gcd(f,g)\cong d,$ meaning they are associates, i.e. differ only by a unit factor) – Bill Dubuque Jan 23 '19 at 16:37
  • The Theorem is simply $\ f R + g R = dR \iff dR \supseteq fR+ gR\supseteq dR\ $ in ideal language, for $, R = F[X]\ $ $\quad $ – Bill Dubuque Jan 23 '19 at 17:14