# Is it always true that in an integral domain $R$, if $x,y\in R$ and $x,y$ coprime then $ux+vy=1$ for some $u,v\in R$?

Is it always true that in an integral domain $R$, if $x,y\in R$ and $x,y$ coprime then $ux+vy=1$ for some $u,v\in R$?

I know the converse is true. I don't know if this direction is true.

If the statement is true. Given an integral domain $R$ and let $x,y \in R$. We let $h$ to be the hcf of $x$ and $y$. Then $x=ha$, $y=hb$. $a,b$ are coprime. Hence $ua+vb=1$ and $ux+vy=h$. Then $R$ is a Bezout domain.

It seems like only when $R$ is a Bezout domain, the proposition is true.

• I just did a quick search and found math.stackexchange.com/questions/121691/… – joeb Feb 17 '17 at 23:09
• How do you define "coprime"? In an arbitrary integral domain, you don't have a gcd. – Jacob Manaker Feb 17 '17 at 23:10
• It is true in Bézout domains (finitely generated ideals are principal). – Bernard Feb 18 '17 at 1:04

## 1 Answer

Even in a Unique Factorization Domain, where (presumably) being coprime means that the only common factors are units, it’s not true. Try the UFD $k[x,y]$, polynomials in two variables over a field. There, $x$ and $y$ are coprime in this sense, but there are no polynomials $g$, $h$ (in the two variables) such that $xg+yh=1$.