The Problem: Suppose $R$ is an integral domain, $a, b\in R$ are nonzero. Suppose $x$ is nonzero, then $x\in(a, b)$ IFF $x=sa-tb$ for some $s, t\in R$.
My Attempt: Certainly this is true if $R$ is an Euclidean domain: if $x\in(a, b)$, then $x\in(d)$ where $d$ is the greatest common divisor of $a$ and $b$. Note that $d$ can be written as a $R$-linear combination $d=ax+by$, hence $x=z(ax+by)$ for some $z$ thus $x=zxa+zyb=sa-tb$, where $s=zx$ and $t=-zy$; conversely, if $x=sa-tb$ then clearly $x\in(a, b)$. But what if $R$ is an integral domain that is NOT an Euclidean domain? Any clarification would be greatly appreciated.
