In Euclidean Domains the Euclidean division algorithm can be used to find $a,b$ such that $ax+by=1$. Let $R$ be a commutative integral domain that is not a Euclidean domain. Assume that $x,y\in R$ satisfy $Rx+Ry=(1)$. Then we know that there exist $a,b\in R$ such that $ax+by=1$.
Not being in a Euclidean domain, is there a process that can be carried out in general (e.g. not in a PID) to find a solution $(a,b)$? Are there specific cases that can be done?
Note: $x,y$ satisfy the "Bezout identity" because $(x)+(y)$ is principal. But I'm not sure if this is enough.
In the cases that $R$ is of finite rank $\mathbb{Z}$ module and of characteristic $0$ ( like the ring of integers of a number field), so a free $\mathbb{Z}$-module) then finding linear combinations $\sum c_i a_i = b\ \ $ ($\ c_i \in R$ unknown, $b$, $a_i\in R$ given) can be reduced to solving a system of linear equations over the integers. This can be done using the Smith normal form .