# Pulling out factors in a Noetherian Domain

Let $R$ be a Noetherian domain (not-necessarily commutative), and let $S$ be a Noetherian subring of $R$. An element $r\in R$ is left $S$-irreducible if, for any $s\in S$ and $r' \in R$ with $sr'=r$, then $s$ is a unit in $S$. These are elements in $R$ which have no non-trivial factors in $S$ on the left.

My question is, when can every element $r\in R$ be uniquely factored as $r=sr_0$ with $s\in S$ and $r_0\in R$ left $S$-irreducible (where uniqueness means up to a unit in $S$). That is, when can every element $r$ have its left $S$-factors pulled out in an essentially unique way?

I am looking for conditions on $R$ and $S$ that imply this. If $R$ contains inverses of non-units in $S$, then these factorizations don't exist, even for $1$. If $R$ is a UFD and non-units in $S$ are non-units in $R$, then these factorizations always exist. If $S$ is a PID, and non-units in $S$ are non-units in $R$, then these factorizations always exist.

This came up while playing with the $n$th Weyl algebra $k[x_1,...x_n,d_1,...d_n]$ and its maximal commutative subalgebra $k[x_1,...x_n]$. This pair has the property in question, but it seems like it should be a consequence of a more general phenomenon.

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