Loosely speaking the question is: If an element of a commutative ring is infinitely divisible by $x$, is it the product of $x$ with an infinitely divisible element?

More preceisely: For a fixed element $x$ of a commutative unital ring $R$ let $I(x)=\bigcap \lbrace x^nR :n\in \mathbb N \rbrace$ be the ideal of elements which are divisible by all powers of $x$.

Is $I(x)= x I(x)$?

This question is due to Graham R. Allan in this article and it appeared in his investigations of embedding the algebra $\mathbb C[[X]]$ of formal power series into Banach algebras (surprisingly, this is indeed possible). Originally, Allan asked this for so-called *stable* elements, which means that the infinite system $a_n - x a_{n+1} =b_n$ is solvable for all sequences $b_n$ in $R$. For stable elements of Banach or Frechet algebras the answer is positive (even more, it is positive for stable elements of algebras having a sub-multiplicative norm - the reason is that for stable elements Baire's theorem for $R^{\mathbb N}$ endowed with the product of the discrete topologies implies something to work with for the given norm).