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 $x$.



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> Is $I(x)= x I(x)$?
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This question is due to Graham R. Allan in this [article](http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-smv129i1p67bwm?q=27d2fc8c-fb81-4dff-a08b-1b5d01e10815%241&qt=IN_PAGE) 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).



A question of Allan on infinite divisibility
What's to stop you from just freely building a counterexample? That is, let the ring be generated by elements $x,a,b\_1,b\_2,\dots,b\_n,\dots$ subject to (only) the relations $a=x^nb\_n$ for all positive integers $n$. Then $a$ is in $I(x)$. Unless I'm making a stupid mistake, the only solutions $q$ of $a=xq$ are finite linear combinations (with integer coefficients adding to 1) of the elements $x^{n-1}b\_n$, and none of those are in $I(x)$.


