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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).

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up vote 4 down vote accepted

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 $I(x)$.

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Looks very good ($x$ should also belong to the generators). Thank you very much. – Jochen Wengenroth Feb 26 '13 at 19:38
Right; I'll edit the answer to add $x$ as a generator. Thanks. – Andreas Blass Feb 26 '13 at 20:02

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