Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question

1 Answer 1

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 $I(x)$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.