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It is easy to see that there are no group divisors of length 2 in a group algebra of a torsion-free group. I saw somewhere mentioned that it is possible to do it for length 3. How?

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  • $\begingroup$ You should correct "group divisors" to "zero divisors" and maybe say that by "length" of an element you mean the cardinal of its support. $\endgroup$
    – YCor
    Jun 10, 2012 at 11:04
  • $\begingroup$ Yves, this is how I read it))) $\endgroup$ Jun 10, 2012 at 11:17
  • $\begingroup$ @Kate: I can read "zero divisors" in the title but "group divisors" in the text. In a first reading I couldn't understand the meaning of the question, and finally understood it after your answer. $\endgroup$
    – YCor
    Jun 10, 2012 at 20:11

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It is open for the group algebra over the field with $2$ elements, see

http://arxiv.org/abs/1202.6645 and http://arxiv.org/abs/1112.1790

and references there. I saw on the website of Mikhailov an unpublished paper, where he contributes this question to I. Rips.

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