Re: JSH: Stop being dense


Keith Ramsay wrote:
sg552@xxxxxxxxxxxxx wrote:
|Possibly stupid question: Is it possible in a ring without identity to
|have two non-zero elements a and b such that ab=a?

Yes. You can have a=b=a^2. Take the ring 2Z[a]/(a^2-a), which
consists of elements of the form (m+na) where m and n are
integers and m is even. The addition and multiplication are given
by (m+na)+(m'+n'a) = (m+m')+(n+n')a and
(m+na)(m'+n'a) = mm'+(mn'+m'n+nn')a.

Even simpler: the even integers modulo 12, usual addition &
multiplication, with
a = b = 4. (Probably a special case of the above ...)

--- Christopher Heckman

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