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Nov 10, 2015 at 20:09 answer added user91132 timeline score: 2
Nov 6, 2015 at 3:00 answer added David Handelman timeline score: 2
Nov 5, 2015 at 1:50 comment added Qiaochu Yuan Such a ring $R$ is necessarily noncommutative. Certainly $R$ cannot be a field. If $R$ is commutative it has a nontrivial maximal ideal $m$, and the finitely generated torsion module $R/m \oplus R/m$ is not cyclic because it has dimension $2$ as an $R/m$-vector space. This argument shows more generally that such a ring $R$ cannot admit a nonzero map into a commutative ring.
Nov 5, 2015 at 1:37 history asked Dr. Evil CC BY-SA 3.0