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Nov 12, 2012 at 21:41 vote accept user27976
Nov 10, 2012 at 14:33 answer added Jeremy Rickard timeline score: 58
Nov 10, 2012 at 14:10 comment added François Brunault Considering the multiplication by $p$ on the nonzero object with $p$ prime, it seems the base ring $R$ of the hypothetical special module can be taken to be either a $\mathbf{Z}/p\mathbf{Z}$ algebra or a $\mathbf{Q}$-algebra.
Nov 10, 2012 at 8:26 comment added Zhen Lin If there is an abelian category $\mathcal{C}$ with only one non-zero object $A$, then its endomorphism ring must fail to have the invariant basis number property: because then $\mathcal{C}(A, A) \cong \mathcal{C}(A, A \times A) \cong \mathcal{C}(A, A) \times \mathcal{C}(A, A)$ as right $\mathcal{C}(A, A)$-modules.
Nov 10, 2012 at 6:34 history edited Eric Wofsey CC BY-SA 3.0
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Nov 10, 2012 at 5:58 history asked user27976 CC BY-SA 3.0