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Nov 3, 2017 at 17:20 comment added Simon Wadsley Surely no because $R$ may be Noetherian and $I$ may be infinite whence the injective envelope of the direct sum of the $Q_i$ is the direct sum of the $Q_i$ which is not the product of the $Q_i$. Or did you mean to insist that $R$ be non-Noetherian.
Nov 3, 2017 at 12:13 comment added Fred Rohrer No, see mathoverflow.net/a/53935.
Nov 3, 2017 at 10:33 history asked Chaitanya CC BY-SA 3.0