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Let R be a non-Noetherian ring. Is its left global dimension lD(R) = sup { id(M) | M is a cyclic R-module }? id(M) denotes the injective dimension of R.

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What would happen if you had an algebra of the form A\oplus B, and considered B as a cyclic module? It seems like it should have injective dimension equal to its injective dimension as a B-module. Then you could give A any properties you wanted which would transfer to A\oplus B (such as higher injective dimension and non-Noetherianness). –  Greg Muller May 11 '10 at 13:36
But Greg's comment doesn't contradict anything in the question -- since there would be other cyclic modules (such as A) which would also have higher injective dimensions. –  Hugh Thomas May 15 '10 at 4:09

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