Let $R$ be a nonNoetherian ring. Is its left global dimension ${\rm{lD}}(R)$ equal to $\sup \{ {\rm{id}}(M) \mid M \text{ is a cyclic $R$module} \}$? Here $\rm{{id}}(M)$ denotes the injective dimension of $M$.

$\begingroup$ 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 Bmodule. Then you could give A any properties you wanted which would transfer to A\oplus B (such as higher injective dimension and nonNoetherianness). $\endgroup$– Greg MullerMay 11, 2010 at 13:36

$\begingroup$ 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. $\endgroup$– Hugh ThomasMay 15, 2010 at 4:09
1 Answer
The answer is no (in general) according to B. L. Osofsky, Global dimension of valuation rings, Corollary 3. For any $1 \leq n \leq \infty$, there are examples where $$\sup \{ {\rm{id}}(M) \mid M \text{ is a cyclic $R$module} \}=1$$ and ${\rm{lD}}(R)=n$.