# Relation between global dimension and sup of injective dimensions of $R/I$

Denote $$d=\operatorname{r.gl.dim}(R)$$ the right global dimension of a ring $$R$$ which is defined by sup over either injective dimension of all modules or projective dimension of all modules.

Set $$\operatorname{InjD}=\sup\{\operatorname{id}(R/I)\mid I\subset R\}$$ where $$I$$ runs through all ideals of $$R$$.

Clearly $$\operatorname{InjD}\leq d$$ as we only runs through modules of the form $$R/I$$.

$$\textbf{Q:}$$ Do I have $$\operatorname{InjD}=d$$ here?

• @JeremyRickard Ah, that was my bad. It should have been $InjD\leq d$. Thanks for the reference. – user45765 Aug 14 at 14:04

Osofsky proved that if $$R$$ is right Noetherian (Theorem C) or right perfect (Theorem B), then $$\text{InjD}=d$$ (using your notation), but that if $$1\leq n\leq\infty$$, then there is a non-Noetherian ring with $$\text{InjD}=1$$ and $$d=n$$.
In the last result, $$\text{InjD}=1$$ can't be improved to $$\text{InjD}=0$$, because in another paper
she proved that $$\text{InjD}=0$$ is equivalent to $$d=0$$.