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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?

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

Osofsky, B. L., Global dimension of valuation rings, Trans. Am. Math. Soc. 127, 136-149 (1967). ZBL0145.27602,

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

Osofsky, B. L., Rings all of whose finitely generated modules are injective, Pac. J. Math. 14, 645-650 (1964). ZBL0145.26601,

she proved that $\text{InjD}=0$ is equivalent to $d=0$.

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