Question: Show gl dim$R=\sup\{pd_RI\}+1$, where I runs through all right ideals of $R$, where $R$ is not Wedderburn.
Thoughts: I know that gl dim$R=1$ iff $R$ is a hereditary ring, but not Wedderburn. So, I am seeing where the question is "coming from", but I am not quite seeing a way to, sort of, break up $R$ in the question. Do we just break it up into its hereditary part and its non-hereditary part? Or, could we maybe play around with short exact sequences..? Any thoughts, ideas, or help is greatly appreciated!
This result was probably first stated in a On the Dimension of Modules and Algebras (III): Global Dimension by Maurice Auslander (as is often the case for theorems on homological dimensions).
The proof relies on two crucial results from homological algebra. Let me state everything for left ideals, you may recover your result for right ideals by just passing to the dual ring.
Theorem 1 (Auslander). Let $R$ be a ring, then $$\operatorname{gl.dim}(R) = \sup \{\operatorname{pd}_R(M) \mid M \text{ cyclic } R \text{-module} \} = \sup \{\operatorname{pd}_R(R/I) \mid I \subseteq R \text{ left ideal} \}.$$
This is a non-trivial result and relies on the Well-Ordering Theorem.
Lemma 2. Let $0 \to U \to P \to M \to 0$ be a short exact sequence with $P$ projective and $U$ non-projective. Then, $\operatorname{pd}(M) = \operatorname{pd}(U) + 1 $.
You can find references for these two results in the paper by Auslander. Both only use basic homological results.
Moreover, from Theorem 1, it follows that $R$ is semisimple if and only if $R/I$ is projective for every left ideal $I \subseteq R$. So if $R$ is not semisimple, there exists an ideal $I \subseteq R$ such that $R/I$ is not projective and therefore $\operatorname{pd}_R(R/I) = \operatorname{pd}_R(I) + 1$ for such $I$. Applying Theorem 1 yields your desired result $$\operatorname{gl.dim}(R) = \sup_I\{\operatorname{pd}_R(I) \} + 1.$$