Let $A= \bigoplus\limits\_{n=0}^{\infty}{A\_n}$ be an $\mathbb{N}$-graded algebra with semisimple $A\_0$. 



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> Question: Do we have that the global dimension of $A$ is equal to $\sup \{i \geq 0 | Ext\_A^i(A\_0,A\_0) \neq 0 \}$?
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Maybe one should ask this question under some mild further restrictions such as $A$ being noetherian and/or $A\_{i+j}=A\_i A\_j$ for all $i,j \geq 0$.


This question was asked in the special case of quadratic algebras here: [Quadratic algebras and Koszul algebras](https://mathoverflow.net/questions/317249/quadratic-algebras-and-koszul-algebras) and a positive answer would have a nice applications.



Global dimension of a graded algebra
Your question is answered in the affirmative in the paper


*Eilenberg, Samuel*, [**Homological dimension and syzygies**](http://dx.doi.org/10.2307/1969977), Ann. Math. (2) 64, 328-336 (1956); Errata 65, 593 (1957). [ZBL0073.26003](https://zbmath.org/?q=an:0073.26003).


According to Proposition 15, the category of graded left $A$-modules has certain nice properties whenever $A\_0$ is a semi-primary ring. Then Theorem 13 can be reformulated as


$$\operatorname{gldim} (A)=\sup \{i \geq 0 | \operatorname{Tor}^A\_i(A\_0,A\_0) \neq 0 \}.$$


It follows from Theorem 11 that


$$\sup \{i \geq 0 | \operatorname{Tor}^A\_i(A\_0,A\_0)\neq 0 \}\\=\sup \{i \geq 0 | \operatorname{Ext}\_A^i(A\_0,A\_0)\neq 0 \}.$$


