Let $A= \bigoplus\limits_{n=0}^{\infty}{A_n}$ be an $\mathbb{N}$-graded algebra with semisimple $A_0$.

Question: Do we have that the global dimension of $A$ is equal to $\sup \{i \geq 0 | Ext_A^i(A_0,A_0) \neq 0 \}$?

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 and a positive answer would have a nice applications.


1 Answer 1


Your question is answered in the affirmative in the paper

Eilenberg, Samuel, Homological dimension and syzygies, Ann. Math. (2) 64, 328-336 (1956); Errata 65, 593 (1957). ZBL0073.26003.

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

$$\sup \{i \geq 0 | \operatorname{Tor}^A_i(A_0,A_0)\neq 0 \}\\=\sup \{i \geq 0 | \operatorname{Ext}_A^i(A_0,A_0)\neq 0 \}.$$

  • $\begingroup$ I wrote this answer at work and didn't realize the reference was behind a paywall. (Why is a paper from 1956 behind a paywall?) $\endgroup$ Dec 12, 2018 at 15:35

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