Would it be $pd_{R}M:=sup\lbrace i\vert Ext^{i}_{R}(M,N)\neq 0:N$ is an $R$-module$\rbrace$ or $pd_{R}M:=sup\lbrace i\vert Ext^{i}_{R}(M,N):N\neq 0$ is a free $R$-module$\rbrace$ or something else entirely?
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My mistake, I have edited what I wrote. – Rhoswyn Nov 28 '18 at 15:43
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We can use the following equivalence to establish the definition:
$\text{Ext}_R^{n}(M,N) = 0$ for all $N$ iff $\text{pd}_R M < n$
This is probably in any textbook discussing homological dimensions, also conveniently in these notes.
From here, a definition easily pops out:
definition: $\text{pd}_R M = \inf\big\{n \mid \text{Ext}_R^{n+1}(M,N) =0 \text{ for all $N$}\big\} = \sup\big\{n \mid \text{Ext}_R^{n}(M,N) \not=0 \text{ for some $N$}\big\}$
Badam Baplan
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Yes you certainly can, and the translation between the inf and sup definitions is immediate. Edited. – Badam Baplan Nov 28 '18 at 16:15