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Is a (say, projective) resolution (of a module) consisting entirely of zero modules considered to have a length (of zero) at all? I think this possibility causes problems in some books.

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closed as off-topic by Ricardo Andrade, Andrey Rekalo, Stefan Kohl, Chris Godsil, David White Nov 28 '13 at 15:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Andrey Rekalo, Stefan Kohl, Chris Godsil, David White
If this question can be reworded to fit the rules in the help center, please edit the question.

I can't see what the point of this question is. Which books? What problems? Why are you worried about this? – Ben Webster May 2 '10 at 15:41
As only the zero module has a projective resolution consisting entirely of zero modules, I cannot see this is worth worrying about. – Robin Chapman May 2 '10 at 15:41
Matsumura (Commutative Ring Theory) defines projective dimension to be the least (nonnegative) integer d where P_d is nonzero for every resolution P. This definition certainly doesn't work for the zero module (its projective dimension will be 1 according to this definition). – ashpool May 3 '10 at 0:34
This is a tiny tea cup, as tea cups go... – Mariano Suárez-Alvarez May 3 '10 at 1:22
It might be a tiny teacup but if one is just learning these things then one does sometimes get confused. Matsumura is usually super-super precise and if there really is a glitch in his definition of PD then that is a little surprising. I got very confused recently about an assertion in a paper that said "k[x,y]/(f) clearly has dimension at least 1" and then later on "...hence f can't be a unit because the zero ring has dimension 0". I contacted the author and they agreed that the logic was faulty and supplied me with a direct proof that f wasn't a unit. It's "empty set theory" butit'simportant – Kevin Buzzard May 3 '10 at 8:28

This is answered in my answer to a related question of yours.

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