0

Why do some modules have finite projective dimension and some don't (if we consider modules over the same ring)? What factors does it depend on?

Kelly
  • 103
  • 2
    Surely it depends on the ring. Over a field every module is projective. Over a PID every finitely generated module has a finite free resolution. Over a polynomial ring we have the chain of syzygies theorem. I'm afraid I don't know enough to say more. – Jyrki Lahtonen Feb 05 '22 at 20:44

1 Answers1

1

It's really not clear to me what kind of answer you are expecting. It's almost like asking "why are some modules finitely generated and some not". The answer should be to convince yourself from examples to see that some modules are "larger" than others.

Your question is not too different. Over any ring $A$, there exist modules with finite projective dimension: Take its regular representation ${}_A A$. On the other hand, there are rings which exhibit infinite projective dimension. By definition, these are those rings $R$ with infinite global dimension $\operatorname{gl.dim}(R) = \infty$. You can try to come up with examples yourself or find plenty of them on the internet. For example, a three-line proof by Mariano Suárez-Álvarez in MSE/23624 shows $\operatorname{proj.dim}_{\mathbb{Z}/(p^2)}(\mathbb{Z}/(p)) = \infty$.

Qi Zhu
  • 7,999