6
$\begingroup$

Is it possible that $\mbox{Tor }^{r+1}(M,N)=0 \ \ \forall N$ yet $\mbox{proj. dim }M>r$?

What I do know is that if $(A,\mathfrak{m})$ is Noetherian local and $M$ is finitely generated over $A$ then $\mbox{Tor }^{r+1}(M,A/\mathfrak{m})=0 $ if and only if $\mbox{proj. dim }M\leq r$.

Generally speaking, is $\mbox{Tor }$ functor as good a tool to measure projective dimension as $\mbox{Ext }$ even when the ring/module is not Noetherian or local?

I suspect we can use $\mbox{Tor }$ to measure projective dimension when ring is Neotherian local and module is finitely generated because flatness and projectivity coincide in such case.

Improve this question
$\endgroup$

3 Answers 3

Reset to default
10
$\begingroup$

Over the integers, the rational numbers Q are flat and so Tor^i(Q,M) = 0 for all M and all i>0. However Q is not projective so has projective dimension 1.

Improve this answer
$\endgroup$
1
  • $\begingroup$ You know $\mathbb{Q}$ has projective dimension 1 and not say greater than 1 by Hailong's answer below. An alternate reason (and to me much easier to see) is the fact that $\mathbb{Z}$ is a PID so it has global dimension 1, so any non-projective $\mathbb{Z}$-module has projective dimension 1. I suppose you could also just write down a projective resolution of $\mathbb{Q}$ $\endgroup$
    – David White
    Commented Jun 28, 2011 at 22:28
11
$\begingroup$

There's another dimension, called flat dimension: $\mathrm{fd}\; M_R = n$ means that $n$ is the smallest integer such that there exists a resolution $$0 \to F_n \to \cdots \to F_1 \to F_0 \to M \to 0$$ where each $F_i$ is a flat module. We have $\mathrm{fd}\; M_R \leq d$ if and only if $\mathrm{Tor}^R_{d+1}(M, N) = 0$ for all $_RN$. Since projective modules are flat, we have $\mathrm{fd}\; M_R \leq \mathrm{pd}\; M_R$, but as Richard Borcherds's answer shows, there are modules for which equality doesn't hold. However, if $R$ is right noetherian and $M_R$ is finitely generated, then $\mathrm{pd}\; M_R = \mathrm{fd}\; M_R$.

Here are some references:

  • Chapter 7 of Noncommutative noetherian rings by McConnell and Robson.
  • Chapter 4 of An introduction to homological algebra by Weibel.
Improve this answer
$\endgroup$
11
$\begingroup$

Since your question is really about projective dimension of flat modules, it is worth noting the following result (see Raynaud-Gruson MR0308104, Cor 3.3.2 or Jensen MR0407091, Thm 5.8) which complements Richard's example:

The projective dimension of a flat module over a commutative Noetherian ring $R$ is bounded by $n+1$ if the cardinality of $R$ is at most $\aleph_n$.

EDIT: It looks like commutativity is not needed!

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ The link to springerlink.com is broken. I'm also unable to find any copy saved on the Wayback Machine. $\endgroup$
    – The Amplitwist
    Commented Jul 29, 2022 at 16:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .