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I'm studying a bit of homological algebra and I'm now studying about the projective dimension of an $R$-module $M$.

This is how it is defined:

Since the category $R-\operatorname{Mod}$ has enough projectives, for any $R$-module $M$ we can write a chain complex like this:

$$\cdots \longrightarrow P_{n} \overset{d_{n}}{\longrightarrow} P_{n-1} \overset{d_{n-1}}{\longrightarrow} \cdots \overset{d_{2}}{\longrightarrow}P_1 \overset{d_{1}}{\longrightarrow}P_0 \overset{\epsilon}{\longrightarrow}M{\longrightarrow} \space 0$$

This induces the following chain complex:

$$0 \longrightarrow \operatorname{Hom}(P_{0},C) \overset{d_{1}'}{\longrightarrow} \operatorname{Hom}(P_{1},C) \overset{d_{2}'}{\longrightarrow} \cdots \overset{d_{2}}{\longrightarrow} \operatorname{Hom}(P_{n},C) \overset{d_{n}'}{\longrightarrow} \operatorname{Hom}(P_{n+1},C) {\longrightarrow} \cdots $$

where $d_{\space n}^{\space '}(f) = f\circ d_{\space n}$.

The n-th homology of this chain complex is defined as $\operatorname{Ext}^n(A,C)$.

The smallest number that $\operatorname{Ext}^n(A,C)=0$ for all $C$ is called the projective dimension of $M$ and is denoted as $\operatorname{pd}(M)$.

Now my question is whether $\operatorname{Ext}^n(A,C) = 0 \implies \operatorname{Ext}^{n+j}(A,C) = 0$ holds for $j \in \mathbb{N}$ or not?

math.n00b
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  • To clarify, you're asking whether $(\forall C, \operatorname{Ext}^n(A,C) = 0) \implies (\forall C, \operatorname{Ext}^{n+1}(A,C))$ is true? The universal quantifiers are important here. – Najib Idrissi Nov 16 '14 at 13:05
  • Well, let's interpret it this way: If $\operatorname{Ext}^n(A,C) = 0$, for an arbitrary $C$, does it imply that $\operatorname{Ext}^{n+j}(A,C) = 0$ as well? That's what I mean. – math.n00b Nov 16 '14 at 13:11
  • The quantifiers are still unclear. – Martin Brandenburg Nov 16 '14 at 13:12
  • Also, the definition of the projective dimension is wrong. – Martin Brandenburg Nov 16 '14 at 13:13
  • @MartinBrandenburg: What is the correct definition? What is unclear exactly? – math.n00b Nov 16 '14 at 13:14
  • Okay so you take some specific $C$, fixed from the beginning, and you ask if $\operatorname{Ext}^{n+1}(A,C) = 0$ when $\operatorname{Ext}^n(A,C) = 0$ (same $C$ in both cases)? This is false. – Najib Idrissi Nov 16 '14 at 13:17
  • The definition of projective dimension can be found in every book on homological algebra. It's the minimal $n$ such that $M$ has a projective resolution of length $n$. It is also the minimal $n$ such that $\mathrm{Ext}^m(M,-)=0$ for all $m>n$. (Not $m \geq n$ or even just $m=n$.) It turns out that this is equivalent to $\mathrm{Ext}^n(M,-) \neq 0$ and $\mathrm{Ext} ^{n+1}(M,-)=0$. This already implies $\mathrm{Ext}^m(M,-)=0$ for all $m>n$. – Martin Brandenburg Nov 16 '14 at 13:17
  • @NajibIdrissi: Yes. $C$ is fixed from the beginning, but it is an arbitrary module. Does it make sense now? Or I'm still unable to convey what I have in mind? – math.n00b Nov 16 '14 at 13:20
  • Then this is wrong as $\hom(\mathbb{Z}/2\mathbb{Z}, \mathbb{Z}) = 0$ but $\operatorname{Ext}(\mathbb{Z}/2\mathbb{Z}, \mathbb{Z}) = \mathbb{Z}/2\mathbb{Z}$. – Najib Idrissi Nov 16 '14 at 13:23
  • @NajibIdrissi: So you're using the convention that $\operatorname{Ext}^0 = \operatorname{Hom}$? What if we exclude $\operatorname{Ext}^0$? It's still wrong? – math.n00b Nov 16 '14 at 13:25
  • It's not a convention, it's a fact... And yes, it's still wrong, but the examples are more contrived because the base ring can't be a PID (in particular not $\mathbb{Z}$). – Najib Idrissi Nov 16 '14 at 13:25
  • @MartinBrandenburg: Would you elaborate on why "this already implies $\mathrm{Ext}^m(M,-)=0$ for all $m > n$"? That's exactly my question. – math.n00b Nov 16 '14 at 13:27
  • @NajibIdrissi: Yes. Now I understand it completely. And I see why it's a fact, not a convention... Thank you. – math.n00b Nov 16 '14 at 13:31

1 Answers1

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The correct definition of the projective dimension of $M$ is the smallest $d$ such that $M$ has a projective resolution of length $d$, or equivalently as the smallest $d$ such that $\mathrm{Ext}^{d+1}(M,-)=0$. (And hence $\mathrm{Ext}^d(M,-) \neq 0$.)

The implication $\mathrm{Ext}^n(M,-)= 0 \Rightarrow \mathrm{Ext}^{n+1}(M,-)=0$ is true. See Weibel's book on homological algebra, Lemma 4.1.6.

However, if $C$ is a fixed module, then $\mathrm{Ext}^n(M,C)=0 \Rightarrow \mathrm{Ext}^{n+1}(M,C)=0$ might fail. For example, we have (for some fixed integer $p>1$) $\mathrm{Ext}^0(\mathbb{Z}/p,\mathbb{Z})=\hom(\mathbb{Z}/p,\mathbb{Z})=0$, but $\mathrm{Ext}^1(\mathbb{Z}/p,\mathbb{Z})=\mathbb{Z}/p$.

Martin Brandenburg
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  • Would you please write down why $\mathrm{Ext}^n(M,-)= 0 \Rightarrow \mathrm{Ext}^{n+1}(M,-)=0$ is true? I just searched on my university's library and we don't have that book in there, so I can't have access to it. – math.n00b Nov 16 '14 at 13:40
  • Google "homological algebra lecture notes" and in the pdfs search for "projective dimension". (I could do that for you and post a link, but you can also do it.) – Martin Brandenburg Nov 16 '14 at 13:42
  • I tried to find that on google and checked several lecture notes about homological algebra, none of them proved why $\mathrm{Ext}^n(M,-)= 0 \Rightarrow \mathrm{Ext}^{n+1}(M,-)=0$ holds. Is the proof long? – math.n00b Nov 16 '14 at 14:04
  • http://www.math.unam.mx/javier/weibel.pdf (sorry Charles) – Martin Brandenburg Nov 16 '14 at 14:22
  • Then it all boils down to why dimension shifting holds... Why $\mathrm{Ext^{n+1}(A,B)} \cong \mathrm{Ext^{1}(M_d,B)}$? – math.n00b Nov 16 '14 at 14:37
  • This is all explained in the book (and in any other book on the subject). – Martin Brandenburg Nov 16 '14 at 19:07