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Let $R$ be a noetherian ring and $M$ a finitely generated $R$-module with projective dimension $n$. Then for every finitely generated $R$-module $N$ we have $\mathrm{Ext}^n(M,N)\neq 0$. Why?

By definition, if the projective dimension is $n$ this means that $\mathrm{Ext}^{n+1}(M,-)=0$ and $\mathrm{Ext}^n(M,-)\neq 0$, so there exists an $N$ such that $\mathrm{Ext}^n(M,N)\neq 0$. Why is this true for every $N$?

I found this theorem on this notes, page 6 proposition 9. Did I misunderstand it? Is there a similar statement?

This result is used on page 41, implication 3 implies 4, and is used in the case $N=R$. Is it true in this case?

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What you say is not true for $N=0$, and $0$ is finitely generated. –  user26857 Oct 10 '12 at 15:09
    
I found this theorem on some notes (the link is in the question), did I misunderstand it? is there a similar statement? –  Chris Oct 10 '12 at 21:07
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it' written in the question, page 6 prop 9 –  Chris Oct 10 '12 at 21:23
    
this theorem is used on page 41 for the implication 3 implies 4 in the special case $N=R$, is it true in this case? –  Chris Oct 10 '12 at 21:48
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3 Answers 3

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Suppose $R$ is local noetherian with maximal ideal $\def\m{\mathfrak m}\m$ and residue field $k$, and let $M$ be a f.g. $R$-module with $\operatorname{pdim}M=n$. Then $\def\Ext{\operatorname{Ext}}\Ext_R^{n}(M,k)\neq0$.

Let $N$ be a non-zero f.g. module. There is a submodule $N'\subseteq N$ such that $N/N'\cong k$ as $R$-modules. From the short exact sequence $0\to N'\to N\to k\to 0$ we get an exact sequence $$\Ext^n_R(M,N)\to\Ext^n_R(M,k)\to\Ext^{n+1}_R(M,N')=0$$ Since the middle term is not zero and the leftmost map is surjective, we see that $\Ext^n_R(M,N)\neq0$.

Remove now the hypothesis that $R$ be local and let again $M$ be finitely generated and of projective dimension $n$. There is a maximal ideal $\def\m{\mathfrak m}$ in $R$ such that $\def\pdim{\operatorname{pdim}}\pdim_RM=\pdim_{R_\m}M_\m$.

Then $\def\Ext{\operatorname{Ext}}\Ext^n_R(M,R)_\m\cong\Ext^n_{R_\m}(M_m,R_m)\neq0$, by the local case.

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this theorem is used on page 41 for the implication 3 implies 4 in the special case $N=R$. Anyway in that theorem we don't have in the hypothesis $R$ local. By the way I think in your answer you want $N\neq0$ –  Chris Oct 10 '12 at 21:49
    
You think or are you sure? :-) –  Mariano Suárez-Alvarez Oct 10 '12 at 21:51
    
Dear Chris, it should be obvious by now that the theorem as stated is false. –  Mariano Suárez-Alvarez Oct 10 '12 at 21:53
    
yeah yeah I understood that. I'm trying to understand now another thing, if we know that the projective dimension of $M$ is less or equal to $n+1$ and we know that $Ext^{n+1}(M,R)=0$ is it true that the projective dimension of $M$ is less or equal to $n$? (Even if $R$ is not local) –  Chris Oct 10 '12 at 21:56
    
Let's see if I understood, in general we cannot use the local case if we have $N$ at the place of $R$ because we don't know that the maximal ideal that gives us the equality for the projective dimensions is in the support of $N$? –  Chris Oct 10 '12 at 22:35
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Take $R=\mathbb Z\times\mathbb Z$, consider the elements $e_1=(1,0)$, $e_2=(0,1)\in R$, and the modules $M=R/(2e_1)$ and $N=Re_2$. Show that the projective dimension of $M$ is $1$ and compute $\operatorname{Ext}_R^1(M,N)$.

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You could take $R=k\times k$, with $k$ a field, $M=Re_1$ and $N=Re_2$, too; here the projective dimension is zero; this is a simpler example, I guess. –  Mariano Suárez-Alvarez Oct 10 '12 at 6:17
    
These two examples have $R$ a non-domain; an example with a domain is with $R=k[X]$, $M=R/(X)$ and $N=R/(X-1)$. –  Mariano Suárez-Alvarez Oct 10 '12 at 6:19
    
I don't understand what these examples are supposed to show, I was asking for a proof that if the projective dimension of $M$ is $n$ then $Ext^n(M,N)\neq0$ for every $N$. I haven't computed those Exts yet, are you claiming they are zero and my statement is false? –  Chris Oct 10 '12 at 7:57
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Because he did not follow the suggestion I wrote in the question itself, and in my comment above! –  Mariano Suárez-Alvarez Oct 10 '12 at 21:03
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I just made sure I picked modules with disjoint support. One easy way to do this (and this gives the two first examples) is to take the spectrum not connected. –  Mariano Suárez-Alvarez Oct 11 '12 at 2:20
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Yes, for $N=R$ the result is true.

Let $R$ be a noetherian ring and $M$ a finitely generated $R$-module with $\mathrm{pd}_R(M)=n$. Then $\mathrm{Ext}_R^n(M,R)\neq 0$.

Proof. First consider $R$ local with maximal ideal $\mathfrak{m}$. Let $F_{\bullet}$ be a minimal free resolution of $M$ and $\varphi_n:F_n\rightarrow F_{n-1}$ the last homomorphism. One knows that the elements of the matrix associated to $\varphi_n$ are all in $\mathfrak{m}$. $\mathrm{Ext}_R^n(M,R)$ is the cokernel of the dual map $\varphi_n^*:F_{n-1}^*\rightarrow F_n^*$ whose matrix is the transpose of the matrix associated to $\varphi_n$. Therefore the elements of the matrix associated to $\varphi_n^*$ are also in $\mathfrak{m}$, and this shows that $\mathrm{Coker}\ \varphi_n^*\neq 0$.

Now it is easy to deduce the general case from the local case via the well known fact that $\mathrm{pd}_R(M)= \sup_{\mathfrak{m}\in\mathrm{Max}(R)}\mathrm{pd}_{R_{\mathfrak{m}}}(M_{\mathfrak{m}})$.

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(You do need the extra fact that the sup is a max) –  Mariano Suárez-Alvarez Oct 11 '12 at 2:16
    
that comes from the fact that $pd_{R_m}\;M_m\leq\;depth\;R_m\leq\;dim\;R_m\leq\;dim\;R$ and from the fact that it is a sup of natural numbers, is that right? –  Chris Oct 11 '12 at 6:01
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