Is it true that:
Let $R$ be a local ring and $\dim R= d$. If $b\subset a$ be two proper ideals of $R$ then for $ n\in {\Bbb N}$, $\varinjlim Ext^d_R(a^n/b^n,R)=0$
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1$\begingroup$ are there any general theorems on what the limit $a^n/b^n$ should look like in a local ring? (I mean, if you take an injective resolution for R then that thing becomes the dth cohomology of some complex. lim is exact (right?) so you might as well take the limit before computing cohomology. (just a shot in the dark) $\endgroup$– Yosemite SamCommented Feb 27, 2012 at 9:27
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2$\begingroup$ I don't know if this is helpful, but if you had $\mathfrak{a}=R$, then you are simply computing here (the cohomology of) $R\Gamma_{\mathfrak{b}} R$ which is just the infinite Koszul complex of $\mathfrak{b}$. $\endgroup$– the LCommented Feb 27, 2012 at 10:11
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5$\begingroup$ It isn't helpful. $\endgroup$– StellaCommented Feb 27, 2012 at 19:34
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1$\begingroup$ Just so I understand, when you are taking your limit, you really mean direct limit, not inverse limit? Can I confirm what the maps are? Is it the natural map $a^n/b^n \to a^{n-1}/b^{n-1}$ defined by the composition: $$a^{n}/b^n \to a^{n-1}/b^n \to a^{n-1}/b^{n-1}?$$ Where the first map is induced by the inclusion $a^n \subseteq a^{n-1}$ and the second by $b^n \subseteq b^{n-1}$? Wouldn't these form an inverse system? Sorry, I just want to understand the problem. Or I suppose you could maps in a direct system by some choice of generators of the ideals perhaps? $\endgroup$– Karl SchwedeCommented Feb 28, 2012 at 0:04
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$\begingroup$ $\varinjlim$ means direct limit and $\varprojlim$ means inverse limit. I think the following short exact sequence can be helpful. $0\rightarrow a^n/b^n\rightarrow R/b^n\rightarrow R/b^n\rightarrow 0$ $\endgroup$– StellaCommented Feb 28, 2012 at 6:03
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2 Answers
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I prove your question. Since the short exact sequence $$0 \to \mathfrak{a}^n/\mathfrak{b}^n \to R/\mathfrak{b}^n \to R/\mathfrak{a}^n \to 0$$ we have the following exact sequence $$\cdots \to \mathrm{Ext}^d_R(R/\mathfrak{a}^n,R) \to \mathrm{Ext}^d_R(R/\mathfrak{b}^n,R) \to \mathrm{Ext}^d_R(\mathfrak{a}^n/\mathfrak{b}^n,R) \to \cdots $$ Passing to the limit we have the exact sequence
$$\cdots \to H^d_{\mathfrak{a}}(R) \to H^d_{\mathfrak{b}}(R) \to \lim \mathrm{Ext}^d_R(\mathfrak{a}^n/\mathfrak{b}^n,R) \to H^{d+1}_{\mathfrak{a}}(R) \cong 0$$
by the Grothendieck vanishing theorem. So it is sufficient to show that if $\mathfrak{b} \subseteq \mathfrak{a}$ then
$$H^d_{\mathfrak{a}}(R) \to H^d_{\mathfrak{b}}(R) $$
is surjecitve. Of course, we may assume that $\mathfrak{a} = \mathfrak{b} + (x)$ for some element $x \in \mathfrak{m}$. By [Brodmann-Sharp: local cohomology, Proposition 8.1.2] we have that
$$\cdots \to H^d_{\mathfrak{a}}(R) \to H^d_{\mathfrak{b}}(R) \to H^d_{\mathfrak{b}}(R_x)$$
Now the claim follows from the fact $\dim R_x < d$ and the Grothendieck vanishing theorem.
answered Mar 5, 2012 at 21:56
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Assume $R$ is a noetherian ring (not necessarily local).
As mentioned in the comments, there is a short exact sequence:
$ 0 \to a^n/b^n \to R/b^n \to R/a^n \to 0$
Let $R \to I$ be an injective resolution. Then we get an exact sequence of complexes
$0 \to Hom_R(R/a^n,I) \to Hom_R(R/b^n,I) \to Hom_R(a^n/b^n,I) \to 0$.
Passing to the limit, and remembering that taking direct limits is an exact operation, we get an exact sequence
$0 \to \varinjlim Hom_R(R/a^n,I) \to \varinjlim Hom_R(R/b^n,I) \to \varinjlim Hom_R(a^n/b^n,I) \to 0$.
Now, the leftmost complex is quasi-isomorphic to $R\Gamma_a I = R\Gamma_a R = K^\infty(a)$.
Similarly, the middle complex is quasi-isomorphic to $K^{\infty}(b)$.
Thus, we get an exact sequence of complexes
$0 \to K^\infty(a) \to K^\infty(b) \to \varinjlim Hom_R(a^n/b^n,I) \to 0$.
Now, if $a$ is generated by $n$ elements, then $K^{\infty}(a)$ lives in degrees $0$ to $n$. Thus, I believe that the question of your $Ext$ being non-zero depends on the minimal number of generators of $a$ and $b$.
