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## lim Ext(a^n/b^n,R)=0

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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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) – Yosemite Sam Feb 27 2012 at 9:27
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 $\mathfrak{b}$. – Liran Shaul Feb 27 2012 at 10:11
It isn't helpful. – anna Feb 27 2012 at 19:34
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? – Karl Schwede Feb 28 2012 at 0:04
$\varinjlim$ means direct limit and $\varprojlim$ means inverse limit. I think the following short exact sequence can be helpful. $0\rightarrow a^n/b^n\rightarrow R/b^n\rightarrow R/b^n\rightarrow 0$ – anna Feb 28 2012 at 6:03

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.

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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 $b$.

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