Question

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}$, $\varinjlim Ext^d_R(a^n/b^n,R)=0$

Answer 1

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.

Answer 2

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 $b$.