There are several constructions of the Prüfer group $\mathbb{Z}/p^\infty$; here are two that are relevant for this question.

- It can be constructed via the short exact sequence $$ 0 \to \mathbb{Z} \to p^{-1} \mathbb{Z} \to \mathbb{Z}/p^\infty \to 0. $$
- It can be constructed as the colimit $$\mathbb{Z}/p^\infty = \operatorname*{colim}_k\mathbb{Z}/p^k\mathbb{Z}$$ where the homomorphisms are induced by multiplication by $p$.

It is not too hard to see that these are isomorphic.

There is a more complicated construction that often appears in topology. Now, let $R$ be a ring, and let $I = (x_0,\ldots,x_{n-1})$ be a finitely-generated ideal inside of $R$, generated by a regular sequence. Then, there is an $R$-module $R/(x_0^\infty,\ldots,x_{n-1}^\infty)$, defined in the following way, by first inverting $x_i$ and taking the cokernel, i.e., by the short exact sequences $$ 0 \to R \to x_0^{-1}R \to R/(x_0^\infty) \to 0 \\ 0 \to R/(x_0^\infty) \to x_1^{-1}R/(x_0^\infty) \to R/(x_0^\infty,x_1^\infty) \to 0 \\ 0 \to R/(x_0^\infty,x_1^\infty) \to x_2^{-1}R/(x_0^\infty,x_1^\infty) \to R/(x_0^\infty,x_1^\infty,x_2^\infty) \to 0 \\ \cdots $$

This is the analogue of construction (1) above.

I would like to propose a different construction, which is the analogue of (2). Let $I^k$ denote the $k$-th power of the ideal $I$. There is a system $$ \cdots \subset I^k \subset I^{k-1} \subset \cdots $$ which gives maps $$ \cdots \to R/I^k \to R/I^{k-1} \to \cdots $$

Now we work in the derived category, and let $DM = \mathbb{R}\text{Hom}_R(M,R)$ be the (derived) dual. The first claim is that $DR/I^k = \Sigma^{-n}R/I^k$. To see this, start with the cofiber sequence $$ R \xrightarrow{x_0} R \to R/(x_0); $$ this is self dual, and $DR/x_0 = \Sigma^{-1}R/(x_1)$. Induction using the cofiber sequences $$ R/(x_0,\ldots,x_{i-1}) \xrightarrow{x_i} R/(x_0,\ldots,x_{i-1}) \to R/(x_0,\ldots,x_i) $$ proves the claim for $R/I$. For $R/I^k$ inductively use the cofiber sequences $$ I^k/I^{k-1} \to R/I^k \to R/I^{k-1} $$ and the fact that $$ \bigoplus_{k \ge 0} I^k/I^{k+1} = R/I[x_1,\ldots,x_n] $$ to see inductively that $DR/I^k = \Sigma^{-n}R/I^k$.

Thus (after desuspension) we can get a sequence $$ \cdots \to R/I^{k-1} \to R/I^k \to \cdots. $$

Define $R/I^\infty$ to be the colimit of this system.

My question is the following:

Is $R/(x_0^\infty,\ldots,x_{n-1}^\infty) \simeq R/I^\infty$?

Here is an idea of how to prove this. Firstly, I believe one can show, by a cofinality argument, that $$ R/I^\infty = \operatorname{colim}_i R/(x_0^i,\ldots,x_{n-1}^i). $$

(Here I mean the colimit over the system of ideals $I_t = (x_0^t,\ldots,x_{n-1}^t)$).

Then to prove the claim note that it is clear that $R/(x_1^\infty) = \operatorname{colim}_i R/(x_0^i)$. Then inductively we have $$ R/(x_0^\infty,\ldots,x_{k}^\infty )= \operatorname{colim}_j(R/(x_0^\infty,\ldots,x_{k-1}^\infty)/x_k^j) \\ = \operatorname{colim}_j(\operatorname{colim}_i R/(x_0^i,\ldots,x_{k-1}^i)/x_k^j) \\ = \operatorname{colim}_j \operatorname{colim}_i R/(x_0^i,\ldots,x_{k-1}^i,x_k^j) \\ = \operatorname{colim}_i R/(x_0^i,\ldots,x_{k-1}^i,x_k^i) $$

By induction this should give the result, I hope.

Disclaimer: I originally asked this on math.se, but did not get a response.