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

  1. It can be constructed via the short exact sequence $$ 0 \to \mathbb{Z} \to p^{-1} \mathbb{Z} \to \mathbb{Z}/p^\infty \to 0. $$
  2. 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.

  • $\begingroup$ The first construction is problematic if $x_0,\dots, x_k$ is not a regular sequence. $\endgroup$ – Fernando Muro Jun 17 '15 at 13:35
  • 1
    $\begingroup$ @FernandoMuro: Yes, the sequence $x_0,\ldots,x_{n-1}$ is assumed to be regular $\endgroup$ – Drew Heard Jun 17 '15 at 14:10
  • $\begingroup$ Are the maps of the direct system $R/I_t\rightarrow R/I_{t+1}$ the maps induced by the multiplication with $x_0\ldots x_{n-1}$ (the direct system whose colimit $R/I^\infty$)? If so, I guess that your question has a positive answer. $\endgroup$ – Aurora Jun 17 '15 at 18:02

I strongly guess that, you mean, the maps of your direct system, $$R/I_t\rightarrow R/I_{t+1},$$ whose colimit is, $$R/I^\infty,$$ are the multiplication by $x_0\ldots x_{n-1}$. So I think that your question has a positive answer. For a complete description I aim to firstly give some definitions. The following definitions are given in the paper Sharp and Zakeri.

Let $R$ be a ring and $U$ be a non-empty subset of $R^n$ where $n\in \mathbb{N}$. Then $U $ is said to be a traingular subset of $R^n$ if it satisfies the following two conditions:

(1) for each $(u_1,\ldots,u_n)\in U$ and $\alpha_1,\ldots,\alpha_n\in \mathbb{N}$ we have $(u_1^{\alpha_1},\ldots,u_n^{\alpha_n})\in U$

(2) for each $(u_1,\ldots,u_n),(v_1,\ldots,v_n)\in U$ there exists $n\times n$ lower triangular matrices $H,K$ with entries in R and $(w_1,\ldots,w_n)\in U$ such that $H[u_1,\ldots,u_n]^\tau=[w_1,\ldots,w_n]^\tau=K[v_1,\ldots,v_n]^\tau$ (here the notation $\tau$ stands for the transpose of matrix).

Now fix a triangular subset $U$ of $R^n$. Let $M$ be an $R$-module. Then one can define an equivalance relation on $U\times M$ such that $((u_1,\ldots,u_n),m)$ is equivalent to $((v_1,\ldots v_n),n)$ if and only if there exists $(w_1,\ldots,w_n)\in U$ and lower triangular matrices $H,K$ such that, $$ H[u_1,\ldots,u_n]^\tau=[w_1,\ldots,w_n]^\tau=K[v_1,\ldots,v_n]^\tau $$ and $\text{det}(H)m-\text{det}(K)n\in (w_1,\ldots,w_{n-1})M$.

By this definition we get an $R$-module denote by $U^{-n}(M)$ which is called the module of generalized fractions. In fact one can verify that this definition generalizes the definition of the localization of an $R$-module at a multiplicative closed subset of $R$.

Now assume that $x_1,\ldots,x_n$ is sequence of elements of $R$. Extend this sequence to an infinite sequence, $x_1,\ldots,x_n,1,1,\ldots,1,\ldots$ ($x_i=1$ for $i\ge n+1$). Then for each $i$ we get the triangular subset, $$U_i:=\{(x_1^{\alpha_1},\ldots,x_i^{\alpha_i});\text{ for some } 0\le j\le i, \alpha_1,\ldots,\alpha_j\in \mathbb{N} \text{ and } ,\alpha_{j+1}=\ldots=\alpha_i=0\},$$ of $R^i$. In particular for each $1\le i\le n$ we get the $R$-module $U_i^{-i}(R)$ of generalized fractions which form the complex, $$ 0\rightarrow R\rightarrow U_1^{-1}(R)\rightarrow \ldots \rightarrow U_i^{-i}(R) \overset{e^i}{\rightarrow}U_{i+1}^{-i-1}(R)\rightarrow\ldots,$$ in which the differential $e^i$ are defined by,

$$ e^i(r/(u_1,\ldots,u_i))=r/(u_1,\ldots,u_i,1).$$

In 2.2 Theorem of the paper On the Generalized Fractions of Sharp and Zakeri it is proved the above complex is isomorphic to the complex which is the same as your first construction, i.e. considering the cokernel of the localization map at element $x_i$ at each step.

Now in your situation we have the sequence $x_0,\ldots,x_{n-1}$ and the above mentioned Theorem states that $U_{n+1}^{-n-1}(R)\cong R/(x_0^\infty,\ldots,x_{n-1}^\infty)$ (with your notation). Now one can prove that the module of generalized fractions $U^{-n-1}_{n+1}(R)$ is nothing but your direct limit $\lim_{\rightarrow} R/(x_0^i,\ldots,x_{n-1}^i)$ in which the maps $R/(x_0^i,\ldots,x_{n-1}^i)\rightarrow R/(x_0^j,\ldots,x_{n-1}^j)$ are multiplication maps by $x_0^{j-i}\ldots x_{n-1}^{j-i}$ (In order to prove this isomorphism you need to use Lemma 2.3 of the paper Modules of Generalized Fractions. Using this lemma you can reduce an arbitrary lower triangular matrix to a diagonal matrix whose diagonal is something like $x_0^t,\ldots,x_{n-1}^t,1$)). In fact your question has still an affirmative answer even if we drop the condition for the sequence to be a regular sequence. The modules you are interested are called the top local cohomology modules supported at the ideal $(x_0,\ldots,x_{n-1})$.

By virtue of the paper Comparison of Certain Complexes of Modules of Generalized Fractions and Čech Complexes, over a Noetherian ring, it is always possible even to compute also the other local cohomology modules by means of the above complex.

  • $\begingroup$ @Drew Heard, You are welcome. Was the answer satisfactory? I asked because you didn't approve it. By the way, I would be so delighted if you could possibly please describe, briefly, the topological view of your question, if it has a topological view. $\endgroup$ – Aurora Jun 21 '15 at 13:55
  • $\begingroup$ I've kept it open in the hope that someone comes along and tells me that the sketch I outlined in the question is correct! I'm still working through the answer you gave; it has quite a few references to sort through. There is a topological nature to the question, which unfortunately I cannot adequately describe in the comments. $\endgroup$ – Drew Heard Jun 24 '15 at 15:40

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