How do we interpret a zero sequence in the context of ideal theory?

Question

We already know and may simply understand the definition of a zero sequence in $\mathbb{Q}$ - it is just a sequence, which converges towards $0$.

Given the context of ideal theory, let $R$ be a ring and $I$ an ideal. In the ring $R^\mathbb{N}=\prod_{n\in\mathbb{N}}R$, which is the repeated direct product of $R$ with itself, a sequence $(x_n)_{n\in\mathbb{N}}$ is called a zero sequence if for every $s\in\mathbb{N}$ there exist a $N\in\mathbb{N}$ (depending on $s$) such that $x_n\in I^s$ for all $n>N$.

How do we interpret/explain such a sequence, for example by using the ring of integers and the ideal containing only even integers? Let us take $s=5$. The ideal $I^s$ will contain all sequences (in this case quintuples) of even integers. How do we interpret $N(s)=N(5)$ and a zero sequence $(x_n)$ in this example?

Why do I need this? This would substantially help to understand the following definition for completion of a ring: Let $R$ be a ring, $I$ an indeal, $I_{ZS}$ the ideal of all zero sequences in $R^\mathbb{N}$, and $S_{CS}$ the subring of $R^\mathbb{N}$ containing all Cauchy sequences. The quotient ring $\hat R_I:=S_{CS}/I_{ZS}$ is called the completion of $R$ with respect to $I$. $S_{CS}/I_{ZS}$ is the residue class ring of $S_{CS}$ modulo $I$.

Answer 1

Consider a commutative ring $R$ with an ideal $I.$ One other way to define the completion $\widehat R_I$ of $R$ with respect to $I$ is as the inverse limit $$\widehat R_I \stackrel{\text{def}} = \varprojlim(R / I^i) \stackrel{\text{def}} = \{(r_i)_{i \geq 0} \,|\, r_i = \sigma_{j, i}(r_j) \text{ whenever } i \leq j \} \subseteq \prod_{i \geq 0} \frac R {I^i},$$ where the map $\sigma_{j, i} : R / I^j \to R / I^i$ is the canonical surjection $r + I^j \mapsto r + I^i.$ Consequently, one can view the elements of $\widehat R_I$ as sequences $(r_i + I^i)_{i \geq 0}$ of elements in $R/I^i$ such that $r_i \equiv r_j \text{ (mod } I^i).$ Observe that if there exists an integer $N \gg 0$ such that an element $r$ of $R$ is in $I^n$ for each integer $n \geq N + 1,$ then we have that $r + I^n = 0 + I^n$ so that $r + I^i = 0 + I^i$ for each integer $1 \leq i \leq n.$ Consequently, any sequence $(r_i + I^i)_{i \geq 0}$ of elements $R/I^i$ that is eventually zero is identically zero in $\widehat R_I.$ But this is no different than taking the quotient of the ring of Cauchy sequences of $\prod_{i \geq 0} R$ by the ideal of sequences that are eventually zero with respect to the $I$-adic topology, so the two definitions of $\widehat R_I$ are compatible with each other.

One way to think about a zero sequence $(r_i + I^i)_{i \geq 0}$ in the $I$-adic topology is that the coset representatives $r_i$ are eventually sufficiently ubiquitous: there exists an integer $N \gg 0$ such that for all integers $n \geq N + 1,$ we have that $r_n + I^n = 0 + I^n,$ hence we have that $r_n$ is in $\cap_{1 \leq i \leq n} I^i.$ Often, it is advantageous to restrict our attention to rings and ideals satisfying $\cap_{n \geq 0} I^n = \{0\},$ in which case $0$ is the only "universal" element of $R$ in the eyes of $I.$ One last observation is that the ideals $I^i$ are sequentially "smaller" in the sense that $I \supseteq I^2 \supseteq I^3 \supseteq \cdots,$ so if there exists a nonzero element $r$ in $I^N$ (for $N \gg 0$), then $r$ must be quite "small" in some sense, too; the notion of smallness is precisely communicated by the $I$-adic topology.

Answer 2

The idea behind the $I$-adic topology on a commutative ring $R$ is the fact that the iterated powers of an ideal $I \subseteq R$ form a descending chain of subsets $$I \supseteq I^2 \supseteq I^3 \supseteq ...$$ They take the role of the descending chain of open / closed balls $$B(0,1) \supseteq B(0,\tfrac{1}{2}) \supseteq B(0,\tfrac{1}{3}) \supseteq ...$$ you know and love for $\Bbb R$.

In $\Bbb R$ a sequence converges to 0, if for every $\varepsilon >0$ there is some $N_\varepsilon$ such that for all $k>N_\varepsilon$ we have $x_k \in B(0,\varepsilon)$. By the Archimedean property it suffices to show that for every $\varepsilon = \tfrac{1}{n}$ ie. $x_k \in B(0,\tfrac{1}{n})$. This hopefully motivates, why we define a zero sequence in the $I$-adic topology in such a way.

Moreover in both cases the topology on the whole space can be derived from this descending chain of subsets by translating it (ie. applying functions of the form $x \mapsto x+t$).