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Let $R$ be a commutative ring and let $I \subset R$ be an ideal. For $n \geq m$, let $\varphi_{m,n}$ denote the canonical ring homomorphism $R / I^n \to R / I^m$. Let $J = \cap_n I^n$. Then $R / J$ is going to end up being the $I$-adic completion of $R$, coming equipped with the obvious natural maps $\lambda_n \colon R / J \to R / I^n$.

My question regards a certain step in establishing that this object is final in the relevant category. Suppose $S$ is a commutative ring coming equipped with maps $\varphi_n \colon S \to R / I^n$ such that for each $n$, $\varphi_{n, n+1} \circ \varphi_{n+1} = \varphi_n$. We want to show that there exists a unique ring homomorphism $\varphi \colon S \to R / J$ such that for each $n$, $\varphi_n$ factorss through $R / J$ appropriately via $\varphi$.

Now, the definition of $\varphi$ is forced on us. Note that for a fixed $s \in S$ and arbitrary $n$, the assumptions on $S$ imply that the coset $\varphi_{n+1}(s)$ of $I^{n+1}$ is contained in the coset $\varphi_{n}(s)$ of $I^n$. So the $\varphi_n(s)$ form a nested sequence of cosets of the corresponding powers of $I$, and it is straightforward to see that the only candidate for $\varphi(s)$ is

$$\varphi(s) = \bigcap_n \varphi_n(s).$$

Assuming that the set on the right is nonempty (moreover, assuming it contains a coset of $J$), it is immediate that it is a coset of $J$: for each $r,r' \in \bigcap_n \varphi_n(s)$, $r - r' \in I^n$ for each $n$, and so $r - r' \in J$.

But why is it guaranteed that $\bigcap_n \varphi_n(s)$ contains a coset of $J$? Intuitively it seems this must be true: the "size" of all the cosets of $J$ are the same. Nevertheless, it seems to me that some work has to be done; for instance, we are aware that the question of the intersection of infinitely many nested sets is a delicate question, say, in topology, where the nonempty intersection of sequences of nested closed sets is equivalent to the ambient space being compact. In this problem, I am concerned because we have made no assumptions on the ring $R$ (for example, if $R$ is Noetherian then this is trivially true).

As a note, this formulation of the $I$-adic completion is taken from Aluffi's text on algebra. And of course, this leads to the construction of the $p$-adic numbers.

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    $\begingroup$ The $I$-adic completion is not isomorphic to $R/J$. For example, take $R = \mathbb{Z}$, $I = (p)$. Then $\bigcap I^n = 0$, but the completion is $\mathbb{Z}_p$, not $\mathbb{Z}/(0) = \mathbb{Z}$. $\endgroup$ – Brandon Carter Feb 16 '15 at 0:15
  • $\begingroup$ Interesting. What is the universal object then, since it is not $R/J$? $\endgroup$ – Doug Feb 18 '15 at 2:57
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    $\begingroup$ It is the inverse limit of the $R/I^n$. $\endgroup$ – Brandon Carter Feb 18 '15 at 3:14
  • $\begingroup$ How does one prove this limit exists (I am familiar with the basic definitions surrounding the inverse limit)? $\endgroup$ – Doug Feb 18 '15 at 4:57
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    $\begingroup$ What do you mean? If you read the Wikipedia page you will see that there is an actual construction of the inverse limit. It is the subset of $\prod R/I^n$ that is compatible with all projections $R/I^n \to R/I^m$ for $n \geq m$. $\endgroup$ – Brandon Carter Feb 18 '15 at 5:29
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Brandon Carter's response in the comments suffices.

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