Let $R$ be a ring, $I$ an ideal. According to Atiyah-Macdonald, if $R$ is Noetherian, then, we have $\hat{I}=\hat{R}I$ where hat denotes $I$-adic completion of $R$ and (I presume) $\hat{I}$ denotes the induced completion on $I$. I don't understand how to arrive at this equality and why the Noetherian hypothesis is necessary. Essentially $\hat{I}$ consists of equivalence classes of Cauchy sequences with elements in $I$. Any element of $\hat{R}I$ is an equivalence class of Cauchy sequences consisting of elements of $I$. I don't see how every Cauchy sequence with elements in $I$ is equivalent to one which can be written as a sum of products of a Cauchy sequence and a constant sequence of an element of $I$.
[Math] I-adic completion of a ring
abstract-algebracommutative-algebraring-theory
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I thought it would be a good exercise to post an answer of my own:
(i) We want to know the completion of the topological ring $A = \bigoplus_{n \in \mathbb N} \mathbb Z / p \mathbb Z$ with respect to the $p$-adic topology, i.e., the topology induced by neighbourhoods of zero of the form $A_n = p^n A$ so that $A=A_0 \supset A_1 \supset A_2\supset \dots$.
From chapter 10 in Atiyah-MacDonald we know that since this is a topological Abelian group with a countable neighbourhood basis of zero such that $A_n \supset A_{n+1}$, the completion is isomorphic to the inverse limit of the inverse system $X_n = A/A_n$ and the transition maps $f_n : A/A_n \to A/A_{n-1}$, $(x \mod p^{n}) \mapsto (x \mod p^{n-1})$. But since $pA = 0$ we get $X_n = A$ and $f_n = id_A$. Now the inverse limit are sequences $\vec{a} \in \bigoplus_n A/A_n = \bigoplus_n A$ such that $id(a_n) = a_{n-1}$, that is, constant sequences. Now clearly, $\varprojlim_n A \cong A$ via the map $(a,a,a, \dots ) \mapsto a$.
(ii) Next we would like to compute the completion of $A$ with respect to the topology induced by the $p$-adic topology on $B = \bigoplus_n \mathbb Z / p^n \mathbb Z$. The map $\alpha : A \to B$ is the inclusion map so that a set $U$ is a neighbourhood in $A$ if and only if $\alpha^{-1}(V) = V \cap A = U$ for $V$ open in $B$. Open sets in $B$ are of the form $p^k B$. Since $B = \bigoplus_n \mathbb Z / p^n \mathbb Z$, $p^k B = \bigoplus_{n=0}^{\infty} p^k \mathbb Z / p^n \mathbb Z$ where for $n \leq k$ the component is zero. We compute the inverse image $\alpha^{-1}(p^kB)$ as follows: Let $(A)_n = \mathbb Z / p \mathbb Z$ be the $n$-th component of $A$. For the $n$-th component of $p^k B$ we get
$$ (p^kB)_n = \begin{cases} 0 & n \leq k \\ p^k \mathbb Z / p^n \mathbb Z & n > k\\ \end{cases} $$
We have $\alpha(A)_n = \alpha_n(\mathbb Z / p \mathbb Z) = p^{n-1} \mathbb Z / p^n \mathbb Z$ so that $$ \alpha^{-1}(p^k B) = \begin{cases} 0 & n \leq k \hspace{0.2cm} (\text{since } \alpha \text{ is injective}) \\ \mathbb Z / p \mathbb Z & n > k \hspace{0.2cm} (\text{since } \mathrm{im}(\alpha ) = p^{n-1} \mathbb Z / p^n \mathbb Z \subset p^{k} \mathbb Z / p^n \mathbb Z)\\ \end{cases}$$
Hence open sets in this topology on $A$ look like $O_k = 0 \oplus \dots \oplus 0 \oplus \mathbb Z / p \mathbb Z \oplus \mathbb Z / p \mathbb Z \dots $ where the first $k$ entries are zero.
For our inverse system this gives us $X_n = A / O_k = \mathbb Z / p \mathbb Z \oplus \dots \oplus \mathbb Z / p \mathbb Z \oplus 0 \oplus \dots $ where the first $k$ entries are non-zero. For the transition maps $t_n: X_n \to X_{n-1}$ this means $(x_1, x_2, \dots,x_{k-1}, x_k, 0 , 0 , \dots) \mapsto (x_1, x_2, \dots,x_{k-1}, 0 , 0 , \dots)$.
For the inverse limit of this system this means that it is all sequences with $x_n \in X_n$, which gives us $\varprojlim_n X_n = \prod_n X_n = \prod_n \mathbb Z / p \mathbb Z$.
Now to conclude that $\varprojlim_n$ is not a right-exact functor on $\mathbb Z - \mathrm{\mathbf{Mod}}$, let $A_k = \alpha^{-1}(p^k B)$. Then the following sequence is exact: $$ 0 \to A_k \hookrightarrow A \xrightarrow{\pi_k} A/A_k \to 0$$
But for the inverse limit we get $$ 0 \to 0 \hookrightarrow A = \bigoplus_n \mathbb Z / p \mathbb Z \xrightarrow{\pi} \prod_n \mathbb Z / p \mathbb Z $$
where $\pi$ cannot be surjective since it is a group homomorphism mapping zero in the $n$-th component to zero.
Yes it is true.in fact you can do better than that: take any $w\in I$, then $\tilde K=\tilde R[\frac{1}{w}]$.(I assume that you really have a valuation field which means that $R$ is local and $I$ is principal and maximal, otherwise you don't get a valuation but see the third paragraph for more general setting)
this is true because you can easily see that components of a Cauchy sequence($a_k$) in $K$ have bounded valuations(the whole sequence is bounded) hence there is a power $n$ of $W$ such that $(w^n a_k)=(b_k)$ is a cauchy sequence in $R$ and hence an element of $\tilde R$. in this way you get a map $\tilde K\to \tilde R[\frac{1}{w}]$ and it is not hard to see that this is an isomorphism.
in fact if you are interested there is a more general theory of this things called tate rings: a Tate ring is a topological ring $R$ and a subring $R_0$ and an ideal $w\in I$ of $R_0$ such that $R=R_0[\frac{1}{w}]$ and the base for the topology is given by $I^n R_0$. in your language $R=K,R_0=R,I=I$ is an example of a tate ring. for a tate you can always define $\tilde{R}=\tilde{R_0}[\frac{1}{w}]$
Best Answer
If $R$ is Noetherian, then $I$ must be finitely generated, say $I = \langle p_1,\ldots, p_n\rangle$. So if an element of $\hat{I}$ is represented by a sum $x = i_1 + i_2 + \cdots$, rewriting $i_m = p_1 i_{m1} + \cdots + p_n i_{mn}$ we can rewrite this as $$x = p_1 \sum_{k_1}i_{1k} + \cdots + p_n\sum_{k_n}i_{nk} \in \hat{R}I$$ which is what you wanted.