I-adic completion of a ring

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 $I$.

• Don't know if this is relevant, but wouldn't $\hat{R}I$ consist of all finite sums of products of.... etc., and not merely single products? – Arturo Magidin Mar 3 '11 at 22:33
• @Arturo: Thanks. That was a typo. – Yan Etor Mar 3 '11 at 22:46

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.
• Thanks. I almost get it. How do we show, $\sum_{k_j}i_{jk}$ is an element of $\hat{R}$? – Yan Etor Mar 4 '11 at 1:19