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 $I$.
I-adic completion of a ring
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abstract-algebra
commutative-algebra
ring-theory
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0@Arturo: Thanks. That was a typo. – 2011-03-03
1 Answers
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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.
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0Never mind. Figured it out. – 2011-03-04