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 $I$.

  • $\begingroup$ 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? $\endgroup$ – Arturo Magidin Mar 3 '11 at 22:33
  • $\begingroup$ @Arturo: Thanks. That was a typo. $\endgroup$ – 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.

  • $\begingroup$ Thanks. I almost get it. How do we show, $\sum_{k_j}i_{jk}$ is an element of $\hat{R}$? $\endgroup$ – Yan Etor Mar 4 '11 at 1:19
  • $\begingroup$ Never mind. Figured it out. $\endgroup$ – Yan Etor Mar 4 '11 at 3:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.