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Let $A$ be a commutative Noetherian ring and $I=(a_1,\cdots,a_n)$ an ideal. Then the $I$-adic completion of $A$ is isomorphic to $A [[ x_1,\cdots,x_n ]]/(x_1-a_1,\cdots,x_n-a_n)$. Now let $e$ be an idempotent of $A$ and apply the above result to the ideal generated by $e$. Then $\hat{A}=A[[x]]/(x-e)$. On the other hand, by definition of the $I$-adic completion we have that $\hat{A}=A/(e)$. Clearly, $A/(e) \neq A[[x]]/(x-e)$. What am i missing?

Here is the way i think about it: To construct the $I$-adic completion, we first start by considering the product $\prod_{i>0} A/I^i$. Since $e^2=e$, for $A=(e)$ this product becomes $\prod_{i>0} A/(e)$. The completion is the subspace of coherent sequences of this product. But each such coherent sequence can be identified with a single element of $A/(e)$.

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  • $\begingroup$ Hi: Does that characterization of the completion being $A[[x]]/(x-e)$ hold outside of local rings? I'm asking because I have no good references, and what I find online always refers to a local Noetherian ring. Thanks for letting me know! $\endgroup$ – rschwieb Apr 2 '13 at 17:29
  • $\begingroup$ @rschwieb: Hi my friend. This characterization is Theorem 8.12, p. 61, in Matsumura's Commutative Ring Theory and his theorem is not necessarily for local rings. Is my characterization $\hat{A}=A/(e)$ correct? $\endgroup$ – Manos Apr 2 '13 at 19:22
  • $\begingroup$ Thanks for the reference! I'm not sure about the $\hat{A}=A/(e)$. I can see what you're thinking, though: the filtration is $A\supseteq (e)\supseteq (e)\supseteq\dots$ and so the direct limit of $A/I^n$ should(?) just be $A/(e)$. While I know the definition of all of these things, my intuition about how they work is not developed. $\endgroup$ – rschwieb Apr 2 '13 at 19:32
  • $\begingroup$ @rschwieb: No problem, your input is appreciated :) Do you agree though that the two characterizations are not equal (and hence i am doing something wrong)? $\endgroup$ – Manos Apr 2 '13 at 19:35
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    $\begingroup$ @rschwieb: Here is something you find useful: math.stackexchange.com/questions/16568/… $\endgroup$ – Manos Apr 2 '13 at 19:42
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In $A[[x]]/(x-e)$ the element $1-e$ is idempotent, but also a unit because it equals $1-x$ (and $(1-x)^{-1} = 1+x+x^2+\dotsc$). Therefore it equals $1$, i.e. we have $e=0$. It follows

$$A[[x]]/(x-e) = A[[x]]/(x-e) / (e) = A/(e)[[x]]/(x) = A/(e).$$

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