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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 $A/(e)$.

vonbrand
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Manos
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  • 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! – rschwieb Apr 02 '13 at 17:29
  • @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? – Manos Apr 02 '13 at 19:22
  • 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. – rschwieb Apr 02 '13 at 19:32
  • @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)? – Manos Apr 02 '13 at 19:35
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    @rschwieb: Here is something you find useful: http://math.stackexchange.com/questions/16568/completion-of-a-noetherian-ring-r-at-a-1-ldots-a-n-is-isomorphic-to-rx-1?rq=1 – Manos Apr 02 '13 at 19:42
  • I'm not knowledgeable enough to see if you are doing something wrong, but I can definitely feel the puzzlement. $A[[x]]/(x-e)$ feels like it should look like $A+(\prod A)e$, and it isn't clear to me why this would look like $A/(e)$ – rschwieb Apr 02 '13 at 19:44
  • Never say "Clearly" without having a proof. ;) – Martin Brandenburg Apr 04 '13 at 16:53

1 Answers1

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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

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

Martin Brandenburg
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