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Suppose $A$ is a Noetherian (not necessarily local) ring and $\mathfrak{m}\subset A$ a maximal ideal. Then is it true that $$\hat{A}_{\hat{\mathfrak{m}}}=\widehat{A _{\mathfrak{m}}},$$ where hats denote completion and subscripts denote localization? If one uses superscripts to denote completion it would be

$$(A^{\mathfrak{m}})_{\mathfrak{m^{\mathfrak{m}}}}=(A _{\mathfrak{m}})^{\mathfrak{m} _{\mathfrak{m}}}.$$

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    $\begingroup$ You should be more careful about what you mean by "completion". If you mean "completion with respect to the filtration by powers of the maximal ideal" then the answer is yes. In fact, $\hat{A}$ is already local so $\hat{A}_{\hat{m}} \cong \hat{A}$. $\endgroup$
    – user91132
    May 9, 2011 at 16:05
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    $\begingroup$ @ashpool: How much time have you spent to try to answer this by your self? $\endgroup$
    – Martin Brandenburg
    May 10, 2011 at 9:01
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    $\begingroup$ @Martin Brandenburg: Sorry, I realized it was a simple problem after I posted it. $\endgroup$
    – ashpool
    May 16, 2011 at 14:01

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It is true. $(\widehat{A}, \widehat{\mathfrak{m}})$ is a Noetherian local ring so your left hand side could be simplified replacing it by $\widehat{A}$. Now let's use the definitions: $\widehat{A} = \varprojlim A/\mathfrak{m}^n$, whereas $\widehat{A_{\mathfrak{m}}} = \varprojlim A_{\mathfrak{m}}/(\mathfrak{m}A_{\mathfrak{m}})^n$. The desired equality is the result of localization being exact so that $A_{\mathfrak{m}}/(\mathfrak{m}A_{\mathfrak{m}})^n = (A/ \mathfrak{m}^n)_{\mathfrak{m}}$ and the fact that in $A/\mathfrak{m}^n$ everything outside the maximal ideal is already invertible, so that $(A/\mathfrak{m}^n)_{\mathfrak{m}} = A/\mathfrak{m}^n$.

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  • $\begingroup$ I would like to point out that this shows $\hat{A}\cong\widehat{A_\mathfrak{m}}$ for any commutative ring $A$ (and maximal ideal $\mathfrak{m}\subset A$): the proof never used Noetherianity. Also, $\hat{A}_{\hat{\mathfrak{m}}}\cong\hat{A}$, since completion with respect to a maximal ideal $\mathfrak{m}$ is always local with maximal ideal $\hat{\mathfrak{m}}$ (again, regardless of Noetherianity). $\endgroup$
    – Elías Guisado Villalgordo
    Sep 18 at 16:48

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