# Does completion commute with localization?

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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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 $\hat{A}_{\hat{m}} \cong \hat{A}$. –  Konstantin Ardakov May 9 '11 at 16:05
@ashpool: How much time have you spent to try to answer this by your self? –  Martin Brandenburg May 10 '11 at 9:01
@Martin Brandenburg: Sorry, I realized it was a simple problem after I posted it. –  ashpool May 16 '11 at 14:01

It is true. $(\widehat{A}, \widehat{m})$ is a Noetherian local ring so your left hand side could be simplified replacing it by $\widehat{A}$. Now let's just use the definitions: $\widehat{A} = \varprojlim A/\mathfrak{m}^n$, whereas $\widehat{A_{\mathfrak{m}}} = \varprojlim A_{m}/(\mathfrak{m}A_{\mathfrak{m}})^n$. 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 $(A/\mathfrak{m^n})_{\mathfrak{m}} = A/\mathfrak{m}^n$.
To add a comment, by Kestutis Cesnavicius's argument, the statement $\widehat{A}=\widehat{A_{\mathfrak{m}}}$ is still true even when $A$ is not noetherian.