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


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



Does completion commute with localization?
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 $(A/\mathfrak{m}^n)\_{\mathfrak{m}} = A/\mathfrak{m}^n$.


