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