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1 vote

Metrizability of $\mathfrak{a}$-adic topology

No, it is not true in general. Take $\mathfrak{a} = A$ and $A$ nonzero to get that $A$ equipped with the $\mathfrak{a}$-adic topology is not Hausdorff and hence not a metric space (since $|A| > 1$). M …
Kestutis Cesnavicius's user avatar
26 votes
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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} = \ …
Kestutis Cesnavicius's user avatar