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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 …
26
votes
Accepted
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} = \ …