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Let $A$ be a ring, $\mathfrak{a}\subset A$ an ideal. Then is the $\mathfrak{a}$-adic topology on $A$ necessarily a metric space? I can see that it is true when $A$ is a DVR, but is it true in general?

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up vote 3 down vote accepted

As explained in the other answers, in general there is a problem of seperatedness. If $\bigcap_{n=1}^{\infty} \mathfrak{a}^n \ne 0$ then the topology is not Hausdorff.

On the positive side however, you have for example Krull's intersection theorem which says that if $A$ is a noetherian integral domain, then for any proper ideal $\mathfrak{a}$, the $\mathfrak{a}$-adic topology is seperated, and hence, metrizable.

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You can always define $d(a,b)=2^{-n}$, where $n$ is the largest integer such that $a-b\in\mathfrak{a}^n$, or $d(a,b)=0$ if $a-b\in\mathfrak{a}^n$ for all $n$. This is a pseudometric, ie it satisfies all the axioms for a metric except that $d(a,b)$ can be zero even when $a\neq b$. Every pseudometric gives a topology, which is Hausdorff iff the pseudometric is a metric. The topology from the above pseudometric is the same as the $\mathfrak{a}$-adic topology.

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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 $|A| > 1$). More generally, your topology will be non-Hausdorff whenever $\bigcap_{n\ge 1} \mathfrak{a}^n \neq 0$ yielding many more counterexamples.

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