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As explained in the other answers, in general there is a problem of seperatednessseparatedness. 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 noetherianNoetherian integral domain, then for any proper ideal $\mathfrak{a}$, the $\mathfrak{a}$-adic topology is seperatedseparated, and hence, metrizable.

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.

As explained in the other answers, in general there is a problem of separatedness. 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 separated, and hence, metrizable.

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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.