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?



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


