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I must be missing something very simple, but suppose that every maximal ideal $\mathfrak{m}$ of a Noetherian ring is closed in the $\mathfrak{a}$-topology on $A$. Then why does this imply that $\mathfrak{a} \subseteq \text{Jac}(A)$, the Jacobson radical of $A$?

I know if there is an ideal $\mathfrak{a}$ not contained in some maximal ideal $\mathfrak{m}$ then for every $i > 0$ we have $$ \mathfrak{m} + \mathfrak{a}^i = A.$$

How does this contradict $\mathfrak{m}$ being closed? I can't see the existence of $x \notin \mathfrak{m}$ with no open neighborhood about it.

user26857
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David Benjamin Lim
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1 Answers1

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An ideal $\mathfrak m$ is closed in the $\mathfrak a$-adic topology iff $\mathfrak m=\bigcap_{i\ge 1} (\mathfrak m+\mathfrak a^i)$. As you observed, this shows that $\mathfrak m+\mathfrak a^i\ne A$ for any $i\ge 1$. If $\mathfrak m$ is maximal this means that $\mathfrak a\subseteq\mathfrak m$.

user26857
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