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Let $A,B$ be Noetherian rings, $A \subseteq B$, such that $B$ is integral over $A$. Given $\mathfrak m\subseteq A$ a maximal ideal, prove that $B/\mathfrak mB$ is an Artinian ring.

I'm really stuck.

Well, I know that $B/\mathfrak m B$ will be integral over $A/(A\cap \mathfrak mB)$. If we manage somehow to prove that both these rings are domains and that $A/(A\cap \mathfrak mB$) is a field, then $B/\mathfrak mB$ is a field (Artinian).

I also considered using the going-up / going-down theorem, but both of them would start from a chain in $A/(A \cap\mathfrak mB), $ and not from $B/\mathfrak mB$.

How does the hypothesis of being Noetherian apply here? Will the fact that every prime has finite height be useful?

Any help? Thank you.

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  • $\begingroup$ What is true is that if $\mathfrak mB$ were also maximal in $B$, then the quotient would indeed be a field. So use the idea that this ideal is not maximal to get a handle on a chain of ideals. $\endgroup$
    – hardmath
    Jul 29, 2020 at 16:42

1 Answer 1

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Since $A \subseteq B$ is integral, you can find a maximal ideal $\mathfrak{n}$ contracting to $\mathfrak{m}$. It follows that $\mathfrak{m}B \cap A = \mathfrak{m}$.

In general, if $A \subseteq B$ is integral and $J$ is an ideal of $B$, then $A/(A \cap J) \subseteq B/J$ is integral.

In your case you get that $B/\mathfrak{m}B$ is an integral extension of the field $A/\mathfrak{m}$. Integral extensions preserve Krull dimension, so $B/\mathfrak{m}B$ is $0$-dimensional. Also $B$ is assumed to be Noetherian, so $B/\mathfrak{m}B$ is Noetherian too. Since Artinian = Noetherian + $0$-dimensional you are done.

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  • $\begingroup$ I can't see why $\mathfrak mB \cap A = \mathfrak m$. Since $1\in B$, we have $\mathfrak m \subseteq \mathfrak mB$. So, $\mathfrak m \subseteq \mathfrak m B\cap A$. If $\mathfrak m B \cap A \neq A$, then its done. But why must $\mathfrak m B \cap A$ be proper? $\endgroup$ Jul 29, 2020 at 22:39
  • $\begingroup$ @math.h For any ideal $J$ of $B$, if $B \cap J = A$ then necessarily $1 \in J$. In words, proper ideals always contract to proper ideals. Meanwhile we saw that $\mathbb{m}B$ is proper, using that the extension is integral. $\endgroup$ Jul 29, 2020 at 22:50

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