# $A,B$ Noetherian rings, $A\subseteq B$ integral extension, $\mathfrak m$ a maximal ideal of $A \implies B/\mathfrak m B$ is Artinian

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.

• 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. Jul 29, 2020 at 16:42

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.
• 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? Jul 29, 2020 at 22:39
• @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. Jul 29, 2020 at 22:50