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.
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.