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I'm struggling with one of the exercises in my course.

It goes something like this: let $A$ be an Artinian ring (that is any descending chain of ideals stabilizes) and let $B$ be a Noetherian ring which is integral over $A,$ then I have to show that $B$ is Artinian as well.

Once again I would appreciate hints much more than a direct answer.

Intuitively (geometrically) it is quite evident since we have something with finite fibers lying over something finite (with Artinian ring of functions). Yet I was unable to formalize it.

Grisha Taroyan
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    A Noetherian ring is Artinian if and only if it has Krull dimension $0$. So, assuming $A$ is Artinian, can you show the $B$ in your situation also has Krull dimension $0$? – metalspringpro Jan 24 '20 at 17:20

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Let $P$ be a prime ideal of $B$. Then, $P \cap A$ is a prime ideal of $A$, hence a maximal ideal since $A$ is Artinian. Also, the quotient ring $B/P$ is an integral extension of the quotient ring $A/(P \cap A)$, and since the former is an integral domain and the latter is a field, the former must also be a field. This means that $P$ must in fact be a maximal ideal.

The above shows more generally that any integral extension of a zero-dimensional ring is again zero-dimensional. Since $B$ is Noetherian and zero-dimensional, it must therefore be Artinian.

Geoffrey Trang
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  • Nice proof, which passes through the same fact that powers ''Going-Up'': if $A \subseteq B$ is an integral extension of domains, then $A$ is a field if and only if $B$ is a field. – Alex Wertheim Jan 24 '20 at 18:16