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Let $R$ be a commutative ring (if necessary we assume it is an integral domain), and $\mathfrak{m}=(f)$ be a maximal ideal that is principal. Is it true that $R/\mathfrak{m}^n$ is a local Artinian for all $n>0$?

I can see that it is local since there can not be any other maximal ideal lying over $\mathfrak{m}^n=(f^n)$. To show it is Artinian, is it possible to show $\mathfrak{m}^l/\mathfrak{m}^n$ are the only possible ideals?

I was reading this note (Proposition 4.4.6. on page 60) about the $p$-adic period ring $B_{dR}$.

CO2
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1 Answers1

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You’ve established the quotient by a power of the maximal ideal is local, and evidentially the unique maximal ideal is also principal.

Its Jacobson radical is nilpotent by definition of this ring, so we are looking at a zero dimensional ring.

A famous exercise says that an ideal maximal among non finitely generated ideals is prime: but if there were such a prime in our ring, it would necessarily be the principal maximal ideal, which is a contradiction. We should conclude the ring is Noetherian since its ideals are all finitely generated.

But then we can apply the well known result that a commutative Noetherian zero dimensional ring is Artinian.

In fact if you look at everything again you see you can replace “principal” with “finitely generated” and get the same result.

rschwieb
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