Deduce that a Noetherian valuation ring is either a field or a Discrete Valuation Ring.

I'm trying to solve this question from a book and I have already proved 1.

1. Let $R$ be a local domain which is not a ﬁeld. Suppose that the maximal ideal $M$ of $R$ is principal and satisﬁes $\cap_{n=1}^{\infty}M^n=0$. Show that every non-zero ideal of $R$ is a power of $M$, and hence that $R$ is a DVR.

2. Deduce that a Noetherian valuation ring is either a field or a DVR.

• Does this answer your question? Equivalent characterizations of discrete valuation rings Jun 26, 2022 at 23:24
• @TrystwithFreedom, No. There are many definitions in that post which others might not have seen Nov 15, 2022 at 6:19

Valuation rings have the property that for all elements $a,b$ we have $a|b$ or $b|a$. It follows easily that every finitely generated ideal is principal. In particular, a noetherian valuation ring has a principal maximal ideal. Besides, Krull's Theorem states that $\bigcap_{n \geq 0} \mathfrak{m}^n=0$ in a local noetherian ring $(R,\mathfrak{m})$. Therefore, 2 follows from 1.

• I guess this assumes the reader knows why the intersection equality in part 1 holds in a Noetherian valuation ring? Mar 14, 2013 at 18:20
• Ah yes, Krull's Theorem. Mar 14, 2013 at 18:32