Suppose R is a Noetherian integral domain. We assume further that R is a local ring of Krull dimension 1. Prove that if M is the unique maximal ideal of R and if every ideal is a power of M, then R is a DVR.

Question

Suppose [latex]R[/latex] is a Noetherian integral domain. We assume further that [latex]R[/latex] is a local ring of Krull dimension 1. Prove that if [latex]M[/latex] is the unique maximal ideal of [latex]R[/latex] and if every ideal is a power of [latex]M[/latex], then [latex]R[/latex] is a DVR.

Accepted Answer

Let [latex]x_{1}, \ldots, x_{n} \in M[/latex] be a minimal list of generators for [latex]M[/latex]. Without loss of generality we assume that [latex]x_{1} otin M^{2}[/latex]. The principal ideal [latex]I[/latex] generated by [latex]x_{1}[/latex] is a power of [latex]M[/latex] by hypothesis: [latex]I=M^{k}[/latex] for some [latex]k \geq 1[/latex]. Since [latex]x_{1} otin M^{2}[/latex], in particular, [latex]x_{1} otin M^{k}[/latex] for any [latex]k>1[/latex]. Therefore, [latex]I=M[/latex]. In particular, [latex]M / M^{2}[/latex] is spanned by [latex]x_{1}[/latex]. Therefore, [latex]\operatorname{dim}_{R / M} M / M^{2}=1[/latex], hence by Problem 225 the problem is finished.

Question 1.226: Suppose R is a Noetherian integral domain. We assume further......

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