Let $R$ be an infinite commutative ring with unity such that every non-zero ideal has finite index. Then $R$ is Noetherian, every non-zero prime ideal is maximal , and I can also show that $R$ is an integral domain. Now also assume that distinct ideals have distinct index; then is it true that $R$ is a UFD, or at least normal (integrally closed in its fraction field) ?

(note: $\dim R \le 1$ , now if $\dim R=0$, then $R$ is an Artinian domain, so a field. So w.l.o.g., assume $R$ has dimension $1$ )

  • $\begingroup$ Such a ring is always regular, hence integrally closed in its fraction field. For each nonzero prime ideal $\mathfrak{p}$ of $R$, there is a bijection between the set of ideals $I$ with $\mathfrak{p}^2 \subseteq \mathfrak{p}$ and the set of $R/\mathfrak{p}$-vector subspaces of $\mathfrak{p}/\mathfrak{p}^2$. If the dimension of this vector space is bigger than $1$, then choosing two different subspaces of dimension $1$ gives two different ideals $I$ with the same index in $R$. Thus, if every pair of ideals has distinct indices, then $R$ is regular at $\mathfrak{p}$. $\endgroup$ – Jason Starr Jan 4 at 17:27
  • $\begingroup$ @JasonStarr: In the second line, do you mean to say set of ideals $I$ with $\mathfrak p^2 \subseteq I \subseteq \mathfrak p$ ? $\endgroup$ – user521337 Jan 4 at 18:31
  • $\begingroup$ @JasonStarr: could you elaborate on why choosing two different sub-spaces of $P/P^2$ of dimension $1$ would give two distinct ideals (that is fine) of same index (why same index ?) ? $\endgroup$ – user521337 Jan 4 at 18:37
  • $\begingroup$ If the prime ideal $\mathfrak{p}$ has finite index, then the residue field $R/\mathfrak{p}$ is a finite field, say $\mathbb{F}_q$. For $I$ satisfying $\mathfrak{p}^2 \subseteq I \subseteq \mathfrak{p}$, the index of $I$ in $R$ is the product of the index of $\mathfrak{p}$ in $R$ and the index of $I$ in $\mathfrak{p}$. This second index equals the product of $q$ and the codimension of $I/\mathfrak{p}^2$ as a subspace of $\mathfrak{p}/\mathfrak{p}^2$. $\endgroup$ – Jason Starr Jan 4 at 18:56
  • $\begingroup$ @JasonStarr: Ah I see it now, silly me ... thanks ... do you think there are non UFD examples ? $\endgroup$ – user521337 Jan 4 at 19:18

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