Timeline for Infinite Noetherian ring of dimension $1$ in which distinct non-zero ideals have distinct and finite index
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 4, 2019 at 19:18 | comment | added | user521337 | @JasonStarr: Ah I see it now, silly me ... thanks ... do you think there are non UFD examples ? | |
| Jan 4, 2019 at 18:56 | comment | added | Jason Starr | 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 $\mathfrak{p}/\mathfrak{p}^2$. | |
| Jan 4, 2019 at 18:37 | comment | added | user521337 | @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 ?) ? | |
| Jan 4, 2019 at 18:31 | comment | added | user521337 | @JasonStarr: In the second line, do you mean to say set of ideals $I$ with $\mathfrak p^2 \subseteq I \subseteq \mathfrak p$ ? | |
| Jan 4, 2019 at 17:27 | comment | added | Jason Starr | 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 $\mathfrak{p}$. | |
| Jan 4, 2019 at 13:37 | history | asked | user521337 | CC BY-SA 4.0 |