Timeline for Is this a characterization of Dedekind domain?
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10 events
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| Mar 26, 2011 at 2:33 | comment | added | KConrad | Tom: I understand the uncertainty. If a Dedekind domain is supposed to be a domain in which the nonzero ideals have unique factorization, then fields seem like Dedekind domains. But I would say a Dedekind domain should be 1-dimensional (they should be something like a smooth curve), so that rules out fields. | |
| Mar 25, 2011 at 14:43 | comment | added | François Brunault | @yeshengkui : yes, if $R$ is a Dedekind domain then for any fractional ideals $I$ and $J$ of $R$, we have $I \oplus J \cong R \oplus IJ$. | |
| Mar 25, 2011 at 12:53 | comment | added | yeshengkui | To Francois: Do the ideals of Dedekind domain satisfy this property? | |
| Mar 25, 2011 at 8:31 | comment | added | François Brunault | I don't know if this works, but did you try to find a counterexample by taking $R$ to be a direct limit of Dedekind domains ? | |
| Mar 24, 2011 at 22:07 | comment | added | Tom Goodwillie | I can never decide whether fields should be considered Dedekind domains. (This somewhat idle question has been debated at MO before.) | |
| Mar 24, 2011 at 19:54 | comment | added | KConrad | Fields fit the condition (just one nonzero ideal) and they are not Dedekind domains. | |
| Mar 24, 2011 at 17:53 | answer | added | Mohan | timeline score: 3 | |
| Mar 24, 2011 at 17:03 | comment | added | Keerthi Madapusi | $\dim R=2$ (and regular in codimension one) with Hilbert-Samuel multiplicity $d-1$. I don't immediately see how to discount this possibility, though I've so far only used a very weak version of your hypothesis. | |
| Mar 24, 2011 at 17:02 | comment | added | Keerthi Madapusi | Suppose that $R$ is a local Noetherian integral domain with maximal ideal $\mathfrak{m}$ and residue field $k$. Your identity implies that $\dim_k(\mathfrak{m}/\mathfrak{m}^2)+\dim_k(\mathfrak{m}^{n-1}/\mathfrak{m}^n)=1+\dim_k(\mathfrak{m}^{n+1}/\mathfrak{m}^n),$ for all $n\in\mathbb{Z}_{>0}$. This means that the Hilbert polynomial of $(R,\mathfrak{m})$ is (on the nose) $H(n)=(d-1)\binom{n+1}{2}+n$, where $d=\dim_k(\mathfrak{m}/\mathfrak{m}^2)$. Since $\dim R=\deg H(n)$, there are only two possibilities: Either $d=1$ and $H(n)=n$, so that $R$ is a d.v.r. and we are done. Or, $d>1$ and... | |
| Mar 24, 2011 at 13:58 | history | asked | yeshengkui | CC BY-SA 2.5 |