Timeline for Finite distributive lattices as lattice of ideals of a finite ring
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 1, 2018 at 14:07 | comment | added | Dominic van der Zypen | But since @LucGuyot has put in a lot of effort, I thought it might be better to clarify this in another question. I feel it's unfair to Luc to unaccept the answer just because I was imprecise. | |
| Mar 1, 2018 at 14:05 | comment | added | Dominic van der Zypen | Sorry for being a bit sloppy! - I was thinking about commutative rings (i.e., both operations are commutative). | |
| Mar 1, 2018 at 4:34 | comment | added | მამუკა ჯიბლაძე | In fact ideal lattices are seldom distributive, it is the lattice of radical ideals that is always distributive. So another interesting question is whether a finite distributive lattice is the lattice of radical ideals of a finite ring; I believe the answer is yes. | |
| Mar 1, 2018 at 2:48 | comment | added | Todd Trimble | Dominic, this is a frustrating thread because you have failed to respond to the questions/issues addressed to you in the comment thread under the question. In addition, you have accepted an answer where is it unclear (in view of discussion by Keith Kearnes) that it really answered the intended question. | |
| Feb 28, 2018 at 22:33 | comment | added | Benjamin Steinberg | @LucGuyot, it is a standard fact that any Artinian ring R is Morita equivalent to a basic Artinian ring. An Artinian ring R is basic if $R/J(R)$ is a direct product of fields. The way you do this is let $P_1,\ldots, P_s$ be a complete list of non-isomorphic projective indecomposable modules and Let $P=P_1\oplus \cdots \oplus P_S$. Then $P$ is a projective generator so $End(P)$ is Morita equivalent to $R$. It is not difficult to check that $End(P)$ is basic. You can find this in standard books on Artin algebras like Auslander et al. | |
| Feb 28, 2018 at 22:05 | comment | added | Luc Guyot | @BenjaminSteinberg It may be obvious, but I fail to prove your claim regarding the reduction to a commutative quotient. If $R$ is a finite ring, I understand that $R/J(R)$ is Morita equivalent to a product of finite fields (Artin-Wedderburn) and hence commutative. But how to produce a finite ring $S$ which is Morita equivalent to $R$ and such that $S/J(S)$ is commutative? | |
| Feb 28, 2018 at 8:04 | vote | accept | Dominic van der Zypen | ||
| Feb 27, 2018 at 20:34 | comment | added | Gerhard Paseman | I suspect there are results on congruence distributive varieties that can help here. You might ask William DeMo or Keith Kearnes directly, as I have not looked at the relevant literature in over a decade. Gerhard "Time Flies When Jumping Primes" Paseman, 2018.02.27. | |
| Feb 27, 2018 at 20:29 | comment | added | Benjamin Steinberg | Ideal generally means two-sided ideal. Since finite rings are artinian and Morita equivalent rings have isomorphic lattices of ideals you can assume the quotient by the radical is commutative (as finite division algebras are commutative) | |
| Feb 27, 2018 at 18:50 | comment | added | Todd Trimble | There are however results that imply that a finite distributive lattice is isomorphic to the lattice of ideals of a ring (not necessarily finite). See for example tandfonline.com/doi/abs/10.1080/00927878008822518 | |
| Feb 27, 2018 at 17:39 | comment | added | Luc Guyot | Do you consider the lattice of (1) right ideals, (2) two-sided ideals of a possibly non-commutative ring, or the lattice of (3) ideals of a commutative ring? Do rings have an identity element? | |
| Feb 27, 2018 at 0:32 | answer | added | Luc Guyot | timeline score: 4 | |
| Feb 26, 2018 at 11:48 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |