Timeline for Finite distributive lattices as lattice of ideals of a finite ring
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Mar 1, 2018 at 15:25 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Removes obsolete notation/definition
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| Mar 1, 2018 at 1:43 | comment | added | Luc Guyot | @KeithKearnes Great, I got it. I have removed the faulty claim. | |
| Mar 1, 2018 at 1:42 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Retracts the newly introduced Claim 2: it is wrong
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| Mar 1, 2018 at 1:23 | comment | added | Keith Kearnes | @LucGuyot: "isn't the Jacobson radical both a maximal and minimal two-sided ideal?" It is maximal, but not minimal. I took $K$ to be a quadratic extension of its prime field so that the $(K,K)$-bimodule $K\otimes_Z K$ would not be simple, hence the radical of the ring would not be minimal. | |
| Mar 1, 2018 at 1:06 | comment | added | Luc Guyot | @KeithKearnes I can't but acknowledge this mistake, thanks again for pointing it out. About your example, isn't the Jacobson radical both a maximal and minimal two-sided ideal? | |
| Mar 1, 2018 at 0:48 | comment | added | Keith Kearnes | @LucGuyot: The problem is that if K is an arbitrary field, then Mn(K) ⊗Z Mn(K)^op need not be a matrix ring over a field. This fails already when n=1. | |
| Feb 28, 2018 at 22:48 | comment | added | Luc Guyot | @KeithKearnes Many thanks for your feedback. I am willing to correct or to remove the faulty parts. But I need to digest your two comments first. | |
| Feb 28, 2018 at 22:13 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Put Claim 2 on hold due to possibly mistake spotted by Keith Kearnes
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| Feb 28, 2018 at 21:35 | comment | added | Keith Kearnes | @LucGuyot: In fact, L IS the ideal lattice of a finite ring. Let $K$ be a finite field that is a quadratic extension of its prime field. Let $R$ be the ring of all $2\times 2$ matrices $\begin{bmatrix} a&m\\0&a\end{bmatrix}$ where $a\in K$, $m\in K\otimes_{\mathbb Z} K$, left/right actions of $K$ on $K\otimes_{\mathbb Z} K$ are determined by $r(p\otimes q)=rp\otimes q$ and $(p\otimes q)r=p\otimes qr$. | |
| Feb 28, 2018 at 21:21 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Settles the case of lattices of two-sided ideals
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| Feb 28, 2018 at 8:04 | vote | accept | Dominic van der Zypen | ||
| Feb 27, 2018 at 20:18 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Reference to R. B. Wirt's PhD thesis on finite non-commutative local rings
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| Feb 27, 2018 at 16:44 | comment | added | Luc Guyot | @ToddTrimble No, he certainly didn't and I shamefully missed that point, thanks. I have now extended my answer. | |
| Feb 27, 2018 at 16:39 | history | undeleted | Luc Guyot | ||
| Feb 27, 2018 at 16:36 | history | edited | Luc Guyot | CC BY-SA 3.0 |
Acknowledge the requirement that rings don't need to be finite
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| Feb 27, 2018 at 16:18 | history | deleted | Luc Guyot | via Vote | |
| Feb 27, 2018 at 15:37 | history | undeleted | Luc Guyot | ||
| Feb 27, 2018 at 1:26 | history | deleted | Luc Guyot | via Vote | |
| Feb 27, 2018 at 0:45 | comment | added | Todd Trimble | But did OP say 'commutative'? | |
| Feb 27, 2018 at 0:32 | history | answered | Luc Guyot | CC BY-SA 3.0 |