Timeline for non commutative polynomial which is zero for all matrix evaluation
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Aug 24, 2022 at 10:02 | history | edited | darij grinberg | CC BY-SA 4.0 |
none of the answers need this condition
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| Aug 24, 2022 at 9:46 | history | edited | darij grinberg | CC BY-SA 4.0 |
require nilpotence in order for the extension to make sense
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| Nov 27, 2015 at 12:43 | vote | accept | thib | ||
| Nov 27, 2015 at 12:40 | vote | accept | thib | ||
| Nov 27, 2015 at 12:43 | |||||
| Nov 27, 2015 at 7:12 | answer | added | user91132 | timeline score: 13 | |
| Nov 27, 2015 at 6:59 | comment | added | Ilya Bogdanov | Still, the Amitsur--Levitski theorem provides the minimal degree of such polynomial for a fixed $m$, and this degree is $2m$. So it provides the answer for the original question. | |
| Nov 27, 2015 at 5:35 | vote | accept | thib | ||
| Nov 27, 2015 at 6:10 | |||||
| Nov 27, 2015 at 5:33 | vote | accept | thib | ||
| Nov 27, 2015 at 5:35 | |||||
| Nov 26, 2015 at 23:52 | answer | added | Emil Jeřábek | timeline score: 14 | |
| Nov 26, 2015 at 22:06 | history | edited | thib | CC BY-SA 3.0 |
added 248 characters in body
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| Nov 26, 2015 at 22:03 | comment | added | Aaron Meyerowitz | As noted, for each $t$ there is a non-zero polynomial in $2t$ variables with all terms of degree $2t$ which is zero for any choice of $m \times m$ matrices, provided $m \le t.$ You gave the example for $t=1.$ But I'd guess that for all $m$ one does get $P=0.$ | |
| Nov 26, 2015 at 22:01 | comment | added | thib | Thanks David for your answer. But I don't want to fix the size of the matrices. I will make some edit. | |
| Nov 26, 2015 at 21:42 | comment | added | David Lampert | Google "Amitsur-Levitzki theorem" and "Hall's identity" | |
| Nov 26, 2015 at 21:30 | review | First posts | |||
| Nov 26, 2015 at 22:04 | |||||
| Nov 26, 2015 at 21:27 | history | asked | thib | CC BY-SA 3.0 |