I have recently learned about nilpotency in a group/ring.I am thinking about doing some work on a non-commutative nilpotent ring but i could not find an example. More precisely, (S,+,.) is nilpotent, where S is nilpotent w.r.t. both of it's operation.
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"nilpotent w.r.t. both operations" does not make any sense. "nilpotent" is defined in terms of repeated multiplication resulting in the additive identity. What would "additively nilpotent" mean? Repeated addition resulting in ...? If you again said "additive identity" then that isn't nilpotency at all, that's just an element of finite order in the additive group. – rschwieb Aug 20 '19 at 18:28
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I said it in a sense that everytime we see a set with two algebraic structure, it amuses me that always one of them is commutative..i just thought what would happen if someone takes both operation noncommutative??? If you know something, please tell me, as i couldn't get any answer from anyone. Everyone is just saying that it for definition. But why is it in the definition if it doesn't need to be there. – Achak0790 Aug 22 '19 at 14:24
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there is already a question addressing this for rings As for other structures, I am not sure, but in all likelihood it is just because it is sufficient to suit most useful applications. – rschwieb Aug 22 '19 at 14:28
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Start with a field of $3$ elements $F_3$ and take the free algebra $F_3\langle x,y\rangle$ in noncommuting indeterminates, then take the quotient by $(x,y)^3$, and let $R$ be the ideal $(x,y)/(x,y)^3$ in the quotient ring.
It's nilpotent because $a^3=0$ for everything in the ring, and it's not commutative because $xy\neq yx$.
It also satisfies $3a=0$ for every $a\in R$, if that is what you meant by "additively nilpotent" (it should actually just be said that everything has additive order less than $3$.)
rschwieb
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Wonderful !!! But please read my comment answer answer my doubt also. Thanks in advance. – Achak0790 Aug 22 '19 at 14:25
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@MichaelRozenberg By definition, if $r,s,t\in (x,y)$, then $rst\in (x,y)^3$. In particular $r^3\in (x,y)^3$. So, nilpotent of order $3$. – rschwieb Aug 23 '19 at 10:50
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And if it's a nil ring of order three, so it's a nilpotent ring of order six by the Nagata-Higman's theorem. What do you think? Thank you! By the way, by this theorem it should be seventh order, but we can prove that an order six is enough. – Michael Rozenberg Aug 23 '19 at 11:47
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@MichaelRozenberg you have me at a disadvantage since I have no idea what you are talking about. I interpreted the question as asking for an example where $R^n={0}$ for a noncommutative ring. That is my educated guess based on the user’s information. You may have some other definition in mind. – rschwieb Aug 23 '19 at 14:01
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@rschwieb $R^n={0}$ it says that any words as $a_1a_2...a_n$ are equal to $0$. This ring named a nilpotent ring. In your example we have a ring with identity $a^3=0$ for any $a\in R$. This ring named a nil ring. – Michael Rozenberg Aug 23 '19 at 14:06
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@MichaelRozenberg No nonzero nil ring has an identity. As i mentioned, this ring by construction satisfies $R^3={0}$. I am not following what objection you have. – rschwieb Aug 23 '19 at 14:33
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@rschwieb Your first statement it's just wrong. A nil ring of the limited order has identity by the definition of this ring. From your second statement I see now what did you mean. Thank you! – Michael Rozenberg Aug 23 '19 at 14:39
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@MichaelRozenberg The identity is an idempotent element. How do you propose to have a nonzero idempotent nilpotent element? – rschwieb Aug 23 '19 at 15:45
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@MichaelRozenberg I’m unsure if that means you now understand my comment about the identity is correct... – rschwieb Aug 23 '19 at 16:39
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@rschwieb It's just another problem. We have a nil ring with identity $a^3=0$.For all associative ring $R$ with this identity we have $R^6={0}$. (also we need if $6a=0$ so $a=0$). – Michael Rozenberg Aug 23 '19 at 16:40
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@MichaelRozenberg well, I agree that logical inconsistency is a problem (that’s what the conditions of a nonzero nilpotent idempotent produce: a contradiction) – rschwieb Aug 23 '19 at 16:45
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Take a set of matrices $n\times n$, $n>3$ such that $\{a_{1n-2},a_{1n-1},a_1n,a_{2n-1},a_{2n},a_{3n}\}\subset\mathbb R$ and $a_{ij}=0$ for any $(i,j)\not\in\{(1,n-2),(1,n-1),(1,n),(2,n-1),(2,n),(3,n)\}.$
Michael Rozenberg
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But it's commutative under addition. I have asked about nilpotency in both operations. – Achak0790 Aug 20 '19 at 17:16
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@Aninda Chakraborty It must be commutative with $+$ because it's a ring. – Michael Rozenberg Aug 20 '19 at 17:20
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Although i don't think it's nilpotent as you can take a_(1,1)=2 and 0 otherwise, then it's not of a finite order. – Achak0790 Aug 20 '19 at 17:22
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Also, why it is necessary to be commutative of atleast one operation, i don't understand that. – Achak0790 Aug 20 '19 at 17:26
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Though i have asked a different question(an example of a nilpotent ring) but it concerns me that if the commutativity is only for the sake of definition or it is actually needed for compatibility of these two operations. – Achak0790 Aug 20 '19 at 17:40
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@Aninda Chakraborty You need associativity for $\cdot$ because otherwise, we not exactly know what is it $S^n$. By the way, antiassociativity of $\cdot$ gives a nilpotent ring: just $S^4=0$. Here $S$ is a ring with a property: $a+a=0\Rightarrow a=0.$ – Michael Rozenberg Aug 20 '19 at 17:48
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No, i'm not talking about this particular question. In general why do we need commutativity of addition in the definition of a ring. And thanks for your example, it's so elementary and beautiful. – Achak0790 Aug 20 '19 at 17:53
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@Aninda Chakraborty By the definition of a ring. Otherwise, it's not a ring. You are welcome! – Michael Rozenberg Aug 20 '19 at 17:55