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An element x in a noncommutative ring R is strongly nilpotent if any chain $x_1=x, x_2, ... $, with $x_{n+1}\in x_n R x_n$ terminates at zero. It becomes clear why this is a good definition once one has shown that the set of all such elements is the semi-prime radical (the intersection of all prime ideals). However, one can ask if this is equivalent to another definition (which at the first glance seems more natural): x is naively strongly nilpotent if for all $a_1,a_2,..$ in $R$ we have $x a_1 x a_2 .. x a_n x =0$ for some $n$. "Naively strongly nilpotent" implies strongly nilpotent; is the converse true in general? [Probably not, but what's a counterexample?]

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2 Answers 2

up vote 5 down vote accepted

Consider the infinite word $W=xx_1xx_2xx_3x....$. Let $R=R(W)$ be the ring which consists of all integral linear combinations of finite subwords of $W$. The product of two subwords $u,v$ is either the concatenation $uv$ if $uv$ is a subword of $W$ or $0$ otherwise. It is clearly an associative ring. The element $x$ in $R(W)$ is not naively strongly nilpotent but is strongly nilpotent. One can also get a finitely generated ring this way. Just consider the word $W=xyxy^2x...$. The ring $R(W)$ is 2-generated, $x$ is strongly nilpotent but not "naive". Here is the proof. The fact that $x$ is not "naively strongly nilpotent" follows from the fact that the words $x,xyx, xyxy^2x,...$ are subwords of $W$ and hence not equal to 0 in $R(W)$. Consider any sequence $t_0=x, t_1\in xR(W)x, t_2\in t_1R(W)t_1,...$, $t_i\ne 0$. Let $t_i=\sum_{j=1}^{n_i} w^{(i)}_j$, where each $w^{(i)}_j$ is a subword of $W$. Then each $w^{(i)}_j$ starts and ends with $x$ and is of length at least 2, each $t_p$, $p > 1$, is a linear combination of words which start with some $w_k^{(1)}$ and end with some $w^{(1)}_l$. But this implies that the lengths of the summands of $t_p$ are bounded (since a long subword of $W$ cannot end with a short word of the form $x...x$), a contradiction. Thus $x$ is strongly nilpotent.

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Thanks a lot! Simple indeed. –  Roman May 16 '12 at 0:13

However Roman may still have a point: the strong nilpotents depend on the naively strong nilpotents in the sense that if there are no naively strong nilpotents, then there are no strong nilpotents.

Proof: If there are no (non-zero) naively strong nilpotents, then there are no nilpotent ideals. Hence for any $x\ne0$, one can always find $0\ne x_{n+1}\in x_nRx_n\subseteq\langle x_n\rangle^2$, $x_0=x$. So $x$ is not strongly nilpotent.

Strong nilpotents $a$ are such that 'fractal'-like words $ax_1ax_2ax_1ax_3ax_1ax_2ax_1a$ are eventually 0. This seems either a deep or an unnatural phenomenon that can be simplified. By unnatural is meant that if one defines a strong-3-nilpotent similarly to a strong nilpotent but with $x_{n+1}\in x_nRx_nRx_n\subseteq\langle x_n\rangle^3$, then the two definitions actually agree.

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