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?]



strong nilpotent elements
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.


