Let $R$ be a commutative ring and let $\mathfrak{a}\subseteq R$ be an ideal. The ideal $\mathfrak{a}$ is called *T-nilpotent* if for every sequence $(r\_i)\_{i\in\mathbb{N}}$ in $\mathfrak{a}$ there exists $n\in\mathbb{N}$ such that $\prod\_{i=0}^nr\_i=0$, and *quasinilpotent* if there exists $n\in\mathbb{N}$ such that for every $r\in\mathfrak{a}$ we have $r^n=0$.


If $\mathfrak{a}$ is nilpotent (i.e., there exists $n\in\mathbb{N}$ with $\mathfrak{a}^n=0$), then it is T-nilpotent and quasinilpotent. If $\mathfrak{a}$ is T-nilpotent or quasinilpotent, then it is nil (i.e., for every $r\in\mathfrak{a}$ there exists $n\in\mathbb{N}$ such that $r^n=0$). 


Conversely, a nil ideal need not be T-nilpotent or quasinilpotent, a T-nilpotent ideal need not be nilpotent, and a quasinilpotent ideal need not be nilpotent.


I guess that a T-nilpotent *and* quasinilpotent ideal need not be nilpotent either. However, I was not able to come up with an example of such an ideal. Thus:



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> What is an example of an ideal in a commutative ring that is T-nilpotent, quasinilpotent, but not nilpotent?
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T-nilpotency and quasinilpotency of ideals
No need to specify the ambient ring since every commutative associative ring $I$ (possibly without unit) is an ideal in a ring (namely $I\oplus\mathbf{Z}$ with multiplicative law $(a\_1,n\_1)(a\_2,n\_2)=(a\_1a\_2+n\_1a\_2+n\_2a\_1,n\_1n\_2)$). 


Now call a X-ring a ring (possibly without unit) satisfying the axioms: associative, commutative, $x+x=x^2=0$ for all $x$.


So the free X-ring $R\_I$ on the generators $(x\_i)\_{i\in I}$ has the basis over the field on 2 elements: $(x\_J)\_{J\in 2^I\_0}$, where $2^I\_0$ denotes the set of nonempty finite subsets of $I$, and $x\_J=\prod\_{j\in J}x\_j$.


If $I$ is finite, then $R\_I$ is finite and nilpotent (of length $\le 2^{|I|}$ but I'm lazy to get the best bound).


Then $R\_I$ is not T-nilpotent when $I$ is infinite. But now let $I$ denote the positive integers, and define $S$ as the quotient modding out by $(x\_ix\_j)$ whenever $2i<j$. It has the basis $(x\_J)$ where $J$ ranges over nonzero finite subset not containing any pair of the form $\{i,j\}$ with $2i<j$.


(a) $S$ is quasinilpotent (with $n=2$ as any X-ring) 


(b) $S$ is not nilpotent (because $x\_n\dots x\_{2n}\neq 0$ for every $n$)


(c) $S$ is T-nilpotent: indeed, consider a sequence $(y\_n)$. Write $y\_1=\sum\_{k=1}^mx\_{J\_k}$, with $J\_k$ nonempty, pairwise distinct and not containing any pair $\{i,j\}$ with $2i<j$; let us show $\prod\_{i=1}^p y\_k=0$ for large $p$. We can assume $m>0$ since otherwise $y\_1=0$. Define $n=\max\bigcup J\_k$. Let $\pi$ map $x\_J$ to itself if $J\subset\{1,\dots,2n\}$ and map $x\_J$ to 0 otherwise, and extend it by linearity; this is a projection, and $y\_1y=y\_1\pi(y)$ for all $y\in S$. Hence $y\_1\dots y\_p=y\_1\pi(y\_2)\dots \pi(y\_p)$. Since this product lies in the (finite) subring generated by $x\_1,\dots,x\_{2n}$, which is nilpotent, it is 0 for large $p$ (say $p\ge 4^n$).


