TITLE: T-nilpotency and quasinilpotency of ideals
QUESTION [5 upvotes]: 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:

What is an example of an ideal in a commutative ring that is T-nilpotent, quasinilpotent, but not nilpotent?

REPLY [2 votes]: Based on Yves answer above, let me give another example (or maybe another description of his example - I am not sure about this).
Let $K$ be a field of characteristic $2$, let $$R=K[(X_i)_{i\in\Bbb N}]/\langle\{X_i^2\mid i\in\Bbb N\}\cup\{X_iX_j\mid i,j\in\Bbb N,2i<j\}\rangle,$$ denote for $i\in\Bbb N$ by $Y_i$ the canonical image of $X_i$ in $R$, and let $\mathfrak a=\langle Y_i\mid i\in\Bbb N\rangle$.
$\mathfrak a$ is not nilpotent: If $\mathfrak a^n=0$, then $Y_n\cdots Y_{2n}=0$, hence there are $i,j\in[n,2n]$ with $2i<j$ - contradiction.
$\mathfrak a$ is quasinilpotent: An element of $\mathfrak a$ has the form $\sum_{i=0}^ra_iY_i$ with $a_i\in R$, and $(\sum_{i=0}^ra_iY_i)^2=\sum_{i=0}^ra_i^2Y_i^2=0$.
$\mathfrak a$ is T-nilpotent: Assume there is a sequence $(r_i)_{i\in\Bbb N}$ in $\mathfrak a$ such that for every $n\in\Bbb N$ we have $\prod_{i=0}^nr_i\neq 0$. Then, there exists such a sequence in which every $r_i$ is a monomial in $(Y_i)_{i\in\Bbb N}$. Without loss of generality we can thus suppose $r_i=Y_{j_i}$ for every $i\in\Bbb N$, where $(j_i)_{i\in\Bbb N}$ is an injective family in $\Bbb N$. Injectivity of this family implies that there exists $k\in\Bbb N$ with $j_k>2j_0$, and thus we get the contradiction $\prod_{i=0}^kr_i=0$.