 Solution

No. Write $x_i=t_i+I$ in the quotient ring $R=L/I$. Its defining relations areConsequently $x_i^{2^{i-1}}=x_1\ne 0$, using the permitted assumption, while $x_i^{2^i}=0$. Thus every $x_i$ is a nonzero nilpotent element.

The principal ideals satisfyEvery inclusion is strict. Otherwise $x_{i+1}=a{x}_i=a{x}_{i+1}^2$ for some $a\in R$, and henceThe fact that one minus a nilpotent element is a unit applies: if $z^m=0$, then $(1-z)(1+z+\cdots +z^{m-1})=1$. Here $z=a{x}_{i+1}$ is nilpotent, so multiplication by this inverse would give $x_{i+1}=0$, a contradiction. The strictly ascending chain of ideals violates the ascending chain condition. Therefore