= Solution
{c}
\b[No.] Write $x_i=t_i+I$ in the <quotient ring> $R=L/I$. Its defining relations are
$$
x_1^2=0,\qquad x_i=x_{i+1}^2\quad(i\geq1).
$$
Consequently $x_i^{2^{i-1}}=x_1\ne0$, using the permitted assumption, while $x_i^{2^i}=0$. Thus every $x_i$ is a nonzero <nilpotent element>.
The <principal ideals> satisfy
$$
(x_1)\subseteq(x_2)\subseteq(x_3)\subseteq\cdots.
$$
Every inclusion is strict. Otherwise $x_{i+1}=a x_i=a x_{i+1}^2$ for some $a\in R$, and hence
$$
x_{i+1}(1-a x_{i+1})=0.
$$
The 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
$$
\boxed{L/I\text{ is not Noetherian}.}
$$
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