Let $R$ be a local ring with a nilpotent maximal ideal $M$ and $I\subseteq M$. Then $ya\in I$ implies $ya=0$ for some $0\neq y\in R$

Question

Let $R$ be a local ring with a nilpotent maximal ideal $M$. If $I\subseteq M$ is a fixed ideal for which a fixed element $a\in R$, $a+I\neq I$. Prove that for any $0\neq y\in R$ and $y\notin I$, $ya\in I$ implies $ya=0$.

My attempt is:

$a+I\neq I$ implies that $a\notin I$. But $ya\in I$ for some $y\in R$ implies that $Ra\subseteq I$ (where $Ra$ is a left ideal generated by $a$), which means that $a\in I$, and this is a contradiction.

Is that reasoning right?

Answer 1

Counterexample : take $R=k[x]/(x^3)$ which is local with nilpotent maximal ideal $(x)$, take $I=(x^2)$, then for $a=y=x$ you have $ay=x^2\in I$, $a+I\neq I$ and $y\notin I$, but $ay\neq 0$