Does $ax\in\mathfrak{m}I$ with $x\in I\setminus\mathfrak{m}I$ and $a \in R$ imply $a\in\mathfrak{m}$ for an invertible fractional $R$-ideal $I$?

Question

Let $R$ be an integral domain, $\mathfrak{m}$ a maximal ideal of $R$, and $I$ an invertible fractional $R$-ideal.

If $x \in I \setminus \mathfrak{m}I$ and $a \not\in \mathfrak{m}$, do we have $ax \not\in \mathfrak{m}I$?

Equivalently, does $ax \in \mathfrak{m}I$ with $x \in I \setminus \mathfrak{m}I$ and $a \in R$ imply $a \in \mathfrak{m}$?

Answer 1

No need any assumption on $I$ like being invertible.

$I/mI$ is an $R/m$-vector space. (If $ax=0$ in a $K$-vector space, then $a=0$ or $x=0$.)