1
$\begingroup$

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 $a \in \mathfrak{m}$?

$\endgroup$
2
$\begingroup$

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 $x=0$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.