If $xR=I$ we can say that $x\in I$?

Question

Probably this is a dumb question, I lack knowledge on abstract algebra. Suppose that $R$ is a ring without unity and $I\subset R$ is a non-trivial ideal.

If $xR=Rx=I$ for some $x\in R$, we can says that $x\in I$? If not, there is some other conditions that ensures that $x\in I$?

Answer 1

The request for a counterexample has already been amply covered by existing answers/comments mentioning subrngs of $\mathbb Z$.

If $xR=Rx=I$ for some $x\in R$, we can says that $x\in I$? If not, there is some other conditions that ensures that $x\in I$?

Trivially $R$ having identity ensures this, but I understand you're probably interested in weaker conditions.

One interesting condition that jumps to mind for me is for $I$ to be a modular ideal. If $I$ is left modular, that means that there exists $e\in R$ such that $re-r\in I$ for all $r\in R$. (Said another way, $R/I$ has a right identity.)

But now look at what that would mean for $I=xR=Rx$: you'd have $xe-x\in I$, but you already know $xe\in I$, so that would mean $x\in I$ as well.

Obviously being right modular would suffice as well, given your assumption that $Rx=I$.

Answer 2

Per Michael Burr's suggestion in the comments, look at $R=2\mathbb Z$, $I=4\mathbb Z$, and $x=2$. $xR=Rx=4\mathbb Z=I$, but obviously, $2\notin4\mathbb Z$.