I originally asked this on [Stack Exchange](https://math.stackexchange.com/questions/3821785/r-i-cong-r-textann-rr-i-but-i-neq-textann-rr-i) but with no luck. Upon doing research with some noncommutative rings, I thought of a curious question. Does there exist a noncommutative unital ring $R$ and left ideal $I$ such that $R/I\cong R/\text{Ann}\_R(R/I)$ as left $R$-modules but $I\neq \text{Ann}\_R(R/I)$? I suspect that such an $R$ and $I$ do exist but are awkward to construct. Note that


$$\text{Ann}\_R(R/I)=\{r\in R:\forall x\in R,\ rx\in I\}\subseteq I$$


and that $\text{Ann}\_R(R/I)$ is always a two-sided ideal. As such, if $I$ is not two-sided then we already get $I\neq \text{Ann}\_R(R/I)$. Any ideas would be appreciated.


My question is equivalent to [this question](https://math.stackexchange.com/questions/878020/example-of-ring-r-with-ideals-i-neq-j-such-that-r-i-cong-r-j-as-modules) with the additional constraint that $J$ must be two-sided.



$R/I\cong R/\text{Ann}_R(R/I)$ but $I\neq\text{Ann}_R(R/I)$
You might as well assume that $\mathrm{Ann}\_R(R/I)=0$ since if $S=R/\mathrm{Ann}\_R(R/I)$ then $R/I\cong S$ as $R$-modules if and only if $S/(I/\mathrm{Ann}\_R(R/I))\cong S$ as $S$-modules.


So now the question is whether you can have a ring $R$ and a non-zero left ideal $I$ such that $R/I\cong R$. Equivalently can there be $r\in R$ such that (i) $\mathrm{ann}\_R(r)=I\neq 0$ and (ii) there is some $s\in R$ with $sr=1$; if $\varphi\colon R/I\to R$ is the isomorphism then take $r=\varphi(1+I)$ and $\varphi(s+I)=1$. To further rephrase we want an element $r$ of $R$ that is a right zero-divisor and a right unit.


This is indeed possible. Here is one example: let $R$ be the $\mathbb{C}$-linear endomorphisms of the polynomial ring $\mathbb{C}[X]$ and let $r$ be the element of $R$ given by multiplication by $X$. It is easy to construct a left inverse for $r$ (in fact one can easily find infinitely many). But if $x$ is the element of $R$ that sends a polynomial to its constant term viewed as the constant polynomial then $xr=0$.


**The punchline** If you just wanted to pull a rabbit out of a hat you could simply say that for $R=\mathrm{End}\_{\mathbb{C}}(\mathbb{C}[X])$ and $r\in R$ given by $r(f)=Xf$, $I=\mathrm{ann}\_R(r)\neq 0$ and there is an isomorphism of left $R$-modules $R/I\to R$ given by $x+I\mapsto xr$. It is surjective since there is $s\in R$ with $sr=1$ (Moreover $\mathrm{Ann}\_R(R/I)=0$).


