Example of ring $R$ with ideals $I\neq J$ such that $R/I \cong R/J$ as modules It's easy to prove that if $I$, $J$ are two-sided ideals and $R/I\cong R/J$ as modules over $R$, then $I=J$. What about left ideals? Is there a simple counterexample?
I believe I've found an answer, but since answering own questions is encouraged, I thought I might post it here. Other examples are obviously welcome.
 A: When I try to find (counter)examples, I don't consider random examples first, but rather try to simplify the general case so that, in the end, examples pop out automatically without any effort. This also works here:
By the universal properties of $R$ and quotient modules, a homomorphism of left $R$-modules $R/I \to R/J$ corresponds to some element $[x] \in R/J$ such that $Ix \subseteq J$.
Hence, an isomorphism $R/I \cong R/J$ is given by two elements $x,y \in R$ such that $Ix \subseteq J$ and $Jy \subseteq I$ and $xy \equiv 1 \bmod I$ and $yx \equiv 1 \bmod J$.
For example, it could happen that $y=x^{-1}$. But then the conditions reduce to $Ix=J$.
Thus, any left ideal $I$ of a ring $R$ with a unit $x$ such that $Ix \neq I$ provides a counterexample.
As mentioned by Marcin Łoś and Andreas Caranti, we may look at $R=\mathrm{End}(V)$ for some vector space $V$ and $I=\{f : f(v)=0\}$ for some vector $v$. If $x \in R^*=\mathrm{GL}(V)$ is chosen such that $v,x(v)$ are linearly independent, then $Ix \neq I$.
A: Yes. 
Let $R$ be ring of endomorphisms of $\mathbb{R}^2$, $I$ and $J$ be annihilators of subspaces spanned by $(1,0)$ and $(0,1)$, respectively. Let $\theta\in R$ be given by $\theta(x,y)=(y,x)$, and $\phi(f)=f\circ \theta$ for $f\in R$. Then $\phi$ is an automorphism of $R$ as a module over itself, and a bijection between $I$ and $J$. Thus, it induces isomorphism between $R/I$ and $R/J$, since $I$ is the kernel of composition of $\phi$ with projection $\pi_J\colon R\rightarrow R/J$
A: I believe a slightly more elementary version of the example of Marcin Łoś consists in taking the ring $R$ of $2 \times 2$ matrices over $\mathbb{R}$, say, and the left ideals
$$
I = \left\{ \begin{bmatrix}a & 0\\ b & 0\end{bmatrix} : a, b \in \mathbb{R} \right\}
, \qquad
J = \left\{ \begin{bmatrix} 0 & a\\0 & b\\\end{bmatrix} : a, b \in \mathbb{R} \right\}.$$
