Maximal left ideals $\leftrightarrow$ simple left modules Suppose $R$ is a ring with unity. This passage in Lang's Algebra discusses the correspondence $$\text{Maximal left ideals of $R$} \leftrightarrow \text{Simple left $R$ modules},$$
where I corresponds to the left module $R/I$, and $M$ corresponds to $R/\text{Ann}_R(m)$ (the annihilator of $m$ in $R$), where $m$ is a generator of $M$. Lang says it is bijective up to isomorphism.
I'd like to understand this correspondence...is it true that $I \cong J$ on the left side implies $R/I \cong R/J$ on the right side? (I know this is not true for arbitrary subgroups of a group, for example). Edit: Commenters have shown that this is not true. I'm now interested mainly in the following:
I'm trying to prove that if $m$ and $n$ are two generators of $M$ simple, then $$\text{Ann}_R(m)\cong \text{Ann}_R(n) \tag{1}$$ as left $R$ submodules of $R$.
So far I haven't been able. We know that there exist $p,q \in R$ such that $m = pn$ and $n = qm$. Then one map from left to right in (1) is right multiplication by $p$, and from right to left we could take right multiplication by $q$. These are definitely homomorphisms...I'd like to show they're isomorphisms. So far I haven't been successful. 
Update: It seems there is a counterexample to the above maps being isomorphisms if we choose generators $1,2$ in $5\mathbb{Z}$. So we'll need to use another map to show the isomorphism (if it's correct).
 A: Consider $2\mathbb{Z}$ and $3\mathbb{Z}$, which are isomorphic as $\mathbb{Z}$-modules, but the quotients $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ are not.
Indeed, Lang's “bijective up to isomorphism” is not really clear.
If $S$ is a simple module, then $R/\operatorname{Ann}_R(m)\cong S$, for every $m\in S$, $m\ne0$. And this is as much as you can say.
A: The expression of a simple right module $S$ as a quotient $R/T$ by projecting $1\in R$ onto a generator of $S$ is such that $T$ depends on the choice of generator of $S$. 
For example, in $M_2(\Bbb R)$, consider the right ideal of matrices whose bottom row is zero. It's generated by both $[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}]$ and $[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}]$, but the annihilator of the former is the right ideal of matrices with top row zero, and the annihilator of the latter is the right ideal of matrices with bottom row zero.
In the quoted passage at least, the author did not express very clearly what was meant by "up to isomorphism." The way you first chose to interpret it does not work (as demonstrated by the examples given in $\Bbb Z$ in other solutions.)
You can find in the literature where two left ideals $L$, $L'$ are called similar if $R/L\cong R/L'$ as left $R$ modules. The $\Bbb Z$ example demonstrates maximal left ideals can be isomorphic without being similar. (And I think they can be similar without being isomorphic, but I don't have an example in mind.)
Similarity is clearly an equivalence relation on left ideals, and it remains so if you restrict it to maximal left ideals. Clearly no matter which generator for $S$ you pick, the resulting quotients of the form $R/L$ are all mutually isomorphic, so the simple left module corresponds to an equivalence class under similarity.
To reiterate: you could say that the equivalence classes of similar maximal left ideals are in bijective correspondence with the (iso)classes of simple left modules, using the scheme that Lang outlines.
A: There's an example (credited to Swan) in Section 17 of the book "Stable Modules and the $D(2)$-Problem" by F.E.A. Johnson (London Math. Soc. Lectures Notes 301 (2003)) of a ring $\Lambda$ (which is a maximal order in a quaternion algebra over $\mathbb{Q}(\zeta+\bar{\zeta})$, where $\zeta=e^{\pi i/8}$) and two maximal left ideals $\Omega_1$ and $\Lambda(a_1)$ of $\Lambda$ that are not isomorphic as left $\Lambda$-modules, but such that $\Lambda/\Omega_1\cong\Lambda/\Lambda(a_1)$ as left $\Lambda$-modules.
Most of this section seems to be accessible on Google books.
