Prove $Rm$ simple $\iff Ann(m)$ is a left ideal of $R$. I'm trying to that prove $Rm$ is a simple ring $\iff Ann(m)$ is a left ideal of $R$.
I did a couple of questions earlier that I believe could be useful to show this:
$(Q1)$ Quotient of cyclic $R$-module is cyclic.
$(Q2)$ Isomorphism and cyclic modules
And I have the following theorem from the book:
$$\text{Let M be a bilateral ideal of a ring R. Then the following is equivalent:}\\
 (1) \text{ M is maximal}\\(2) \text{The ring }\frac{R}{M}\neq 0\text{ and is simple} $$
I believe the former theorem could be applied changing bilateral idea by left ideal, right?.
Then if $Rm$ simple by $(Q2)$ is $\frac{R}{Ann(m)}\simeq Rm$ (I would like to prove this using left $R-$modules, but is not explicit in $Q2$). By the last isomorphism is $\frac{R}{Ann(m)}$ simple and by the theorem follows $Ann(m)$ left maximal (instead of bilateral using left).
Conversely, if $Ann(m)$ is a left maximal, the ring $\frac{R}{Ann(m)}$ is simple, and by the ismorphism follows $Rm$ simple.
Now, what's left and I'm couldn't do is prove the former isomorphism: $\displaystyle\frac{R}{Ann(m)}\simeq Rm$. How could I do this?, and was the proof above right?.
 A: 
I'm trying to that prove $Rm$ is a simple ring $\iff$ $Ann(m)$ is a left ideal of $R$.

Thanks for including your work. It tells us where a couple problems are. First, I think you might be obfuscating simple rings with simple modules. Let's also be clear that (from context) you mean the annihilator of something called $m$ apparently in a left $R$ module.
You mention "if $Ann(m)$ is a left maximal, the ring $R/Ann(m)$ is simple," but $R/Ann(m)$ won't be a ring unless $Ann(m)$ is a two-sided ideal, and it usually won't be. But $R/Ann(m)$ certainly is a left module. Another thing is that the left annihilator of $m$ will always be a left ideal of $R$, regardless of what $Rm$ looks like. So there is something more wrong with the statement.
You can say this:

$Rm$ is a simple left $R$ module $\iff$ $Ann(m)$ is a maximal left ideal of $R$

The proof is to notice that $Rm\cong R/Ann(m)$ by a natural map: right multiplication by $m$. If $Ann(m)$ is a maximal left ideal, then by submodule correspondence, the module $Rm$ is simple.
It's also true that $R/I$ is a simple ring iff $I$ is a maximal two-sided ideal. You can't tie this to the annihilator of an element in a module, though, since the annihilator is not generally a two-sided ideal.
For commutative rings, of course, $Ann(m)$ will be an ideal, and the two cases merge.
