Simple module is isomorphic to $R/M$ where $M$ is a maximal ideal In Michael Artin's Algebra textbook page 484 Chapter 12 Exercise 1.6:
A module is called simple if it is not the zero module and if it has no proper submodule.
(a) Prove that any simple module is isomorphic to R/M, where M is a maximal ideal.  
My questions are:  


*

*Is $R/M$ a left module over $R$? or over itself $R/M$?  

*When we defined a module $N$ we defined a map $R\times N\to N$ where $R$ is a ring and $N$ is an Abelian group. Now let the ring $R$ be $M$ as given in question, where $M$ is a maximal ideal of $R$. So the map becomes 
$M\times N\to N$. As far as I know, there no equivalent notion of ideal in module. The reason I said that is because from the definition of quotient module $V/W$, $W$ is just a submodule of $V$, not an "ideal" submodule of $V$. Is my reasoning correct?   

*If my reasoning in 2. above is wrong, then how can we interpret a maximal ideal in a language of module? Is there a definition of maximal ideal in module distinct to definition of a maximal ideal in ring?  

*To prove the question, I consulted other sources in the internet. There is a proof that says "Any submodule of $R/M$ is an ideal, since it would be an additive group that is closed under multiplication by elements of $R$ and therefore $R/M$." Since I am still not clear of the definition of ideal in module, I am confused by the meaning of the above statement, especially what does it mean by "closed under multiplication by elements of R and therefore $R/M$"?  


Thank you very much for any helps. I really appreciate it.
 A: An right $R$ module is, as you said, an abelian group $N$ and the map from $N\times R\to N$ satisfying special properties like distributivity etc. In other words, $nr$ is another element of $N$ for every $n\in N$, and $r\in R$.
If $N'$ is an additive subgroup of $N$, we say that it's a submodule if $n'r\in N'$ for every $n'\in N'$ and $r\in R$. In other words, it looks like $N'R\subseteq N'$ (which didn't necessarily have to happen!).
Now $R$ can be considered as a right module over itself, since the ring multiplication $R\times R\to R$ satisfies all the axioms that are necessary to make $R$ into a right module. Since we have already defined what a submodule is, we can ask what the submodules of the right module $R$ are. They an additive subgroup $I$ of $R$ is a submodule if $IR\subseteq I$. 
You'll notice that looks like the definition of a right ideal: and that's exactly what it is! The term "right ideal" is just another term for a submodule of the right module $R$. Similarly, $R$ can be considered as a left module over itself, and you can ask about its left submodules (=left ideals). Nobody says "ideals in a module" because ideals are reserved as a term for submodules of $R$.
If you already know that for any submodule $N'$ of $N$ you can form the quotient module $N/N'$, then it will be no surprise that if $M$ is a right ideal (=right submodule!) of $R$, then you can form the quotient module $R/M$.
The proof of the title fact is very easy if you know the isomorphism theorems for modules. If $S$ is a right $R$ module (nonzero, of course), pick a nonzero $s\in S$ and make a map from $R\to S$ by $r\mapsto sr$. Call this map $\phi$. This is clearly a nonzero map to $S$. Since the image is a submodule of $S$ and $S$ is simple, it must be all of $S$. By the isomorphism theorem, $R/\ker(\phi)\cong Im(\phi)=S$. By the correspondence of submodules, the absence of submodules of $S$ corresponds to an absence of submodules between $R$ and $\ker(\phi)$, and so $\ker(\phi)$ is a maximal right ideal of $R$.
A: Please forgive if its a double post, I'm really a beginner in this Module Theory. So please mention if there is any flaw in argument.
Lemma: Let $f:V\to W$ be a module homomorphism, where $V,W$ are both $R$ modules. Let $V',W'$ be respectively submodules of $V,W$. Then image of $V'$ and pre-image of $W'$ are also submodules of their repective space.
Proof: I'm only showing image part. We've to show $f(V')$ is submodule of $W$. Its clealry additive subgroup of $W$ since $f$ is a homomorphism and $V'$ was additive subgroup of $V$. Now take $f(v')\in f(V')$. For all $r\in R, rf(v')=f(rv')\in f(V')$ since $rv'\in V'$. So, $f(V')$ is a submodule of $W$.
Now for original problem. let $S$ be a simple $R$ module. So, for any $s\in S, Rs$ be a submodule of $S$, asper condition it has be $S$. $S$ any element of $S$ can be written as $sr$ for a fixed $s\in S$. This gives a surjection $R\to S$, given by $\phi_{s}:r\to sr$, which is a homomorphism. So exist an ideal $I$ of $R$ such $S$ is isomorphic to $R/I$. So if $I$ is maximal then done, if not exist a proper ideal $J$ of $R$ containing $I$. But then $J/I$ is proper ideal of $R/I$ and so submodule of it too. Now we know image of $R/I$ is a proper submodule of $S$ since there is a isomorphism $R/I \to S$, contradiction, so $I$ is maximal ideal
