TITLE: Simple module is isomorphic to $R/M$ where $M$ is a maximal ideal
QUESTION [15 upvotes]: 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.

REPLY [1 votes]: 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
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