Abstract Algebra – Simple Module is Isomorphic to R/M Where M is a Maximal Ideal

abstract-algebramodules

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:

  1. Is $R/M$ a left module over $R$? or over itself $R/M$?
  2. 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?
  3. 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?
  4. 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.

Best Answer

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 $R$.

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