Proof of Bland's Example 13: Simple Modules and Quotients of Maximal Modules

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Chapter 1, Section 1.4 Modules ... ...

I need help with the proving a statement Bland makes in Example 13 ... ...

Example 13 reads as follows:View attachment 6387In the above text from Bland, we read the following:

" ... If $$N$$ is a maximal submodule of $$M$$, then it follows that $$M/N$$ is a simple $$R$$-module ... ... "I do not understand why this is true ... can anyone help with a formal proof of this statement ...
Hope someone can help ...

Peter

 

[Euge]

Let $\Sigma$ be a submodule of $M/N$. It must be shown that $\Sigma$ is either zero or the whole module $M/N$. Its pre-image $S$ under the natural projection $M\to M/N$ is a submodule of $M$ containing $N$. Maximality of $N$ implies $S = N$ or $S = M$. If $S = N$, then $\Sigma$ is zero; if $S = M$, then $\Sigma = M/N$.

 

[Math Amateur]

Let $\Sigma$ be a submodule of $M/N$. It must be shown that $\Sigma$ is either zero or the whole module $M/N$. Its pre-image $S$ under the natural projection $M\to M/N$ is a submodule of $M$ containing $N$. Maximality of $N$ implies $S = N$ or $S = M$. If $S = N$, then $\Sigma$ is zero; if $S = M$, then $\Sigma = M/N$.

Thanks for the help, Euge ...

But ... can you help a bit further ...

You write:

" ... ... Its pre-image $S$ under the natural projection $M\to M/N$ is a submodule of $M$ containing $N$. ... ... "Can you explain why this is true ... can you indicate how this is proved ...

Peter

 

[Euge]

You actually learned this already from the lattice isomorphism theorem. Let $\pi$ denote the natural projection. The module $N$ is contained in $S$, for if $n\in N$, then $\pi(n) = 0\in \Sigma$. Therefore $n\in \pi^{-1}(\Sigma) = S$. Since $n$ was arbitrary, $S$ contains $N$.

 

[Math Amateur]

[TIKZ]n[/TIKZ]

You actually learned this already from the lattice isomorphism theorem. Let $\pi$ denote the natural projection. The module $N$ is contained in $S$, for if $n\in N$, then $\pi(n) = 0\in \Sigma$. Therefore $n\in \pi^{-1}(\Sigma) = S$. Since $n$ was arbitrary, $S$ contains $N$.

Thanks Euge ... appreciate your help ...

Peter