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

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Example Modules
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
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
 
Physics news on Phys.org
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 $\Sigma = M/N$.
 
Euge said:
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
 
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 $N$.
 
[TIKZ]n[/TIKZ]
Euge said:
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
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top