MHB How Does the Correspondence Theorem Prove Maximal Submodules and Simplicity?

[Math Amateur]

I am reading Joseph J. Rotman's book: Advanced Modern Algebra (AMA) and I am currently focused on Section 7.1 Chain Conditions (for modules) ...

I need some help in order to gain a full understanding of some remarks made in AMA on page 526 on modules in the context of chain conditions and composition series for modules ... ...

The remarks read as follows:View attachment 4928My questions on this text are as follows:Question 1

In the above text we read:

" ... ... The Correspondence Theorem shows that a submodule $$N$$ of a left $$R$$-module is a maximal submodule if and only if $$M/N$$ is simple ... ... "

Can someone please explain exactly how the Correspondence Theorem leads to this result ... ?Question 2

In the above text we read:

" ... ... a left $$R$$-module is simple if and only if it is isomorphic to $$R/I$$ for some maximal left ideal $$I$$ ... ... "

Can someone please demonstrate, formally and rigorously why this is true ...?

Hope someone can help ...

Peter=================================================

The above post refers to the Correspondence Theorem for Modules which in Rotman's Advanced Modern Algebra is Theorem 6.22 ... I am therefore providing the text of Theorem 6.22 as follows:

https://www.physicsforums.com/attachments/4929

 

[Deveno]

Suppose $N$ is a maximal submodule of $M$. Consider the canonical projection map ($R$-module homomorphism):

$\pi: M \to M/N$ which sends $m \mapsto m + N$.

Suppose $S$ is a submodule of $M/N$. What the correspondence theorem tells us is that $S = \pi(M')$ for some submodule:

$N \subseteq M' \subseteq M$.

If we wished, we could write: $S = M'/N$.

Since $N$ is maximal, either $M' = N$, in which case $S = N/N = \{0_M + N\} = \{0_{M/N}\}$, or:

$M' = M$, in which case $S = M/N$.

Since we set no prior restrictions on $S$, this means that $\{0_{M/N}\}$ and $M/N$ are the *only* submodules of $M/N$, that is: $M/N$ is simple.

For the converse assertion, suppose $M/N$ is simple. Then any submodule $M'$ of $M$ which contains $N$ is (by the correspondence theorem!) a pre-image under $\pi$ of a submodule of $M/N$. Since $M/N$ is simple, there are only $2$ such submodules: $M/N$ and $\{0_{M/N}\}$. Hence either:

$M' = \pi^{-1}(M/N) = M$, or:

$M' = \pi^{-1}(\{0_{M/N}\}) = N$, proving the maximality of $M'$.

Digest this, and we'll tackle part two.

 

[Math Amateur]

Suppose $N$ is a maximal submodule of $M$. Consider the canonical projection map ($R$-module homomorphism):

$\pi: M \to M/N$ which sends $m \mapsto m + N$.

Suppose $S$ is a submodule of $M/N$. What the correspondence theorem tells us is that $S = \pi(M')$ for some submodule:

$N \subseteq M' \subseteq M$.

If we wished, we could write: $S = M'/N$.

Since $N$ is maximal, either $M' = N$, in which case $S = N/N = \{0_M + N\} = \{0_{M/N}\}$, or:

$M' = M$, in which case $S = M/N$.

Since we set no prior restrictions on $S$, this means that $\{0_{M/N}\}$ and $M/N$ are the *only* submodules of $M/N$, that is: $M/N$ is simple.

For the converse assertion, suppose $M/N$ is simple. Then any submodule $M'$ of $M$ which contains $N$ is (by the correspondence theorem!) a pre-image under $\pi$ of a submodule of $M/N$. Since $M/N$ is simple, there are only $2$ such submodules: $M/N$ and $\{0_{M/N}\}$. Hence either:

$M' = \pi^{-1}(M/N) = M$, or:

$M' = \pi^{-1}(\{0_{M/N}\}) = N$, proving the maximality of $M'$.

Digest this, and we'll tackle part two.

Thanks Deveno ...

Appreciate the help ...

Working through your post now ...

Peter

 

[Math Amateur]

Suppose $N$ is a maximal submodule of $M$. Consider the canonical projection map ($R$-module homomorphism):

$\pi: M \to M/N$ which sends $m \mapsto m + N$.

Suppose $S$ is a submodule of $M/N$. What the correspondence theorem tells us is that $S = \pi(M')$ for some submodule:

$N \subseteq M' \subseteq M$.

If we wished, we could write: $S = M'/N$.

Since $N$ is maximal, either $M' = N$, in which case $S = N/N = \{0_M + N\} = \{0_{M/N}\}$, or:

$M' = M$, in which case $S = M/N$.

Since we set no prior restrictions on $S$, this means that $\{0_{M/N}\}$ and $M/N$ are the *only* submodules of $M/N$, that is: $M/N$ is simple.

For the converse assertion, suppose $M/N$ is simple. Then any submodule $M'$ of $M$ which contains $N$ is (by the correspondence theorem!) a pre-image under $\pi$ of a submodule of $M/N$. Since $M/N$ is simple, there are only $2$ such submodules: $M/N$ and $\{0_{M/N}\}$. Hence either:

$M' = \pi^{-1}(M/N) = M$, or:

$M' = \pi^{-1}(\{0_{M/N}\}) = N$, proving the maximality of $M'$.

Digest this, and we'll tackle part two.

Thanks Deveno ... well, that is now clear ... and it not only answers Question 1 for me ... but now I think I understand what puzzled me from the beginning of my study of composition series of modules ... namely some remarks by Cohn in Introduction to Ring Theory (Springer Undergraduate Mathematics Series) that I was trying to understand ... hence your post has been most helpful ...

The relevant section from Cohn's book comes from Section 2.2 Chain Conditions (page 61) ... ... and reads as follows:View attachment 4937In the above text we read:

" ... Given a chain of submodules of $$M$$, we may be able to insert terms to obtain a refinement of our chain, i.e. a chain with more links; when this is not possible, our original chain is said to be a composition series. Thus a series

$$C_1 \subseteq C_2 \subseteq C_3 \ ... \ ... \ \subseteq C_r $$ ... ... ... (2.13)

is a composition series for $$M$$ iff $$C_r = M$$ and each factor $$C_i / C_{i - 1}$$ is a simple module. ... ... "So Cohn is essentially stating the proposition:A series of modules is such that $$C_r = M$$ and each factor $$C_i / C_{i - 1}$$ is a simple module if and only if no refinement of the series is possible ... ...

I was previously struggling to see exactly why this is the case ... and was trying to formulate a formal and rigorous proof ...

Given your post I think I can now see the rough form of a formal and rigorous proof of what Cohn is saying ...

... ... ...

Now you have essentially shown that

$$M/N$$ is simple if and only if the submodule $$N$$ of $$M$$ is maximal ...

which in terms of Cohn's series of submodules is

$$C_i / C_{i - 1}$$ is simple if and only if $$C_{i - 1}$$ is maximal in $$C_i$$

BUT ... $$C_{i - 1}$$ maximal in $$C_i$$ means there is NO submodule D containing $$C_{i - 1} \text{ in } C_i$$ (see Figure 1 below) ... and this means the series of submodules of $$M$$ cannot be refined ...

Can someone please critique my analysis ... possibly pointing out errors or shortcomings ...View attachment 4938Hope someone can help ...

Peter=====================================================

*** EDIT ***

To ensure that MHB readers can appreciate and understand Cohn's definitions and notation on this topic I am providing his introduction to the section on chain conditions which includes (2.13) ... ...https://www.physicsforums.com/attachments/4939

 

[Deveno]

...
Now you have essentially shown that

$$M/N$$ is simple if and only if the submodule $$N$$ of $$M$$ is maximal ...

which in terms of Cohn's series of submodules is

$$C_i / C_{i - 1}$$ is simple if and only if $$C_{i - 1}$$ is maximal in $$C_i$$

BUT ... $$C_{i - 1}$$ maximal in $$C_i$$ means there is NO submodule D containing $$C_{i - 1} \text{ in } C_i$$ (see Figure 1 below) ... and this means the series of submodules of $$M$$ cannot be refined ...

Can someone please critique my analysis ... possibly pointing out errors or shortcomings ...

Hope someone can help ...

Peter
... ...

What you have looks good except for one small (but important) detail-it should read:

"There is no PROPER submodule $D$ of $C_i$ PROPERLY containing $C_{i-1}$..."

for as stated, both $C_i$ and $C_{i-1}$ fulfill your condition.