MHB Chains of Modules and Composition Series

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings, on Page 61 we find a definition of a refinement of a chain and a definition of a composition series.

The relevant text on page 61 is as follows:View attachment 3181

In the above text, Cohn indicates that a refinement of a chain (added links) is a composition series for a module $$M$$, but then goes on to to characterise a composition series for a module $$M$$ as a chain in which $$C_r = M$$ for some positive integer $$r$$, and for which $$C_i/C_{i-1}$$ is a simple module for each $$i$$.

So then, is Cohn saying that if a refinement is not possible, then it follows that $$C_r =M$$ for some $$r$$ and $$C_i/C_{i-1}$$ is a simple module for each $$i$$? If so, why/how is this the case?

Peter

 

[mathbalarka]

Claim For any $R$-module $R$, and a maximal submodule $M$ of $R$, $R/M$ is simple

Since every submodule of an $R$ module is a module over an ideal of $R$, we can produce an ideal $I$ of $R$ such that $M$ is an $I$-module. Consider the canonical map $\varphi :R \to R/I$. For any submodule $N$ of $R/M$, we can produce an ideal $J$ of $R$ such that $N$ is a $J$-module. $\varphi^{-1}(J)$ contains $I$, but is a proper ideal of $R$. Hence the minimality of $I$ is contradicted, thus $R/I$ is simple, hence so is the $R/I$-module $R/M$. $\blacksquare$

Let $M$ be a module and $0 = M_0 \subset M_1 \subset M_2 \subset \, \cdots \, \subset M_{n - 1} \subset M_n = M$ be a composition series for $M$. By Cohn's definition, you see that this is a chain which can't be "expanded", i.e., $M_{i-1}$ is maximal in $M_i$. By the theorem above, the composition factors $M_i/M_{i-1}$ are simple. $\blacksquare$

 

[Math Amateur]

Claim For any $R$-module $R$, and a maximal submodule $M$ of $R$, $R/M$ is simple

Since every submodule of an $R$ module is a module over an ideal of $R$, we can produce an ideal $I$ of $R$ such that $M$ is an $I$-module. Consider the canonical map $\varphi :R \to R/I$. For any submodule $N$ of $R/M$, we can produce an ideal $J$ of $R$ such that $N$ is a $J$-module. $\varphi^{-1}(J)$ contains $I$, but is a proper ideal of $R$. Hence the minimality of $I$ is contradicted, thus $R/I$ is simple, hence so is the $R/I$-module $R/M$. $\blacksquare$

Let $M$ be a module and $0 = M_0 \subset M_1 \subset M_2 \subset \, \cdots \, \subset M_{n - 1} \subset M_n = M$ be a composition series for $M$. By Cohn's definition, you see that this is a chain which can't be "expanded", i.e., $M_{i-1}$ is maximal in $M_i$. By the theorem above, the composition factors $M_i/M_{i-1}$ are simple. $\blacksquare$

Hi Mathbalarka,

Thanks for the help!

In your post you write:" … …every submodule of an $R$ module is a module over an ideal of $R$, … … "

Can you give more details on this proposition … indeed can you show that this is the case … that is, prove that this is true … …

Would appreciate your help on this matter.

Peter

 

[Deveno]

Hi Mathbalarka,

Thanks for the help!

In your post you write:" … …every submodule of an $R$ module is a module over an ideal of $R$, … … "

Can you give more details on this proposition … indeed can you show that this is the case … that is, prove that this is true … …

Would appreciate your help on this matter.

Peter

Hmm...let's see what we can do. Suppose $M$ is our (right) $R$-module, and $N$ our submodule. We need to somehow produce an ideal $I$ of $R$, so that $N$ is an $I$-module.

My first thought is to set:

$I = \{a \in R: na \in N,\ \forall n \in N\}$.

Does this work?

Suppose $a,b \in I$. Then for any $n \in N$, we have $na,nb \in N$. Since $N$ is an abelian group under the module addition of $M$, and thus closed under addition, we have $na + nb = n(a+b) \in N$. Since this is true of every $n \in N$, we have $a+b \in I$.

Note that if $a \in I$, then for any $n \in N$, we have $na \in N$ (since submodules are closed under the $R$-action), and since an abelian group contains all additive inverses, we have:

$-na = n(-a) \in N$, so that $-a \in I$ whenever $a \in I$. Thus $I$ is an additive subgroup of $R$.

Now suppose we have $r \in R$, and $a \in I$. Since $N$ is an $R$-submodule, it is closed under the (right) $R$-action, that is:

$n \in N \implies nr \in N$ for any $r \in R$.

So if $a \in I$, then for any $n \in N,\ na \in N$, and thus $(na)r = n(ar) \in N$, thus $ar \in I$.

This establishes that $I$ is at least a right ideal of $R$.

To see that $I$ is also a left ideal of $R$, note that if $n \in N$, then $nr \in N$ for any $r \in R$, whence (by the definition of $I$), $n(ra) = (nr)a \in N$, so that $ra \in I$, for any $r \in R$, and $a \in I$.

So $I$ is indeed a (two-sided) ideal of $R$.

So we have an ideal of $R$, now, and by the way we defined it, we see that the $I$-action on $N$ is just the $R$-action restricted to $I$. That it satisfies the rules:

$n(a+b) = na + nb$
$(n+n')a = na + n'a$ is clear because these hold for any $n,n' \in N$ and $a,b \in R$.

Similarly, the property:

$(na)b = n(ab)$ is also inherited from the $R$-module structure of $M$.

So the only "real" thing to verify, is that we have closure (of addition, and scalar multiplication). The first follows from the fact that $N$ is an additive subgroup of $M$. The second follows from the way we defined $I$.

A similar proof holds if we take $M$ to be a left $R$-module.

I've never seen this proved before, I just "made it up as I went along". In other words, it just sort of "flows" from the definitions.