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This is a second followup question to this question I asked a couple of days ago (here is the first followup question). After resolving the issues I raised in both of the linked questions I proceeded to prove the Jordan-Holder theorem which says that

Jordan-Holder Theorem. Let $M$ be an $R$-module with a composition series $\left\{0 \right\} = M_{t} < \ldots < M_{0} = M$. Then:

  1. Every series of sub-modules can be refined to a composition series.
  2. Each pair of composition series $\mathcal{N},\mathcal{M}$ of $M$ has the same length and factors (up to order).

I was able to prove the first part, but unfortunately not the second. Here Is what I have tried:

Proof for $\mathbf{1}$ - I argue by induction that for every $n\in \mathbb{N}$ if $M$ is an $R$-module with a composition series of length $n$ then every series of submodules (not necessarily a composition series) of $M$ can be refined to a composition series (of $M$) of length $n$. Base case - for $n=1$ if $M$ has a composition series of length $1$ then it must be of the form $\left\{0 \right\} = M_{1} \leq M_{0} = M$ and hence $M$ is a simple module for which the claim holds. Induction step - Assume $n \in \mathbb{N}$ is such that for every $R$-module $M$ for which there exists a composition series of length $\ell \leq n$ the claim is satisfied (i.e each series of submodules of $M$ can be refined to a composition series of length $\ell$). Let $M$ be a $R$-module with a composition series $\left\{0\right\} = M_{n+1} < \ldots < M_{1} <M_{0} = M$ and let $\left\{0\right\} = N_{k} < \ldots < N_{1} < N_{0} = M$ be a series of submodules. By the claim proved in the followup question there exists a composition series $\mathcal{T}$ of $M$ such that $\mathcal{T}$ goes through $N_{1}$, $\mathcal{T}$ has length $n+1$, and all of its quotients are isomorphic to $M_{i}/M_{i+1}$. Denote it by $$\mathcal{T} \;:\; \left\{0 \right\} = T_{n+1} < \ldots < T_{r} = N_{1} < \ldots <T_{0} = M$$ where $1\leq r \leq n+1$. Since $1 \leq r \leq n+1$ It follows that $n+1 - r \leq n$, and therefore $N_{1}$ is an $R$-module with a composition series $\left\{0 \right\} = T_{n+1} < \ldots < T_{r} = N_{1}$ of length $\ell = \left(n+1 \right) - r \leq n$. By the induction hypothesis the series of submodules $\left\{0\right\} = N_{k} < \ldots < N_{1} $ can be refined to a composition series of $N_{1}$ of length $\ell$. Deonte it by $$\mathcal{S} \; : \; \left\{0\right\} = S_{n+1} < \ldots < S_{r} = N_{1}$$ Therefore the following series $$\mathcal{X} \; : \; \left\{0\right\} = S_{n+1} < \ldots < S_{r} = N_{1} = T_{r} < \ldots <T_{0} = M$$ is a series obtained by refining $\left\{0\right\} = N_{k} < \ldots < N_{1} < N_{0} = M$ which satisfies the properties that; it is of length $n+1$ and all of its factories are simple. Hence $\mathcal{X}$ is a composition series of $M$ obtained by refining $\left\{0\right\} = N_{k} < \ldots < N_{1} < N_{0}$ which proves $1$ in the theorem.

As said above I am unable to prove the second part and would very much appreciate any help in doing so. Thank you!

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    $\begingroup$ Here is a possible proof (page 6, 19, but I haven't checked that). $\endgroup$ Commented Nov 13 at 13:13
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    $\begingroup$ It seems to me that Chapter 6 of Atiyah and MacDonald's book ‘Introduction to Commutative Algebra’ contains the proof of this result, but I may be mistaken (I don't have a copy to hand at the moment). $\endgroup$ Commented Nov 13 at 13:25
  • $\begingroup$ Is this not covered in nearly all the books on modules? The argument is very similar to the one given for the analogous result on composition series of finite groups. The differences come from the facts that there the finiteness is never in doubt. OTOH on the group side you need to check the normality of certain subgroups - an aspect totally absent from the category of modules. $\endgroup$ Commented Nov 18 at 5:12

2 Answers 2

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$\DeclareMathOperator{\len}{len}$

Theorem (Jordan-Holder).

Let $R$ be a ring. Let $M$ be an $R$-module with $\len_R(M)=n<\infty$.

  • There exists a composition series $0 = M_0\subset\cdots\subset M_n = M$ of length $n$.
  • If $0 = N_0 \subset\cdots\subset N_r = M$ is another composition series, then $r=n$ and there is a permutation $\sigma$ of $r=n$ such that $M_i/M_{i-1} \cong N_{\sigma(i)}/M_{\sigma(i)-1}$.

Proof.

There must exist a chain $0= M_0\subset\cdots\subset M_n=M$ with $M_i\neq M_{i+1}$ by the definition of length. I claim that $M_{i+1}/M_i$ is simple for every $i$: if not, then we could lift a proper nonzero submodule $S\subset M_{i+1}/M_i$ to $M_i \subset \overline{S} \subset M_{i+1}$ with $M_i\neq \overline{S}$ and $\overline{S}\neq M_{i+1}$, demonstrating $\len_R(M) > n$. This proves the first claim.

To prove the second claim, we induct on $\len_R(M)$. If $\len_R(M)=0$, then $M=0$; if $\len_R(M)=1$, then $M$ is simple, and in either case the statement is trivial. If $M_{n-1}=N_{r-1}$, then we have the result by the inductive hypothesis; else $M_{n-1}+N_{r-1} = M$, giving $M/M_{n-1} \cong N_{r-1}/(M_{n-1}\cap N_{r-1})$ and $M/N_{r-1} \cong M_{n-1}/(M_{n-1}\cap N_{r-1})$. Letting $$0\subset K_1\subset \cdots\subset K_s \subset (M_{n-1}\cap N_{r-1})$$ be a chain of maximal length for $M_{n-1}\cap N_{r-1}$, we see that the two chains $$ 0\subset\cdots\subset K_s \subset (M_{n-1}\cap N_{r-1}) \subset M_{n-1} \subset M$$ and $$ 0\subset\cdots\subset K_s \subset (M_{n-1}\cap N_{r-1}) \subset N_{r-1} \subset M$$ are of the same length with the same subquotients up to swapping the final two. On the other hand, by the induction hypothesis applied to $M_{n-1}$, the first chain is equivalent to $0\subset M_1\subset\cdots\subset M_{n-1}\subset M$ and by the induction hypothesis applied to $N_{r-1}$, the second chain is equivalent to $0\subset N_1\subset\cdots\subset N_{r-1}\subset M$. Therefore all four of these chains are of the same length with the same subquotients up to permutation and we're done. $\blacksquare$

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Let me first prove the following claim:

Each pair of composition series $\mathcal{N}, \mathcal{M}$ of $M$ have the same length.

To prove this claim, we need the following lemma:

If $N$ is a submodule of $M$ with $N\neq M$, then $\ell(N)<\ell(M)$, where $\ell(N)$ denotes the least length of all composition series of $N$, similar for $\ell(M)$.

Proof. Suppose that $$M=M_0\supseteq M_1\supseteq \cdots \supseteq M_{\ell(M)}=\{0\}$$ is a composition series of $M$, then $$N=N\cap M_0\supseteq N\cap M_1\supseteq \cdots N\cap M_{\ell(M)}=\{0\}.\tag{1}$$ Note that$$(N\cap M_i)/(N\cap M_{i+1})\cong ((N\cap M_i)+M_{i+1})/M_{i+1}.$$ By definition, $M_i/M_{i+1}$ is a simple module. Since $((N\cap M_i)+M_{i+1})/M_{i+1}$ is a submodule of $M_i/M_{i+1}$, it is either a simple module or trivial. Thus, $\ell(N)\leq \ell(M)$. Now, assume $\ell(N)=\ell(M)$, then $$(N\cap M_i)+M_{i+1}=M_i, \forall\ 0\leq i\leq n-1.\tag{2}$$ We prove by backward induction that $N\cap M_i=M_i, \forall\ 0\leq i\leq \ell(M)=\ell(N)$. Clearly we have $N\cap M_{\ell(M)}=N\cap \{0\}=M_{\ell(M)}$. For $i\leq \ell(M)-1$, we have $$N\cap M_i=(N\cap M_i)+(N\cap M_{i+1})=(N\cap M_i)+M_{i+1}\stackrel{(2)}{=}M_i.$$ In particular, this gives $N\cap M=M$, i.e., $N=M$, which is a contradiction. Thus, $\ell(N)<\ell(M)$. $\square$

Now, we prove the claim:

Proof. Let $\mathcal{N}$ be the following composition series: $$M=M_0'\supseteq M_1'\supseteq \cdots \supseteq M_{k}'=\{0\}.$$ We get $$\ell(M)=\ell(M_0')>\ell(M_1')>\cdots>\ell(M_k')=0$$ by the lemma above. In particular, this shows that $k\leq \ell(M)$. But $\ell(M)$ is the least length of all composition series of $M$, so $k\geq \ell(M)$. Thus, $k=\ell(M)$. Since $\mathcal{N}$ is arbitrary, we are done. $\square$

Then we prove that

Each pair of composition series $\mathcal{N}, \mathcal{M}$ of $M$ have the same factors (up to order).

Proof. We prove it by induction. If $\ell(M)=0$, then this claim is vacuously true. Now, assume $\ell(M)>0$ and this claim is true for all modules $N$ with $\ell(N)<\ell(M)$. Given two composition series $$M=M_0\supseteq M_1\supseteq \cdots \supseteq M_{\ell(M)}=\{0\}$$and $$M=N_0\supseteq N_1\supseteq \cdots \supseteq N_{\ell(M)}=\{0\}.$$ If $M_1=N_1$, then we can reduce it to the composition series of $M_1$ and $N_1$ and use the induction hypothesis (since $\ell(M_1),\ell(N_1)<\ell(M))$. If $M_1\neq N_1$, then $M=M_1+N_1$. In this case, $$M/M_1=(M_1+N_1)/M_1\cong N_1/(M_1\cap N_1), M/N_1=(M_1+N_1)/N_1\cong M_1/(M_1\cap N_1).$$ Note that $\ell(M_1), \ell(N_1), \ell(M_1\cap N_1)<\ell(M)$, which means we can use induction for them. Let $$I_1=\{\text{composition factors of}\ M_1\}, I_2=\{\text{composition factors of}\ N_1\}, I_3=\{\text{composition factors of}\ M_1\cap N_1\}.$$ Then there is a one-to-one correspondence between $$I_1\cup \{M/M_1\}\ \text{and}\ I_3\cup \{M_1/(M_1\cap N_1)\}\cup \{M/M_1\}.$$ Note that $I_3\cup \{M_1/(M_1\cap N_1)\}\cup \{M/M_1\}$ is one-to-one correspondence to $$I_3\cup \{M/N_1\}\cup \{N_1/(M_1\cap N_1)\}.$$ Finally, $I_3\cup \{M/N_1\}\cup \{N_1/(M_1\cap N_1)\}$ is one-to-one correspondence to $$I_2\cup \{M/N_1\}$$ and we are done. $\square$

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  • $\begingroup$ what a great answer, thanks a lot! $\endgroup$ Commented Nov 16 at 8:27
  • $\begingroup$ @Shavit You are welcome :) $\endgroup$ Commented Nov 16 at 8:31

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