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I saw the following definitions in the book 'Module Theory - An approach to linear algebra' by T.S. Blyth:

Let $R$ be a unitary ring and let $M$ be an $R$ module. The author defines a tower of $R$-submodules of $M$ as a finite decreasing(strictly) chain of modules $T$ (of the following form) $$T:M=M_0\supset M_1\supset M_2\supset\ldots\supset M_r=\{0\}$$ Suppose that $T_1$ and $T_2$ are such towers of $R$-submodules of $M$: $$T_1:M=M_0\supset M_1\supset M_2\supset\ldots\supset M_s=\{0\}$$ $$T_2:M=N_0\supset N_1\supset N_2\supset\ldots\supset N_t=\{0\}$$ Then, $T_1$ and $T_2$ are defined to be equivalent if $s=t$ and there is a permutation $\sigma$ of $\{1,2,\ldots,s-1\}$ such that for all $i=0,1,\ldots, s-1$, $\frac{N_i}{N_{i+1}}\cong\frac{M_{\sigma(i)}}{M_{\sigma(i)+1}}$. The author goes on to prove the Schreier's refinement theorem (which states that given two towers, one can always find refinements of them that are equivalent).

Question: What is the motivation behind the definition of equivalence on towers and why is it interesting/useful?

Any insights are appreciated. Thank you.

RichardAshcroft
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    Well, the maximum height of a module tower tells you about the size of a module in an interesting way, for starters. I view module towers similar to how I view divisibility towers (for a natural number $n>0$, a divisibility tower is a sequence of naturals $n_1\mid n_2\mid\ldots\mid n_k$ with $n_k=n$, and $n_i<n_{i+1}$). For divisibility towers, the maximum size of a tower tells you about the factorization of $n$. – Rushabh Mehta Nov 30 '21 at 20:52
  • @DonThousand Thank you. The definition of a tower indeed is natural and interesting to me. My question was meant to be (I apologize for not making it clear enough) about the definition of equivalence on towers. It didn't seem natural to me off the top of my head. If you could expand on that, it'd be great. I'll make an edit in this regard. – RichardAshcroft Nov 30 '21 at 20:58
  • @DonThousand On second thought, the divisibility tower analogy is really nice and sort of addresses my question. Thanks again. – RichardAshcroft Nov 30 '21 at 21:04

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