$Length of a matrix module$

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$Length of a matrix module$

Let $R$ be a finite principal ideal ring and $M \subseteq R^m$ an $R$-module. Let $A$ denote the $R$-module of all $m \times n$ matrices over $R$, with $m \leq n,$ whose column space (the $R$-module generated by their columns) is a submodule of $M$. Then I would like to prove the claim that $\lambda_R(A) = n\lambda_R(M)$, where $\lambda_R$ denotes length as an $R$-module.

In the case that $M$ is free (say of rank $k$), we can argue by considering a basis $B= \{m_1, \dots, m_k\}$ of $M$ and prove the claim by showing that the matrices $M_{ij} = [0 \mid 0 \mid \cdots \mid m_j \mid \cdots \mid 0]$,

whose $i$-th column is $m_j$ and all other columns are $0$, form a basis of $A$, by using the linear independence and spanning properties of $B$ to imply the same properties for the set $\{M_{ij}\}.$ However, I can't see how to extend this argument to the case that $M$ is not free: even if we consider a good enough generating set $B$ of $M$, we cannot use the linear independence to imply the linear independence of $\{M_{ij}\}$.

Is the claim wrong when $M$ is not free? And if it's true, how could we prove it?

Algebra Abstract Algebra Linear Algebra

Elviegem

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First of all, let us clarify a detail in the definition of A, "the R-module of (m x n)-matrices whose column space is a submodule of M". As a module, this (A) can be identified with the product space $M \times ... \times M = M^n$. Indeed, any column of a matrix in A must be an element of M (otherwise the space spanned by the columns is not a submodule of M), and conversely, any n-tuple of elements in M will span a subspace of M (their span being by definition the smallest submodule that contains all these n elements, possibly shown to be equal to the set of all their linear combinations).

Thanks to the Jordan–Hölder theorem, the length of a module is exactly the length of any composition series of that module. So you can prove the property following roughly the same idea as what you mentioned for the case of a free module, with "base" replaced with composition series.

Indeed, if $\{o\}=N_0\subsetneq N_1\subsetneq...\subsetneq N_k=M$ is a composition series of $M$, with $k=\lambda_R(M)$, then $N_0^{n}\subsetneq N_1\times N_0^{n-1}\subsetneq ...\subsetneq N_k\times N_0^{n-1}\subsetneq N_k\times N_1\times N_0^{n-2}\subsetneq ...\subsetneq N_k^{n-1}\times N_{k-1} \subsetneq N_k^n = M^n = A$ is a composition series of $A$ (with the identification $M^n=A$ mentioned in the first paragraph): It is easy to see that the quotient of any two consecutive spaces in this sequence is simple (as a module), namely, isomorphic to the respective $N_{j} / N_{j-1}$.
This sequence has $1+n\cdot k$ terms, so $\lambda_R(A)=n\,k=n\,\lambda_R(M)$, as expected.

Erdos

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$Length of a matrix module$

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