Let $M \subseteq \mathbb{Z}_{k_1} \times \cdots \times \mathbb{Z}_{k_n}$ be a $\mathbb{Z}$-module. What can we say about a generating set of $M$? Is there always a basis? Is there a generating set of size $n$?
Generating sets of $\mathbb{Z}_{k_1} \times \cdots \times \mathbb{Z}_{k_n}$
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0I've seen it in books when they treat the fundamental theorem of finitely generated abelian groups; I believe it is generally called "independence" (without the "linear"). – 2011-09-05
1 Answers
I'm assuming that $\mathbb{Z}_{k_i}$ means $\mathbb{Z}/k_i \mathbb{Z}$.
First, an $R$-module has a basis iff it is a free $R$-module. But any submodule $M$ of $\mathbb{M} := \mathbb{Z}/k_1 \mathbb{Z} \times \ldots \mathbb{Z}/k_n \mathbb{Z}$ is a torsion module, so it cannot have a basis (except in the trivial case $M = 0$ when the empty set is a basis).
But yes, each submodule $M$ can be generated by $n$ elements. Here is a somewhat cheap and dirty way to see this: the module $\mathbb{M}$ itself has an $n$ element generating set, hence so does every quotient module. But the submodules of $\mathbb{M}$ are isomorphic to the quotient modules of the dual $\mathbb{M}^{\vee} = \operatorname{Hom}(\mathbb{M},\mathbb{C}^{\times})$ which is isomorphic to $\mathbb{M}$.
The "better" way to derive this is via the structure theory of modules over a PID: invariant factors and such...
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0Thanks. Can I always pick vectors in row echelon form to form such a generating set? – 2011-09-05