# Generating sets of $\mathbb{Z}_{k_1} \times \cdots \times \mathbb{Z}_{k_n}$

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 $n$?

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Sorry, I was trying to explain the scalar multiplication. The operation is indeed componentwise addition. – Thomas Sep 5 '11 at 21:01
If by "basis" you mean a generating set $\{a_1,\ldots,a_m\}$ such that for all integers $\alpha_1,\ldots,\alpha_m$, if $\alpha_1a_1+\cdots+\alpha_ma_m = 0$, then $\alpha_1=\cdots=\alpha_m=0$, then "no", there is only a basis if $k_i=0$ for all $i$. If you only require that $\alpha_ia_i=0$ hold, then "yes", but the size of the basis is not unique (e.g., $\mathbb{Z}_{15}$ can be generated by a single element, or by two linearly independent elements, if we write $\mathbb{Z}_{15}=\mathbb{Z}_3\times\mathbb{Z}_5)$. – Arturo Magidin Sep 5 '11 at 21:02
Note that any such $M$ is a finitely generated abelian group, so it is always isomorphic to a product of cyclic groups. – Arturo Magidin Sep 5 '11 at 21:03
Is there a name of this second definition of "linearly independent"? – Thomas Sep 5 '11 at 21:05
I'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"). – Arturo Magidin Sep 5 '11 at 21:07

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 $\mathbb{M}$.