Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that $M$ is a submodule of $\mathbb{Z}^n$, but $\mathbb{Z}^n/M$ is not free. We can write

$\mathbb{Z}^n/M \simeq \mathbb{Z}/(q_1)\oplus\cdots\oplus \mathbb{Z}/(q_r) \oplus \mathbb{Z}^t$

for some $r$ and $t$ and $q_1|q_2|\cdots|q_r$.

[Question] I'd like to construct a submodule $M'\subset M$ such that $\mathbb{Z}^n/M'$ is free, specifically, $\mathbb{Z}^n/M' \simeq \mathbb{Z}^{r+t}$. Is this always possible? How should I define $M'$?

(I know how to construct $M'\supset M$ such that it removes the torsion part of $\mathbb{Z}^n/M$, i.e., $\mathbb{Z}^n/M'\simeq \mathbb{Z}^t$. One can simply let $M' = \{x: \text{there exists }a\neq 0\text{ such that } ax\in M\}$. But I am not sure how to shrink $M$ to make $\mathbb{Z}^n/M$ free as desired above.)

Thanks.

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

I assume that $M$ is a submodule of ${\Bbb Z}^n$, and that the $q_i$s are defined so that $q_1>1$ and $q_r>0$. Here is an algorithm for constructing $M'$.

If $M$ has generators $a_1$, $\dots$, $a_m$, and if we pick a basis $b_1$, $\dots$, $b_n$ of ${\Bbb Z}^n$, we can represent each $a_i$ as $\sum_j s_{ij} b_j$, where $s_{i1}$, $\dots$, $s_{in}\in{\Bbb Z}$, so that the generating set of $M$ can be represented by the $m$ by $n$ matrix $S=(s_{ij})$ over $\Bbb Z$. Then, $S$ can be reduced to Smith normal form by left- and right-multiplying by invertible square matrices over $\Bbb Z$. This is equivalent to picking a new generating set for $M$ and a new basis for ${\Bbb Z}^n$.

Once $S$ is in Smith normal form it is a (possibly rectangular) diagonal matrix with $1$, $\dots$, $1$, $q_1$, $\dots$, $q_r$, $0$, $\dots$, $0$ along the diagonal. Supposing that there are $u$ $1$s, $M'$ will be generated by the first $u$ elements of the new basis for ${\Bbb Z}^n$, which are the same as the first $u$ elements of the new generating set for $M$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.