Simplifying an expression for elements in a $\mathbb{Z}$-module

Question

I'm considering the following $\mathbb{Z}$-module

$$ K= \Big\{ \sum_{j=1}^k m_j \cdot \frac{p_j}{q_j}\cdot \frac{1}{D^{n_j}}: m_1,...,m_k\in \mathbb{Z},n_1,...,n_k\in \mathbb{Z}_{\geq 0} \Big\}, $$

where $\frac{p_1}{q_1},...,\frac{p_k}{q_k}\in \mathbb{Q}$ are reduced fractions and $D\in \mathbb{N}$. My question is under which conditions can I find $q\in \mathbb{N}$ such that I can write any element in $K$ as $\frac{m}{q}\cdot \frac{1}{D^n}$ for some $m\in \mathbb{Z}$ and $n\in \mathbb{Z}$.

I am pretty sure that any term in $K$ can be written as $\frac{m}{q}\cdot \frac{1}{D^n}$, with $q=\text{lcm}(q_1,...,q_k)$. It also seems like if $p_{i}\cdot \frac{q}{q_i}$ and $p_{j}\cdot \frac{q}{q_j}$ are co-prime, then I can express $\frac{1}{q}\cdot \frac{1}{D^n}$ and be done. Are there necessary and sufficient conditions for when this holds in general?

My algebra skills are pretty weak, so I would appreciate suggestions, remarks and examples of when this might not hold.

Answer 1

The number $q=\text{lcm}(q_1,...,q_k)$ fits. Indeed, it is well-known that any finitely generated Abelian group is a direct sum of cyclic groups. It follows that any finitely generated subgroup of $\mathbb Q$ is cyclic. Put $$ K_0= \Big\{ \sum_{j=1}^k m_j \cdot \frac{p_j}{q_j}: m_1,...,m_k\in \mathbb{Z}\Big\}.$$ Then $K_0$ is a finitely generated subgroup of $\mathbb Q$, and so $K_0$ is cyclic. Therefore $K_0$ is generated by some its element, which has a form $\frac pq$ for some $p\in\mathbb Z$. So $K_0=\frac pq\mathbb Z$ and $$K=\bigcup_{n=0}^\infty \frac 1{D^n}K_0=\bigcup_{n=0}^\infty \frac 1{D^n}\cdot\frac pq\mathbb Z.$$

Answer 2

This was a comment which got too long, as it is really only a variation on the theme of Alex Ravsky's answer.

First, in the definition of $K$, I think there is a typo -- the final factor $\frac{1}{D^{n_k}}$ I think should read $\frac{1}{D^{n_j}}$?

That said, in fact it makes no difference: if we set $f=\max\{n_j:1 \leq j\leq k\}$, then one can replace $\frac{m_j}{D^{n_j}}$ by $\frac{m_jD^{f-n_j}}{D^f}$, that is, if $x = \sum_{j=1}^k \frac{m_j}{D^{n_j}}.\frac{p_j}{q_j}$, and $m_j' = m_j D^{f-n_j}$ we have $$ x = \sum_{j=1}^k \frac{m_j}{D^{n_j}}.\frac{p_j}{q_j} = \frac{1}{D^f}\sum_{j=1}^k m_j'\frac{p_j}{q_j}. $$

Thus if we let $K_0 = \{\sum_{j=1}^k m_j\frac{p_j}{q_j}: m_j \in \mathbb Z, 1 \leq j \leq k\}$ as in Ravsky's answer, it follows immediately that $K = \bigcup_{f \geq 0} D^{-f}K_0$. Thus to show that any element of $K$ is of the required form, it suffices to show that any element of $K_0$ can be written as $\frac{a}{q}$ for some $q \in \mathbb Z_{>0}$.

For this, let $q = \mathrm{l.c.m.}\{q_j:1 \leq j \leq k\}$ and $r_j = q/q_j$. Then $p_j/q_j = (p_jr_j)/q$, and any element of $K_0$ can therefore be written in the form $q^{-1}\sum_{j=1}^k m_j(p_jr_j)$ for some integers $m_j$, ($1\leq j \leq k$). But if we let $$ I = \{\sum_{j=1}^k m_j.(p_jr_j): m_j\in \mathbb Z, 1\leq j \leq k\} $$ then clearly $I$ is an ideal in $\mathbb Z$ so that, by the division algorithm it is a principal ideal -- in fact $I= d\mathbb Z$ where $$ d = \mathrm{g.c.d.}\{p_jr_j:1 \leq j\leq k\}. $$

Thus $K = \bigcup_{n \geq 0}\mathbb Z.\frac{d}{q}.D^{-n}$.

[Aside: If you know a little more number theory, then the number $\frac{d}{q}$ is in effect the greatest common divisor of $\{p_j/q_j:1 \leq j \leq k\}$, since $K_0$ is a "fractional ideal" of $\mathbb Z$, which is therefore principal, and $d/q$ is a generator, and $K$ is a fractional ideal of $\mathbb Z_{(D)}$, the localization of $\mathbb Z$ in which $D$ becomes a unit.]