I am having a little trouble understanding the following:
"If $p_1, \ldots, p_k$ be the list of distinct primes dividing the product $mn,$ then we can factor $m$ and $n$ as $m=p_1^{r_1} \cdots p_k^{r_k}$ and $n=p_1^{s_1} \cdots p_k^{s_k},$ where $r_i$ and $s_i$ are nonnegative integers and may be zero in at most one of the decompositions for $m$ and $n.$"
The italics are not mine, they are the authors. What is confusing me is indeed the `at most one' part. Is it saying:
1) that we may have $p^0$ only one time in either the $m$ decomposition OR the $n$ decomposition (but certainly not in both decompositions)? For example, $m=p_1^0$ and $n=p_1^1$.
2) that there can be a $p^0$ only one time in EACH of them (but basically appearing in both of them)? For example, $m=p_1^0p_2^1$ and $n=p_1^1p_2^0.$
3) that there can be a $p^0$ only one time per prime? I'm guessing that this is not the case that the authoer intended. That much seems obvious because if we had a distinct $p^0$ for both $m$ and $n$, then why bother mentioning the $p$ at all? For example, $m=2^0 \cdot p^1$ and $n = 2^0 \cdot p^2.$
If anyone wants any further context, just let me know. I'm working with the Fundamental Theorem of Finite Abelian Groups and factoring $\Bbb{Z}_{mn}$ in various direct products $\Bbb{Z}_{k_1} \times \cdots \times \Bbb{Z}_{k_t}.$
If $p_1, \cdot, p_k$ is the list of all distinct prime divisors of $mn$, then $mn$ may be written as $mn = p_1^{t_1} \cdot p_2^{t_2} \cdot \ldots \cdot p_k^{t_k}$, where $t_1, t_2, \ldots, t_k \in \Bbb N$ are uniquely determined. And since $p_i \mid mn = p_1^{t_1} \cdot p_2^{t_2} \cdot \ldots \cdot p_k^{t_k}$ iff $t_i > 0$, all the $t_i$ are actually non-zero. Now, $m$ and $n$ may also be written as $m = p_1^{r_1} \cdot \ldots \cdot p_k^{r_k}$, $n = p_1^{s_1} \cdot \ldots \cdot p_k^{s_k}$, where $r_1, \ldots, r_k, s_1, \ldots, s_k \in \Bbb N$ are uniquely determined and furthermore $mn = (p_1^{r_1} \cdot \ldots \cdot p_k^{r_k})\cdot ( p_1^{s_1} \cdot \ldots \cdot p_k^{s_k}) = (p_1^{r_1}\cdot p_1^{s_1}) \cdot \ldots \cdot (p_k^{r_k}\cdot p_1^{s_k}) = p_1^{r_1 + s_1} \cdot \ldots \cdot p_k^{r_k + s_k}$. So, by the uniqueness of the $t_i$, we get $r_i + s_i = t_i \neq 0$ for all $i \in \{1, \ldots, k\}$ and therefore each pair $(r_i,s_i)$ has to contain at least one non-zero element.
edit: I assume $m,n \in \mathbb N$, if $m,n \in \mathbb Z$, you just have to add $(-1)^{r_0}$, $(-1)^{s_0}$ and $(-1)^{t_0}$ as a factor, respectively.