Direct Product and Direct sum of a Category (Integer divide) 
For the following category, give a definition of the direct product and direct sum; determine if they exist; if they do, give an explicit description, and give an explicit description of the direct product and the direct sum over an empty set:

The category $D \mathbb{Z}$.  In this category, the objects are the integers, and there are no morphisms from $m$ to $n$ unless $m$ divides $n$, in which case there is only one morphism.


$$Ob(D \mathbb{Z}) = \mathbb{Z} $$
\begin{equation*}
\text{Mor}(m,n) = \begin{cases}
\{*_{m,n}\} & \text{if } \; m|n \\
\emptyset & \text{otherwise}. \\
\end{cases}
\end{equation*}
Where we use the notation $m\mid n$ means $m$ divides $n$ if there exists $k \in \mathbb{Z}$ s.t. $mk=n$.
I think the category $D \mathbb{Z}$ is the set of ideals created by $d \in D$
E.g. $D=3$ then $D \mathbb{Z} = \{..., -6,-3,0,3,6, ...\}$
The direct product is the usual multiplication $(\times)$ 
$$if \; m|n \Rightarrow m|\lambda \times n, (\forall \lambda) $$
And direct is sum the usual addition $(+)$
$$if \; m|a \; and \; m|b \Rightarrow m|a+b$$
There is no direct product nor direct sum when we do not have an $m|n$ as there is no morphism.
When we do have $m|n$ then the direct product and direct sum over an empty set are vacuously empty.
 A: (Direct sum) In the category $D\Bbb Z$, for $a,b\in \Bbb Z$, 
the direct sum is 
$$
a \coprod b := \operatorname{lcm}(a,b),
$$
the least common multiple of $a,b$ (which we'll take to be $\ge 0$). 
Clearly, $a\!\mid\!\operatorname{lcm}(a,b)$ and $b\!\mid\!\operatorname{lcm}(a,b)$.
That is, there are morphisms
$$
a\longrightarrow \operatorname{lcm}(a,b) \longleftarrow b.
$$
Suppose now that $x$ is such that
$$
a\longrightarrow x \longleftarrow b,
$$
which means $a\!\mid\! x$ and $b \!\mid\! x$. Then by definition of "$\operatorname{lcm}$", we have $\operatorname{lcm}(a,b)\!\mid\! x$, which is to say, there is a (necessarily unique) morphism
$$
\operatorname{lcm}(a,b)\longrightarrow x,
$$
such that 


*

*the (unique) morphism $a\to x$ equals the composition of the morphisms $\operatorname{lcm}(a,b)\to a\to x$, and

*the (unique) morphism $b\to x$ equals the composition of the morphisms $\operatorname{lcm}(a,b)\to b\to x$.


These two bullet points are trivial: "divides" is transitive, and for any $y,z\in \Bbb Z$ there is at most one morphism $y\to z$.
(Direct sum of the empty set) 
$$
\coprod \emptyset = 0.\quad\text{(Why?)}
$$
(Direct product, direct product of the empty set) Exercise.
