Let $A$ be a finite subset of $\mathbb Z^2$. Let $\mathbb ZA$ be the subgroup of $\mathbb Z^2$ generated by $A$. Let $\mathbb R_{+}, \mathbb Q_{+}$ denote the set of non-negative real and rational numbers respectively. Then I know that for some $a_1,...,a_n \in \mathbb ZA \cap \mathbb Q_{+} A$ , we have $\mathbb ZA \cap \mathbb Q_{+} A=\mathbb ZA \cap \mathbb R_{+} A=\mathbb Na_1+\cdots+\mathbb Na_n$ .
Here $\mathbb N$ is the set of non-negative integers.
My question is:
If $A=\{ (0,4); (4,0); (1,3); (3,1)\}\subseteq \mathbb Z^2$, then how to find a finite set of generators for $\mathbb ZA \cap \mathbb Q_{+} A$ ?
i.e. how to find finitely many $a_1,...,a_n\in \mathbb ZA \cap \mathbb Q_{+} A$ such that $\mathbb ZA \cap \mathbb Q_{+} A=\mathbb Na_1+\cdots+\mathbb Na_n$ ?
NOTE: Just to clarify, $\mathbb Q_{+}A$ is the set of all finite $\mathbb Q_{+}$-linear combination of elements of $A$ (and similarly for $\mathbb R_{+}A$ ... )
Note that for all $(x,y)\in A$ you have $x+y\equiv0\pmod{4}$, so the same holds for all $(x,y)\in\Bbb{Z}A$. Moreover you have $$(4,0),\ (3,1),\ (2,2),\ (1,3),\ (0,4)\in\Bbb{Z}A,$$ and of course $$\Bbb{Q}_+^2\supseteq\Bbb{Q}^+A\supseteq\Bbb{Q}_+(4,0)+\Bbb{Q}_+(0,4)=\Bbb{Q}_+^2,$$ and from this it follows that $$\Bbb{Z}A\cap\Bbb{Q}_+A=\Bbb{N}(4,0)+\Bbb{N}(3,1)+\Bbb{N}(2,2)+\Bbb{N}(1,3)+\Bbb{N}(0,4),$$ and it is not hard to see that this generating set is minimal. I leave the details of the above to you.