A semigroup $\mathbb{N}\mathcal{A}$ is saturated in $M$ if and only if $\mathbb{N}\mathcal{A}=\text{Cone}(\mathcal{A})\cap M$ I'll first recall some definitions here for convenience. Given a finite set $S$ in a real vector space $V$, $$\text{Cone}(S)=\{\sum\limits_{u\in S} \lambda_u u\mid \lambda_u\geq 0\}.$$ An affine semigroup $S\subset M$ is called saturated if for all $k\in \mathbb{N}$ and $m\in M$, $km\in S$ implies $m\in S$.
The following is exercise 1.3.4 from Cox, Little, and Schenck's Toric Varieties: Let $\mathcal{A}\subseteq M$ be a finite set. Prove that the semigroup $\mathbb{N}\mathcal{A}$ is saturated in $M$ if and only if $\mathbb{N}\mathcal{A}=\text{Cone}(\mathcal{A})\cap M$. (Hint: Apply (1.2.2) to $\text{Cone}(\mathcal{A})\subset M_{\mathbb{R}})$.
First of all I think the hint might have a typo, as (1.2.2) doesn't seem to have anything to do with this problem. But anyway, here's what I've got so far:
One inclusion is obvious, so we'll do the other. Let $\mathcal{A}=\{m_1,\dots,m_n\}$ and let $s\in \text{Cone}(\mathcal{A})\cap M$. We can write $s=\sum\limits_{i=1}^n a_im_i$ with $a_i\geq 0$. Now write $a_i=\lfloor a_i\rfloor + \delta_i$, so that $s=\sum\limits_{i=1}^n\lfloor a_i\rfloor m_i + \sum\limits_{i=1}^n \delta_i m_i$. Rearranging we get $$s-\sum\limits_{i=1}^n\delta_i m_i=\sum\limits_{i=1}^n \lfloor a_i\rfloor m_i\in\mathbb{N}\mathcal{A}.$$ Now, if we could ensure the $\delta_i$ were rational, then we'd be done. Indeed, we could multiply this equation through by some large natural number to kill all the denominators, then use the saturatedness of $\mathbb{N}\mathcal{A}$ to conclude that $s\in \mathbb{N}\mathcal{A}$. Can such an assumption be made?
 A: Nathan Lowry: in case you still need clarification on this. Expressing $s$ as a linear combination of the $m_i$ is a linear system of equations with the $m_i$ as matrix, $s$ as right-hand side, and the $a_i$ as indeterminates. Whether a linear system (defined over the rationals) has a solution does not depend on whether you look for a rational or a real solution; if a (real) solution exists at all, any natural method you use (Gauss, Cramer, etc) will give you a rational one.
A: The assumption may fail for an arbitrary choice of the $a_i$’s, but you can modify them. The problem is reduced to: let $A\in M_{r\times n}(\mathbb{Z})$, $b\in M_{r\times1}(\mathbb{Z})$ and $a=(a_1,\cdots,a_n)^T\in M_{n\times 1}(\mathbb{R}_{\geq 0})$ such that $Aa=b$, then find $a’=(a_1’,\cdots,a_n’)^T \in M_{n\times 1}(\mathbb{Q}_{\geq0}) $ with $Aa’=b$.
One can write $a=x_0+\mu_1e_1+\cdots+\mu_re_r$ (see for example this post), where $x_0\in M_{n\times 1}(\mathbb{Q})$ is a solution to $Ax=b$, $\mu_i\in\mathbb{R}$ and
$e_1,\cdots, e_r\in M_{n\times 1}(\mathbb{Q}) $ is a basis for the $\mathbb{R}$-solution space of $Ax=0$. If $a_i>0$ for all $i$, then we can modify $\mu_i$ to $\mu_i’\in\mathbb{Q}$ such that $a’= x_0+\mu_1’e_1+\cdots+\mu_r’e_r \in M_{n\times 1}(\mathbb{Q}_{\geq0})$.
Otherwise, we may assume $a_1=\cdots=a_l=0$ and $a_j>0$ for $l<j\leq n$. Then $(\mu_1,\cdots,\mu_r)^T$ is a solution to $Cy=d$ where $C\in M_{l\times r}(\mathbb{Z})$ and $d\in M_{l\times 1}(\mathbb{Z})$. Again,
one can write $(\mu_1,\cdots,\mu_r)^T =y_0+\lambda_1 f_1+\cdots+\lambda_s f_s$ where $y_0\in M_{r\times 1}(\mathbb{Q})$ is a solution to $Cy=d$, $\lambda_i\in\mathbb{R}$ and
$f_1,\cdots, f_s \in M_{r\times 1}(\mathbb{Q}) $ is a basis for the $\mathbb{R}$-solution space of $Cx=0$. Then modify $\lambda_i$ as above.
