Let $X$ be a finite free $\mathbf{Z}$-module of rank $n$ and let $X_+ \subset X$ be a cone i.e. a subset of $X$ such that for any $x,y \in X_+$ and $a,b \in \mathbf{N}$ we have $ax+by \in X_+$.
Is it true that that $X_+$ is finitely generated in the sense that there exist elements $x_1,\dots,x_d \in X_+$ such that any element in $X_+$ is a $\mathbf{N}$-linear combination of $x_1,\dots,x_d$ ?
If this is not true does it help if I assume that there are elements $e_1,\dots,e_n \in X_+$ which form a basis of $X \otimes_{\mathbf{Z}} \mathbf{Q}$. Of course $e_1,\dots,e_n$ need not be generators of $X_+$.
Is this is not true are there some conditions that ensure that the result is true ?
It's not true in general that $X_+$ is finitely generated. A simple counterexample is to choose an irrational number $s > 0$ and take $X_+ = \{(a,b) \in \mathbb N^2 \mid 0 < \frac{b}{a} < s\}$. More generally, if $X_+$ is contained in the first quadrant (say), if $r,R \in [0,\infty]$ are the infimum and supremum of the slopes $\frac{b}{a}$ for $(a,b) \in X_+$, and if $X_+$ is finitely generated, then $r$ is the minimum slope of a generator and $R$ is the maximum slope of a generator.