How to show that this cone doesn't have a generating set that forms a basis of $\mathbb Z^2$? Consider the cone $\sigma=\mathbb R_{\ge0} e_1 +\mathbb R_{\ge0}(e_1+2e_2)$ in $\mathbb R^2$. I am trying to show that $\sigma$ is not a smooth cone. That is I want to show that $\sigma$ doesn't have a generating set that forms a $\mathbb Z$ - basis of $\mathbb Z^2$.
Clearly $e_1,e_1+2e_2$ doesn't form a $\mathbb Z$ - basis of $\mathbb Z^2$. But how do I show that it has no such generating set? I understand intuitively that checking only on generators on the boundary of the cone is enough but how do I say that more concretely?

Here's a concrete way to explain it: show that there is no linear combination (over $\Bbb Z$) of $e_1$ and $e_1 + 2e_2$ that produces the vector $e_2 \in \Bbb Z^2$.
