Let $C=\{\lambda(1,0)+\mu(1,\sqrt2):\lambda,\mu\in\mathbb R_{\geq 0}\}\subset\mathbb R^2$. I want to show that $C\cap\mathbb Z^2$ is not a finitely generated monoid.
Let $(n,m)\in C\cap\mathbb Z^2$. Then $$m=\sqrt 2\mu; n=\lambda+\mu.$$ Therefore, $\mu=\frac{m}{\sqrt2}$ and $\lambda=n-\frac{m}{\sqrt2}$. As $\lambda\geq 0$, we have $\frac{m}{n}\leq\sqrt2$. Therefore we have $$C\cap\mathbb Z^2=\{(n,m)\in\mathbb N^2:0\leq\frac{m}{n}\leq\sqrt2\}\cup\{(0,0)\}.$$ Here $\mathbb N$ denotes the set of non negative integers. But from here I cannot show that $C\cap\mathbb Z^2$ is not finitely generated. This is an exercise in a book. As a hint it is given that if $C\cap\mathbb Z^2$ is finitely generated then $\sqrt2$ must be rational.
Can anyone please help me how to show this?
Thank you.
HINT:
If $v=(x, y)$ is a non-zero vector in the cone $C$ then $x>0$ and $y\ge 0$. Moreover, if $v$, $v_1$, $\ldots$, $v_N$ are nonzero vectors in $C$ such that $v$ is a positive linear combination of $v_i$ then $$\frac{y}{x}\le \max \frac{y_i}{x_i}$$
If the $v_i$ are in $C\cap \mathbb{Q}^2$ then $\max \frac{y_i}{x_i}< \sqrt{2}$, since all of them are in $C$ and the slopes are rationals.
Now: for every $r< \sqrt{2}$, $r \in \mathbb{Q}$ there exists $v \in C \cap \mathbb{Z}^2$ so that $\frac{y}{x}= r$. Therefore, for every $v_1$, $\ldots$, $v_n$ in $C \cap \mathbb{Q}^2$ there exists $v \in C \cap \mathbb{Z}^2$ that is not in the cone spanned by the $v_i$. We conclude that the monoid $C \cap \mathbb{Z}^2$ is not finitely generated.
$\bf{Added:}$ Let $C$ be a finitely generated proper cone (proper meaning it does not contain any lines) in $\mathbb{R}^N$ with non-void interior. Say $C$ is generated by $u_1$, $\ldots$, $u_m$ ( a minimal system of generators). Then this system is uniquely determined ( as a set of elements in the projective space).
Let's denote for a vector $v\ne 0$ in $\mathbb{R}^N$ its image in $P(\mathbb{R}^N) = \mathbb{P}^{N-1}$
Now, consider any $v_1$, $\ldots$, $v_n$ inside $C$ such that the set $\{[v_1], \ldots, [v_n]\}$ does Not contain $\{[u_1], \ldots, [u_m]\}$. The cone $C'$ generated by $v_1, \ldots, v_n$ is closed and strictly contained in $C$. Let $k$ be such that $[u_k]\not \in \{[v_1], \ldots, [v_n]\}$. There exists a vector $v\in C \backslash C'$ close to $u_k$, $v$ with rational coordinates ( here is where we use $C$ has full dimension, that is, $\{u_1, \ldots, u_n\}$ contain a basis of the full space. Multiply $v$ by some positive integer and get a vector with integer coordinates in $C \backslash C'$.
Now, assume that not all of the $[u_1]$, $\ldots$, $[u_m]$ have rational projective coordinates. Then $C\cap \mathbb{Z}^N$ will not be finitely generated. Similarly $C\cap \mathbb{Q}^N$ is not finitely generated.