Intersection of subgroup of $\mathbb{Z}^{d}$ with $\mathbb{N}^{d}$ is a finitely generated monoid?

Question

Let $K \subset \mathbb{Z}^{d}$ be a subgroup. Is it necessarily true that the intersection $\mathbb{N}^{d} \cap K$ is a finitely generated monoid?

Edit: More generally, Jason Starr shows in this answer that the intersection in $\mathbb{Z}^{d}$ of finitely many finitely generated subsemigroups is again finitely generated.

Keywords: affine semigroup, monoid

Answer 1

HINT: Yes, it is finitely generated. Let $K \cap \mathbb{N}^d = S$. Now, $S$ is closed under addition, and also under substraction, whenever defined. That is: if $s_1$, $s_2$ are in $S$ and $s_2 - s_1 \in \mathbb{N}^d$, then $s_2 - s_1 \in S$.

Let $S_1 =S\backslash \{0\}$. There exists a finite subset $B \subset S_1$ so that for every $s \in S_1$ there exists $b \in B$ so that $b \prec s$. This is Dickson lemma (link). It is not very hard now to show that $B$ generates $S_1$ ( use some induction).

Note: Not every sub-monoid of $\mathbb{N}^2$ is finitely generated. Indeed, consider the monoid generated by all $(1,n)$, with $n\ge 0$.