Subsemigroup of integers is finite type?

Question

I think every subsemigroup of semi-group $(\mathbb N,+)$ is generated by finite many integers (this is not true for $\mathbb N^2$). Remind: subsemigroup $=$ subset stable for $+$.
Can you give short proof of reference?
Motivation is finite generasion for semi-group in toric varieties but algebraic geometry is not necessary for to answer my elementary purly algebra question.

Answer 1

You may already know

Lemma. If $a,b\in\Bbb N$, then $a\Bbb N+b\Bbb N$ contains all sufficiently large multiples of $\gcd(a,b)$.

Proof. Let $d=\gcd(a,b)$. We know $d=xa+yb$ with $x,y\in\Bbb Z$. More precisely, we can find such $x,y$ with $0\le x<b$ and $-a<y\le 0$. Trivially $2ab=ab+ba\in a\Bbb N+b\Bbb N$. If $kd\in a\Bbb N+b\Bbb N$, say $kd=na+mb$, and $kd\ge 2ab$ then we cannot have both $n<b$ and $m<a$. If $m\ge a\ge -y$ then $(k+1)d=(n+x)a+(m+y)b$ shows $(k+1)d\in a\Bbb N+b\Bbb N$. Otherwise, $n\ge b$ and $(k+1)d=(n+x-b)a+(m+y+a)b$ does it. By induction, all multiples of $d$ from $2ab$ on are in $a\Bbb N+b\Bbb N$. (As mentioned in the comments, with a bit more care the bound $2ab$ can be lowered substantially, but we don't need this here) $_\square$

Corollary 1. If $A$ is a subsemigroup of $\Bbb N$ and contains almost all all sufficiently large multiples of $a\in\Bbb N$ and all sufficiently large multiples of $b\in\Bbb N$, then it contains all sufficiently large multiples of $\gcd(a,b)$.

Proof. Assume $ka\in\Bbb N$ for all $k>k_a$ and $kb\in\Bbb N$ for all $k>k_b$. Let $p$ be a prime $>\max\{k_a,b\}$ and $q$ a prime $>\max\{k_b,a,p\}$. Then $pa\Bbb N+qb\Bbb N\subseteq A$ and hence all sufficiently large multiples of $\gcd(pa,qb)=\gcd(a,b)$ are in $A$. $_\square$

By induction this immediately generalizes the lemma to

Corollary 2. If $a_1,\ldots,a_n\in\Bbb N$ then $a_1\Bbb N+\ldots +a_n\Bbb N$ contains all sufficiently large multiples of $\gcd(a_1,\ldots,a_n)$. $_\square$

Now let $A$ be a subsemigroup of $\Bbb N$. Let $d$ be the $\gcd$ of all elements of $A$. Then we can find finitely many elements $a_1,\ldots,a_n$ of $A$ with $\gcd(a_1,\ldots,a_n)=d$. By the above, we know that all sufficiently large multiples of $d$ are in the subsemigroup generatd by $a_1,\ldots,a_n$. Add the finitely many elements of $A$ below that threshold to see that $A$ is finitely generated.

Edit (evgeniamerkulova). For dummies like me who had difficulty, notice that from 2nd line of proof of lemma starting with If $kd\in a\Bbb N+b\Bbb N$...Hagen is proving implication $kd\in a\Bbb N+b\Bbb N\implies (k+1)d\in a\Bbb N+b\Bbb N$