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Do you know if all submonoids of $\mathbb{N}^k$ are finitely generated? If not, can you give me a counter-example?

EDIT : I mean $\mathbb{N}^k$ as a submonoid of $(\mathbb{Z}^k,+)$.

I already know that every submonoid of $\mathbb{N}$ is finitely generated. But I don't even know if the result is true with $k > 1$... I couldn't find a counter-example by my own.

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Submonoids of $\mathbb{N}^k$ with what operation? –  mjqxxxx Dec 2 '13 at 22:46
    
and what did you try already? –  Ittay Weiss Dec 2 '13 at 22:46
    
@mjqxxxx: I would guess addition... –  Najib Idrissi Dec 2 '13 at 22:48
    
I edited my post, answering your questions. –  Plop Dec 2 '13 at 22:58

1 Answer 1

up vote 4 down vote accepted

I assume that you mean $\mathbb{N} = \{0,1,\dots\}$ and the structure is addition, and then the product structure on $\mathbb{N}^k$ (that is "if $M$ and $N$ are monoids then you get the product structure on $M \times N$"; the wording is a bit ambiguous here).

Let $k=2$ so we consider $\mathbb{N}^2$. Let $M$ be the submonoid generated by all the $e_i = (1,i)$ for $i \geq 1$. In other words: $$M = \{(0,0)\} \cup \{(a,b) : b \geq a > 0 \}$$

Then $M$ is not finitely generated, because none of the $e_i$ can be written as a sum of two nonzero elements in $M$, so they would appear in every set of generator.

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