## Finitely generated monoids are finitely presented?

I saw in the answer of this post that any finitely generated monoids are finitely presented in the sense that there is a coequalizer diagram $P_1\rightrightarrows P_0\rightarrow M$ with $P_1$ and $P_0$ free commutative and finitely generated.

My question is:

Can we make $P_1$ maps to the kernel of $P_0\rightarrow M$? (to be more, precise, can we find a presentation of the form $P_1\rightarrow P_0\rightarrow M$ like the case for abelian groups?)

And if a monoid is finitely presented for one presentation, is it the same for other presentations? Thanks

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You need the monoids to be commutative; otherwise, take any f.g. group that is not f.p., for example $\mathbb{Z}\wr\mathbb{Z}$ or $\text{SL}_3(\mathbb{Z}[t]$). – Tom Church Aug 25 2010 at 5:26

Every finitely generated commutative monoid is indeed finitely presented. Of course the presentation depends on the generating set. But the minimal presentation will be always finite (provided you choose finite generating sets). See http://www.math.vanderbilt.edu/~msapir/ftp/pub/survey/survey.pdf

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Nobody has answered the actual question, though of course the preface to the question should say "any finitely generated commutative monoid is finitely presented in the sense that there is a coequalizer diagram $P_1 \stackrel{\to}{\to} P_0 \to M$ with $P_1$ and $P_0$ free commutative and finitely generated."

I believe the answer to the question is no, we cannot always find a presentation $P_1 \to P_0 \to M$ where $P_1$ maps to the kernel of $P_0 \to M$.

However, I don't have a proof.

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Why is this an answer instead of an up-vote of Tom Church's comment from... 3 years ago? – Vidit Nanda 2 days ago
He is saying that the question was never answered. The question seems to be whether the finite presentation can be made to involve only relations of the form w=1 like for groups. – Benjamin Steinberg 2 days ago

The answer is no if the OP means all relations should be consequences of the relations of the form w=1 with 1 the identity. Let S be the multiplicative monoid of the two-element field. If F is a free commutative monoid with generators A and if F maps onto S via f, then if we set B to be the generators mapping to 1 under f, we have a word represents the identity iff it is a product of letters from B. But if I add all such relations I will get as the quotient a free commutative monoid on A-B not S.

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