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
 A: 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 
A: 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. 
A: 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.
