Say I have a commutative monoid $M$ that is generated by three elements $A,B,C$, where I have that $A+C=2B$. I want to write this a free (does that even mean anything?) monoid $\mathbb N^3$ with generators $X$, $Y$, $Z$ quotiented out by some equivalence relation $\sim$ that yields $[X]+[Z]=2[Y]$. Is there a way to do this?
If I was working with groups, I could just take the free abelian group generated by $X$, $Y$, and $Z$, and then quotient out by the subgroup generated by $ X+Z-2Y$, but I don't have an inverse operation so I can't take that approach. Furthermore, I don't think there even exists a submonoid of $\mathbb N^3$ that when I quotient out $\mathbb N^3$ by I would get such a relation.
Indeed, suppose such a submonoid $P$ existed, then there is a projection map $\pi:\mathbb N^3\rightarrow \mathbb N^3/P$, and we would have that $\pi^{-1}(0)=P$. However, I think this map is somehow injective, because as far as I can see there is no linear combination of $[X]$, $[Y]$ and $[Z]$ over $\mathbb N$ that equals zero. If this is true, then no such such submonoid could exist as $P=\{0\}$, and clearly that quotient would just be $\mathbb N^3$.
I then suspect that there must be a way to write a congruence relation on $\mathbb N^3$ which yields this space, but I can't think of one that would work. So perhaps instead we take $\mathbb N^4$, with generators $X,Y,Z,Y^{-1}$, and then the submonoid $P$ generated by $\{X+Z+2Y^{-1}, Y+Y^{-1},Y^{-1}\}$. Then we would have that $X+Z\sim 2Y$, as:
\begin{align*}
(X+Z)+2Y^{-1}+2Y=2Y+(X+Z+2Y^{-1})
\end{align*}
but this then makes $Y\sim 0$, as $Y+(Y^{-1})=0+(Y+Y^{-1})$, so this doesn't work either.
Overall, I am just very confused. I know very little about monoids, and my motivation comes from trying to verify an isomorphism that my professor wrote down in our Toric Varieties course. Basically, we have the monoid $S\subset \mathbb Z^3$ generated by $e_1,e_2,-e_1+e_2$, where $e_1,e_2,e_3$ are the standard basis for the lattice, and she claimed that:
\begin{align}
\mathbb C[S]\cong \mathbb C[x,y,z]/\langle xz-y^2\rangle
\end{align}
I could get a well defined surjective map from the quotient polynomial to the $\mathbb C$-algebra induced by $S$, but showing that this map was either $a)$ injective, or $b)$ had a well defined inverse seems to hinge on saying something about $S$ as a quotient of a free monoid modulo some relations. Making this precise has been quite difficult.
Best Answer
For your original question: We quotient by the equivalence relation defined in this way. Say an equivalence relation is "good" if $xRy$ implies $(x+z)R(y+z)$ and $(a + c)R(b+b)$. Consider all the good equivalence relations and take their intersection $\approx$. Since intersections of equivalence relations are still equivalence relations, we get to quotient by $\approx$. You can prove as an exercise that addition is well-defined on the quotient, and it gives you an equivalence relation.
Generally, given an algebraic structure $A$ (say monoid), you quotient not by some substructure, but a congruence, defined as a substructure of $A\times A$ satisfying reflexivity, symmetry and transitivity. Note that this is more than an equivalence, because it is not just a subset but a substructure. So it also needs to be closed under algebraic operations.
In the case of groups, given a subgroup $R \subseteq G \times G$, $(x,y) \in R$ if and only if $(x y^{-1}, y y^{-1}) \in R$, since $(y,y)\in R$ by reflexivity, and $R$ is closed under multiplication and inverse. So we just need to know the subset $\{t \mid (t, 1) \in R\}$ to completely determine $R$. Exercise. Prove that $R$ is a congruence iff this subset is a normal subgroup od $G$. This justifies our quotienting by normal subgroups.