In the Algebra by Serge Lang, he constructed a Grothendieck group of commutative monoid $M$, namely $K(M)$:(page 39-40)

$M$ is a commutative monoid. Let $F_{ab} (M)$ be the free abelian group generated by $M$, and denote the generator of $F_{ab} (M)$ corresponding to an element $x \in M$ by $[x]$. Let $B$ be the subgroup generated by all elements of type $$[x+y]-[x]-[y]$$ where $x, y \in M$. Let $K(M) = F_{ab}(M)/B$, and $\gamma : M \rightarrow K(M)$, which obtain by composing the injection of $M$ into $F_{ab}(M)$ given by $x \mapsto [x]$, and the canonical map. Then any monoid homomorphism $f : M \rightarrow A $ of $M$ to an abelian group $A$, we have a unique group homomorphism $f_* : K(M) \rightarrow A$ satisfies $$f= f_*\circ \gamma\,.$$ Then $K(M)$ is the Grothendieck group.

My question is why he constructed such a $B = \langle[x+y]-[x]-[y]\rangle$. And how does the $F_{ab}(M)/B$ look like?

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    $\begingroup$ What is $M$? How does the binary operation $+$ work on $M$? Is $M$ a commutative monoid? If $M$ is an abelian group, then $K(M)$ is isomorphic to $M$. And I assume $\gamma:M\to K(M)$, right? How is $\gamma$ defined? Is $\gamma(x):=[x]$ for all $x\in M$? $\endgroup$ – Batominovski Dec 9 '18 at 13:17
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    $\begingroup$ @Batominovski I don't have the book handy, but in view of the rest of the question I'm pretty sure $M$ is intended to be a commutative semigroup. $\endgroup$ – Andreas Blass Dec 9 '18 at 13:22
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    $\begingroup$ The motivation for having $[x+y]-([x]+[y])$ in the kernel of $F_{ab}(M)\to K(M)$ comes from $K(M)$ should be a group completion of $M$, so it should preserve the $+$ that we already have on $M$. The elements of $F_{ab}(M)/B$ is therefore just an equivalence class of formal differences. $\endgroup$ – user10354138 Dec 9 '18 at 13:23
  • $\begingroup$ To see the reason for constructing $B$ (and taking the quotient of $F_{ab}(M)$ by it) try to prove that $\gamma$ (defined in the obvious way, sending $x$ to the coset of $[x]$ with respect to $B$) is a homomorphism of semigroups. You'll find that the proof depends on having (at least) this particular $B$. $\endgroup$ – Andreas Blass Dec 9 '18 at 13:24

Will this construction of $K(M)$ work better for you? I assume that $M$ is a commutative semigroup; that is, $M$ does not necessarily have the zero element. In a sense, $K(M)$ is a generalization of how we construct $\mathbb{Z}$ from the natural numbers, or $\mathbb{Q}_{\neq 0}$ from the nonzero integers.

Equip $M\times M$ with the following equivalence relation $\sim$ defined as follows: for $x,y,z,w\in M$, $$(x,y)\sim (z,w)\text{ if and only if there exists }m\in M\text{ such that }x+w+m=z+y+m\,.$$ Let $G:=(M\times M)/\sim$ be the set of equivalence classes of $M\times M$ with respect to $\sim$. Note that $A$ is an abelian group with addition $+$ defined by $$\big[(x,y)\big]_G+\big[(z,w)\big]_G:=\big[(x+z,y+w)\big]_G\text{ for all }x,y,z,w\in M\,,$$ where $\big[(x,y)\big]_G$ denote the equivalence class in $G$ that contain $(x,y)\in M\times M$. The zero element of $G$ is just the class $$0_G:=\big[(x,x)\big]_G\,(\text{ for any }x\in M)\,.$$ For each $x,y\in M$, the inverse of $\big[(x,y)\big]_G$ in $G$ is simply $\big[(y,x)\big]_G$. Then, $K(M)$ is isomorphic to $G$, and the map $\gamma:M\to K(M)$ is precisely the map sending $x\in M$ to $$\gamma(x):=\big[(x+y,y)\big]_G\,(\text{ for any }y\in M)\,.$$ This map is injective if and only if $M$ has the cancellative property (i.e., for $x,y,z\in M$, $x+y=x+z$ implies $y=z$). The map is bijective if and only if $M$ is an abelian group.

To show that $G$ indeed satisfies the universal property, let $f:M\to A$ be a semigroup homomorphism from $M$ to an abelian group $A$. Define $f_*:G\to A$ via $$\big[(x,y)\big]_G\mapsto f(x)-f(y)$$ for all $x,y\in M$. Show that $f_*$ is a group homomorphism, and that $f_*\circ \gamma=f$. If $g:G\to A$ is another group homomorphism such that $g\circ \gamma=f$, then $$g\Big(\big[(x+y,y)\big]_G\Big)=(g\circ\gamma)(x)=f(x)\,,$$ for any $x,y\in M$. That is, for all $x,y,z\in M$, we have $$\begin{align}g\Big(\big[(x,y)\big]_G\Big)+f(y)&=g\Big(\big[(x,y)\big]_G\Big)+g\Big(\big[(y+z,z)\big]_G\Big)\\&=g\Big(\big[(x+y+z,y+z)\big]_G\big)=f(x)\,.\end{align}$$ This proves that $$g\Big(\big[(x,y)\big]_G\Big)=f(x)-f(y)=f_*\Big(\big[(x,y)\big]_G\Big)$$ for $x,y\in M$, implying that $g=f_*$.

To make your visualization of $K(M)$ a bit stronger, here are examples.

  1. Consider $M:=(\mathbb{Z},\max)$ (that is $x+y$ is defined to be $\max\{x,y\}$ here). Then, show that $K(M)$ is the trivial abelian group $0$.

  2. Consider $M:=(\mathbb{Z}_{>0},\gcd)$ (that is, $x+y$ is defined to be $\gcd(x,y)$ here). Then, show that $K(M)$ is the trivial abelian group $0$.

  3. Consider $M:=(\mathbb{Z}_{\geq 0},+)$. Then, show that $K(M)\cong(\mathbb{Z},+)$.

  4. Consider $M:=(\mathbb{C}_{\neq 0},*)$, where $x*y$ is defined to be $|xy|$. Prove that $K(M)\cong (\mathbb{R}_{>0},\cdot)$.

  5. Consider $M:=(\mathbb{Z}_{>0},\cdot)$. Show that $K(M)\cong(\mathbb{Q}_{>0},\cdot)$.


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