# Question about construction of The Grothendieck group.

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?

• 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 $x\in M$? – Batominovski Dec 9 '18 at 13:17
• @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. – Andreas Blass Dec 9 '18 at 13:22
• 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. – user10354138 Dec 9 '18 at 13:23
• 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 $B$. – 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 $$K(M)\cong(\mathbb{Q}_{>0},\cdot)$$.