2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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 $K(M)\cong(\mathbb{Q}_{>0},\cdot)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.