I am currently studying the Grothendieck group and its construction from a commutative monoid $M$. I am troubled by the following question that came to my mind in recent days. Please help me.
My question is the following:
Let $M$ be a commutative monoid, written multiplicatively, with the identity element $1$. Suppose there exists an element $0 \in M$ such that $m \cdot 0 = 0$, for all $m \in M$, then what will be the Grothendieck group $\mathrm{K}(M)$ of $M$?
For instance, if we take $M = \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers including $0$, then $M$ is a commutative monoid under multiplication with identity element $1$. Then what is $\mathrm{K}(M)$? Is it trivial?
Best Answer
The way we construct Grothendieck group from a commutative monoid $M$ is as follows: $K(M)$ is defined as $(M\times M)/\sim$ where $(a,b)\sim (c,d)$ if $a\cdot d\cdot k=b\cdot c\cdot k$ (written in multiplicative way) for some $k\in M$.
Now you consider $M$ with special element $0$ such that $m\cdot 0=0$ for any $m$. In such situation clearly $(a,b)\sim (c,d)$ for any $a,b,c,d$, because we can simply pick $k=0$. And therefore $K(M)$ is the trivial group.
As an alternative proof, you can look at the universal property of $K(M)$. So $K(M)$ comes with a monoid homomorphism $i:M\to K(M)$ and every monoid homomorphism $M\to A$ to an abelian group $A$ has to factorize through it. So let $f:M\to A$ be such homomorphism. Then
$$f(x)=f(x)\cdot 1=f(x)\cdot f(0)\cdot f(0)^{-1}=f(x\cdot 0)\cdot f(0)^{-1}=f(0)\cdot f(0)^{-1}=1$$
and so $f$ is trivial. In particular it uniquely factorizes through the trivial group, and so by the universal property $K(M)$ has to be trivial.