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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?

Shaun
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Ratanjit
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  • Yes, it is trivial. This is stated on the Wikipedia article, for example. – anon Apr 12 '23 at 21:12
  • Can you please give me the reason behind it? @runway44 – Ratanjit Apr 12 '23 at 21:14
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    The $0$ element, like any other element, represents an invertible element in the Grothendieck group. If you cancel $0$ from both sides of $0\cdot m=0$, you get $m=1$, for all $m$. – anon Apr 12 '23 at 21:14
  • In general these things happen more often are sometimes called Eilenberg swindle. A classic example is also if you include $\infty$ in $\mathbb{N}$ and do the group construction you will get that it collapses. More generally, similar to localizing rings, your map into the completed group is injective on every element that suffices cancelation $a+c=b+c \implies a=b$. – Felix Apr 13 '23 at 06:42

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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.

freakish
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