Grothendieck-Group, formal difference

Question

My question concerns the Grothendieck completion of a abelian Monoid A and the formal difference. I quote from a book: Instead of writing elements $G(A)$ (thats the Grothendieck-Group), it is traditonal to write them as formal differences $a_1-a_2$. In general, it is necesseray to "add on c" when defining $G(A)$. In other words $a_1-a_2=b_1-b_2$ in $G(A)$ does not necessarily imply that $a_1+b_2=a_2+b_1$ in $A$.

So i don't really understand why that last part. Can anyone please give a short argument why this statement holds?

Thanks in advance!

Answer 1

Let $M$ be a commutative monoid. The Grothendieck group of $M$ is constructed as the quotient of the monoid $M \times M$ under the congruence $\sim$ on $M \times M$ defined by $$ (m_1, m_2) \sim (n_1, n_2) \text{ if and only if, for some $c \in M$, $m_1 + n_2 + c = m_2 + n_1 + c$} $$ This element $c$ is really necessary, as shown by the following example. Let $M = \{0,1\}$ with $0+0 = 0$ and $0+1 = 1+0 = 1+1 = 1$. In this case, the relation $\sim$ on $M\times M$ is $(0,0) \sim (0,1) \sim (1,0) \sim (1,1)$ (take $c = 1$ to see this) and thus $G(M)$ is the trivial group. However, if you consider the relation $\equiv$ defined by $$ (m_1, m_2) \equiv (n_1, n_2) \text{ if and only if $m_1 + n_2 = m_2 + n_1$} $$ you will get $(0,1) \equiv (1,1)$, $(1,0) \equiv (1,1)$, but $(0,1) \not\equiv (1, 0)$ and thus $\equiv$ is not even transitive.

Answer 2

It is necessary to add this behaviour in case your monoid is not cancellative, meaning that there is no cancellation law.

Imagine you have $a,b,c\in A$ such that $a+ c = b+c$ but $a\neq b$ (think of the multiplicative monoid of a commutative ring that is not a domain). Then $a-b = (a+c) - (b+c) = 0 = 0 - 0$ in $G(A)$ but $a+0 \neq b+0$ in $A$.

This kind of condition was already well-known before the definition of a Grothendieck group in the case of a localization of a non-integral commutative ring.