There is a way to construct Grothendieck group from the given commutative monoid which can be found in many books with a chapter on K-theory( for example 'K-Theory' by Atiyah or 'Topology and Analysis' by Booss and Bleecker). This way begins with the diagonal homomorphism $\Delta : A\to A\times A $ with image $\Delta (A)$, then B consists of all the cosets of $\Delta A$, $B:=\{(a_{1},a_{2})+\Delta (A): a_{i}\in A, i=1,2 \}$ where $(a_{1},a_{2})+\Delta (A):=\{(a_{1}+a,a_{2}+a),a\in A \}$, forms a group which is defined to be the Grothendieck group contstructed from the commutative monoid A. And this construction should be conanically isomorphic to other constructions. Here is my question:
This construction is inspired by the construction of integers out of the commutative monoid of non-negative integers, but there is something wrong with the contruction above. For example the coset of $(1,1)$ should be the set $\{ (1,1),(2,2),(3,3),...\}$ and the coset of $(2,2)$ should be the set $\{(2,2),(3,3),(4,4,)...\}$, they are not the same element in group B, but they should be the same element in the Grothendieck group, because they all represent the number $0$. Where am I wrong? Thanks for your help.
I think when the commutative monoid is not a group, one must modify the definition of what it means for two elements of $A \times A$ to be equivalent. For the commutative monoid $A$, one defines $(a_1,a_2)$ and $(a_1',a_2')$ in $A \times A$ to be equivalent if the sets $(a_1,a_2) + \Delta(A)$ and $(a_1',a_2') + \Delta(A)$ have nonempty intersection.