The idea is to start with a commutative monoid and turn it into a group. That is, you start with something that fails to be a group because it lacks inverses, and fix it, using the fact it is commutative.
So first I need a monoid, M.
will do.
0+0=0
0+1=1
1+0=1
1+1=1
The operation is logical-or so it is associative and has 0 as its unit. Thats the monoid taken care off. Obviously it is commutative.
Now for F_{ab}(M), the free abelian group generated by M. I'll write elements of F_{ab}(M) like this 0^{n}1^{m} where n and m are integers. Example:
0^{5}1^{6}+0^{-2}1^{3}
=0^{3}1^{9}
In my text book the generator of F_{ab}(M) corresponding to an element x ∈ M is written [x]. So [0] = 0^{1}1^{0} and [1] = 0^{0}1^{1}
Next I need B the subgroup of F_{ab}(M) generated by all elements of type [x+y]-[x]-[y]. Now you can see why I wanted to start small. I only have four generators to consider.
[0+0]-[0]-[0] = [0]-[0]-[0] = 0^{-1}1^{0}
[0+1]-[0]-[1] = [1]-[0]-[1] = 0^{-1}1^{0}
[1+0]-[1]-[0] = [1]-[1]-[0] =
0^{-1}1^{0}
[1+1]-[1]-[1] = [1]-[1]-[1] =
0^{0}1^{-1}
Since B contains 0^{-1}1^{0} it must contain its inverse 0^{1}1^{0}. Similarly 0^{0}1^{-1} ∈ B so we must have 0^{0}1^{1} ∈ B
B has both of the generators of F_{ab}(M) so it must be the whole of F_{ab}(M).
The final step is to construct the Grothendieck group as the quotient of F_{ab}(M) by B. Since F_{ab}(M) = B this is easy, it is isomorphic to the group with one element. That is a little disappointing, my monoid shrank!
On the other hand it is quite illuminating. My text book goes on to talk about cancellation laws. My monoid doesn't let me cancel 1's. I was half expecting the Grothendieck construction to fix the fact that it isn't a group by adding inverses, but it does the obvious thing and gets rid of things that don't cancel.
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