How to show directly that two elements become equal in Grothendieck group?

Question

Consider commutative semigroup S and its Grothendieck completion group G(S).Suppose I insist on defining G(S) as free abelian group on basis $[a]$ (with $a\in S$) divided out by the relations $[a+b]-[a]-[b]$. How do I show with that definition that if the images of $a,b\in S$ become equal in G(S), then necessarily there existed $c\in S$ with a+c=b+c ? I know it true because I can prove it with other construction of grothendieck group, but I should like direct proof with above free abelian group construction.

Answer 1

Well your notation is a little bit confusing. Since we are considering free abelian groups, we should avoid additive notations for the commutative semigroup S. Let's talk of $(S, \ast )$. If $F(S)$ is the free abelian group with basis $S$, we can assume $S \subseteq F(S)$. So the set of relations is $$R = \{ (a \ast b) -a - b : a,b \in S \}.$$ Then, the Grothendieck completion is the abelian group $$G(S) = F(S)/ \langle R \rangle.$$ If $a \in S$, I'll denote the coset $a + \langle R \rangle$ by $[a]$.

The proof isn't hard. You'll just need to remember that in free abelian groups the expression in terms of a basis is unique. You can find one proof, with an ugly notation, in Rotman's 'Advanced Modern Algebra'. I'll reproduce below the proof on Magum's book 'An Algebraic Introduction to K-theory'.

Suppose that $[a]=[b]$. Then, $a-b \in \langle R \rangle$. That is, $$a-b = \sum_{i=1}^n n_i((a_i \ast b_i) - a_i - b_i),$$ where $n_i = 1$ or $-1$ and $a_i, b_i \in S$. Bringing terms with negative coefficients to the other side, $$ a + \sum_{n_i = -1}(a_i \ast b_i) + \sum_{n_i = 1}(a_i + b_i) = b + \sum_{n_i = 1}(a_i \ast b_i) + \sum_{n_i = -1}(a_i + b_i).$$ Since $S$ is a basis of $F(S)$, the terms on one side of the equation are a permutation of those on the other side. Since $(S, \ast )$ is commutative, it follow that, in S, $$ a \ast \prod_{n_i = -1}(a_i \ast b_i) \ast \prod_{n_i = 1}(a_i \ast b_i) = b \ast \prod_{n_i = 1}(a_i \ast b_i) \ast \prod_{n_i = -1}(a_i \ast b_i). $$ So $a \ast c = b \ast c$, where $$ c = \prod_{i = 1}^n (a_i \ast b_i).$$