Why the Grothendieck group/ring?

[alg] [cat.theory]
Given an additive category \mathcal{C} we can form the Grothendieck group \mathcal{K}(\mathcal{C}). If further \mathcal{C} is a tensor category then \mathcal{K}(\mathcal{C}) is also a ring, the Grothendieck ring.

If \mathcal{C} is semisimple then any equality in \mathcal{K}(\mathcal{C}) corresponds to an isomorphism in \mathcal{C}.

Advertisements

[alg]
Subgroups of finitely-generated abelian groups are finitely-generated.

Not necessarily true for finitely-generated non-abelian groups.

[alg]
If R is a commutative ring then the matrix ring M_n(R) is an associative algebra.

Fails if R is non-commutative: for then we need not have r(AB) = A(rB).