# $K_0(R)$ is generated by $[R]$?

Let $$R$$ be a unital associative ring. Let $$F$$ be the free abelian group on the set of all isomorphism classes $$[P]$$ of f.g. projective $$R$$-modules $$P$$. Let $$K_0(R)$$ be the quotient of $$F$$ modulo the subgroup $$S$$ spanned by $$[P]+[Q]-[P\oplus Q]$$ for all projective $$P$$ and $$Q$$.

a) Suppose every f.g. projective of $$R$$ is free.

b) Suppose $$R$$ has the invariant basis property (i.e. the rank of a free module is unique).

If $$R$$ has property $$a),b)$$, then it is obvious that $$K_0(R)$$ is generated by $$[R]$$ as $$\mathbb{Z}$$-algebra. (Here I assumed $$K_0(R)$$ is already endowed with ring structure via tensor product.)

We say that an $$R$$-module $$M$$ is stably free if $$M\oplus F=G$$ for some f.g. free modules $$F,G$$ (thus, $$[M]$$ is the difference of 2 free modules).

Question: The book claims that if $$a)$$ is replaced by every f.g. projective mod being stably free and $$b)$$ holds, then $$K_0(R)$$ is generated by $$[R]$$. How do I deduce this when $$M$$ is stably free? In other words, I need to deduce $$M$$ free.

Ref. Milnor's Algebraic K-theory pg 5.(Question 1 and 2)

• What have you tried? – Pedro Tamaroff Nov 24 '18 at 20:36
• @PedroTamaroff I think I am stuck at the first step as I do not see how to deduce a non-trivial free module morphism to $M$. I would expect $G/F$ to be free module. However, I am very uncertain about this step. Then from here, I would expect localization kicks in to deduce the free module level isomorphism on every point and here I need use invariant basis property. Hence it induces isomorphism which concludes $M$ being free. – user45765 Nov 24 '18 at 20:38
• @PedroTamaroff I think I might be heading towards wrong way of thinking as I am wondering whether every projective being free here. This is a fairly stringent condition on the ring property. Say $R$ being PID fits. – user45765 Nov 24 '18 at 20:40

Assumption (b) is irrelevant, and you can conclude that $$K_0(R)$$ is generated by $$[R]$$ as a $$\mathbb{Z}$$-module, not just as a $$\mathbb{Z}$$-algebra. For any $$[M]\in K_0(R)$$, there are finitely generated free modules $$F$$ and $$G$$ such that $$M\oplus F= G$$ and so $$[M]=[G]-[F]$$. There are $$m,n\in\mathbb{N}$$ such that $$G\cong R^m$$ and $$F\cong R^n$$, and so $$[M]=[G]-[F]=[R^m]-[R^n]=m[R]-n[R]=(m-n)[R]$$ is an integer multiple of $$[R]$$.
Note in particular that proving $$[M]$$ is equal to a multiple of $$[R]$$ in $$K_0(R)$$ does not mean that $$M$$ itself is a free module.
• Maybe this is a dumb question. If $R$ does not have IBN, then say $R^m\cong R^{m'}$ and $R^n\cong R^{n'}$. So $(m-n)[R]=(m'-n')[R]$? How do I know $m-n=m'-n'$? Thanks. – user45765 Nov 24 '18 at 21:07
• You don't. This doesn't change the fact that $[R]$ generates $K_0(R)$ as an abelian group. – Eric Wofsey Nov 24 '18 at 21:08