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)
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.