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)

  • $\begingroup$ What have you tried? $\endgroup$
    – Pedro
    Nov 24, 2018 at 20:36
  • $\begingroup$ @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. $\endgroup$
    – user45765
    Nov 24, 2018 at 20:38
  • $\begingroup$ @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. $\endgroup$
    – user45765
    Nov 24, 2018 at 20:40

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – user45765
    Nov 24, 2018 at 21:07
  • 4
    $\begingroup$ You don't. This doesn't change the fact that $[R]$ generates $K_0(R)$ as an abelian group. $\endgroup$ Nov 24, 2018 at 21:08

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