# Defining the Rank of a Projective Module

I am trying to understand the definition of rank for a projective module over a noncommutative ring. The definition I am using is:

A sufficient condition for the rank of a free module over a ring $R$ to be uniquely defined is the existence of a homomorphism $\phi:R \to k$ into a skew-field $k$. In this case the concept of the rank of a module can be extended to projective modules as follows. The homomorphism $\phi$ induces a homomorphism of the groups of projective classes $\phi:K_0 R \to K_0 k \approx k$, and the rank of a projective module $P$ is by definition the image of a representative of $P$ in ${\bf Z}$.

What I can't see is how this homomorphism is defined, and why $K_0 k \approx k$. Can anyone spell this out please?

• Is $\phi$ supposed to be injective ? Otherwise you can always take $\phi: R \to R/m$ for any maximal ideal $m$ of $R$. – Ralph May 2 '13 at 20:21
• The usual definition of $K_0$ implies $K_0(K)\simeq \mathbb{Z}$ when $K$ is a field. – Ehsaan May 3 at 23:46

If $$R$$ is a ring and $$R^m\simeq R^n$$ as $$R$$-modules, it does not follow that $$m=n$$. An easy example is when $$R$$ is the endomorphism ring of an infinite-dimensional vector space $$V=\mathbb{C}\oplus \mathbb{C}\oplus\mathbb{C}\oplus\cdots$$: indeed, since $$V\simeq V\oplus V$$ as $$\mathbb{C}$$-vector spaces, you can show $$R\simeq R^2$$ as right $$R$$-modules.
A ring $$R$$ has the invariant basis number property (IBN) if $$R^m\simeq R^n$$ as $$R$$-modules implies $$m=n$$. This is a very fluid property: if $$R\rightarrow K$$ is a unital ring homomorphism and $$K$$ has IBN, then $$R$$ has IBN. In particular, since all fields have IBN (linear algebra), and any commutative ring has $$R\rightarrow R/m$$ for a maximal ideal $$m$$, it follows that all commutative rings have IBN. It's one of my personal favorite exercises to show that all noetherian rings have IBN.
Since the map $$n\mapsto R^n$$ induces a map $$\mathbb{Z}\rightarrow K_0(R)$$, you can view the failure of IBN as the non-injectivity of this map. Since every module over a field $$K$$ is free, this map is both injective and surjective, and you get $$K_0(K)\simeq \mathbb{Z}$$.
Your definition of rank seems highly dependent on the choice of map $$R\rightarrow K$$, so let me give you the one I'm familiar with.
You can define rank for commutative connected (noetherian?) rings $$R$$ as follows. If $$P$$ is a finitely-generated projective $$R$$-module, it might not be free. But if $$p$$ is a prime ideal, then $$P_p:=P\otimes R_p$$ is finitely-generated projective over the local ring $$R_p$$, hence it is actually free: so $$P_p\simeq R_p^n$$ for some $$n$$. But a priori, $$n=n(p)$$ may depend on $$p$$, so we get a map $$n=n_P : \mathrm{Spec}(R)\rightarrow \mathbb{N}.$$ It can be shown that this map is continuous when $$\mathrm{Spec}(R)$$ has teh Zariski topology, so since $$\mathbb{N}$$ is discrete, it follows that $$n_P$$ is constant on the connected components of $$R$$. So if $$R$$ has no idempotents, then $$\mathrm{Spec}(R)$$ is connected, so the map $$n_P$$ must be constant: this constant is called the rank of $$P$$.