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?

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.