3
$\begingroup$

Let $A$ be an algebra over some field $k$. Let $K_P(A)$ be the Grothendieck group of the category of projective $A$-modules and $K_F(A)$ the category of finite dimensional $A$-modules. I've been told there are examples where $K_P(A)$ and $K_F(A)$ have different rank, but I've never seen an example.

Does anyone have such an example?

Improve this question
$\endgroup$
3
  • $\begingroup$ What does PIMS stand for? $\endgroup$
    – Theo Buehler
    Commented Feb 19, 2011 at 1:25
  • $\begingroup$ projective indecomposable modules $\endgroup$
    – David Hill
    Commented Feb 19, 2011 at 1:45
  • 1
    $\begingroup$ I guess the first use of PIM was as an abbreviation for "principal indecomposable module", before the notion of projective module became widespread in this kind of algebra. Fortunately "principal" and "projective" both start with the same letter, though it's actually a little more natural to write "indecomposable projective module". $\endgroup$
    – Jim Humphreys
    Commented Feb 20, 2011 at 20:16

3 Answers 3

Reset to default
4
$\begingroup$

Let $A=k[x]$. Then $K_P(A)$ has rank 1 (I assume that you consider only finitely generated projective modules) and $K_F(A)$ has infinite rank.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Do you know one where both are finite dimensional? $\endgroup$
    – David Hill
    Commented Feb 19, 2011 at 1:43
3
$\begingroup$

If k is an algebraically closed field, then A = { f(x)/g(x) : g(0)⋅g(1) ≠ 0; f,g in k[x] } ≤ k(x) is a commutative ring with exactly two maximal ideals, (x) and (x−1). It has two simple modules and both happen to be one dimensional as k-vector spaces. It is a PID, so it has only one projective indecomposable module, which is of course free and cyclic as an A-module, but infinite dimensional as a k-vector space.

Hence the rank of KP(A) is 1, and the rank of KF(A) is 2.

This is just a modification of Victor Ostrik's example to cut down on the number of simple modules.

Improve this answer
$\endgroup$
2
$\begingroup$

An example where both groups are of finite rank but different is the Weyl algebra. There are no finite dimensional modules, so one of the groups is trivial, and $K_0(A)=\mathbb Z$.

(I'm considering the $K_0$ if f.g. projectives...)

Improve this answer
$\endgroup$
7
  • $\begingroup$ Is there anything in between? In some sense, this is the same example with the reciprocal of the rank. $\endgroup$
    – David Hill
    Commented Feb 19, 2011 at 2:10
  • $\begingroup$ «this is the same example with the reciprocal of the rank»? I don't know what you mean by that. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 19, 2011 at 2:13
  • $\begingroup$ In Victor's example, there were infinitely many simples but one projective. In your example, there is 1 projective and no simples. The algebras are related by x↔\partial/\partial x so it's like you're taking reciprocals $\endgroup$
    – David Hill
    Commented Feb 19, 2011 at 2:19
  • $\begingroup$ The Weyl algebra is generated by $x$ and $d/dx$. Also, it has many, many simple modules, but all of them infinite-dimensional. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 19, 2011 at 2:21
  • $\begingroup$ Also, there are many non-isomorphic projectives, but they collapse in $K_0$. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 19, 2011 at 2:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .