7

Suppose $M$ is a finitely generated left module over a ring $R.$

We define the rank of $M$ as the minimal number of generators of $M.$

If in addition $M$ is free, then we define the free-rank of $M$ as minimal cardinality of a basis of $M.$

It is clear that $\text{rank}(M)\leq\text{free-rank}(M)$.

I want to know an example when $\text{rank}(M)<\text{free-rank}(M)$.

If $R$ is commutative, then they are equal; so $R$ must be non-commutative.

Sean Lawton
  • 8,384
Andres Abella
  • 73
  • 1
    You want $R$ to have the IBN-property in order for the free-rank to be well-defined. On the other hand in such a situation you would have a surjection $R^m \twoheadrightarrow R^n$ for some $m<n$ and therefore $R^m = P \oplus R^n$ since $R^n$ is projective. It's not quite there yet since $P$ is only projective, not necessarily free, but this smells a lot like a violation of IBN – Johannes Hahn Oct 21 '15 at 17:49

1 Answers1

8

There are abelian groups $A$ such that $A\cong A\oplus A \oplus A$ but $A\not\cong A\oplus A$.

Let $E=\operatorname{End}(A)$. The functor $F=\operatorname{Hom}(A,-)$ is an equivalence from the category of finite direct sums of copies of $A$ to the category of finitely generated free $E$-modules.

So $E\oplus E$ is a module with free rank $2$ but is a direct summand of $E\cong E\oplus E\oplus E$, and so has rank $1$.

For an example where the ring has the IBN property:

For any $n>2$ there is an abelian group $A_n$ such that $A_n^n\cong A_n$ but $A_n^k\not\cong A_n$ for $1<k<n$. Let $E_n=\operatorname{End}(A_n)$ and $E=\prod_{n>2}E_n$.

Then $E^2$ has free rank $2$ but rank $1$.

Jeremy Rickard
  • 33,905