Exercise IV.2.12 of Hungerford's Algebra asks to show the following:

If $F$ is a free module over a ring with identity such that $F$ has a basis of finite cardinality $n > 1$ and another basis of cardinality $n + 1$, then $F$ has a basis of cardinality $m$ for every $m > n\ (m \in \mathbb N)$.

This is easy to prove using induction and the fact that if $M$ is an $R$-module with a basis of size $n$, then $M\simeq\bigoplus_{k=1}^nR$ as $R$-modules.

My question is,

for each $n\geq 1$ is there a ring $R$ with identity and a free $R$-module $M$ such that $M$ has a basis of size $m$ for all $m\geq n$ and $M$ does not have a basis of size $k$ for all $k<n$?

In another exercise of the same section the case $n=1$ is established as follows:

Let $K$ be a ring with identity and $F$ a free $K$-module with an infinite denumerable basis $\{ e_1,e_2,\ldots\}$. Put $R =$ Hom$_K(F,F)$. Then the author shows that $R$ has a basis of size $2$ as an $R$-module; namely $\{f_1,f_2\}$, where $f_1(e_{2n})=e_n, f(e_{2n-1})=0,f_2(e_{2n})=0,f(e_{2n-1})=e_n$, and of course $R$ has a basis of size $1$; $\{1_R\}$.

But I don't know what to do when $n\geq 2$.

I thank beforehand any help.

  • $\begingroup$ Does $R \times R$ solve the problem for $n=2$? $\endgroup$ – N. S. Mar 20 '13 at 0:11
  • $\begingroup$ The case $n=2$ would be a ring $R$ such that $R\oplus R\simeq \bigoplus_{k=1}^nR$ for all $n\geq 2$ but $R\ncong R\oplus R$ $\endgroup$ – Camilo Arosemena-Serrato Mar 20 '13 at 0:14

The example you gave is the canonical example of a ring without Invariant Basis Number, and aside from that, I only know about one other family of rings produced with Leavitt path algebras.

I think I'm remembering right that their key feature that set them apart from the example you gave was that they could produce this $R^n\cong R^m$ behavior for prescribed $m,n\in \Bbb N$.

Check them out in Abrams and Anh's paper!

Added: Hmph, I guess I forgot that link wasn't accessible to everyone! Anyhow, I found a slideshow Gene made that will also do the trick.


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.