1

I was reading this webpage http://ysharifi.wordpress.com/2011/10/28/examples-of-projective-modules/, and there was something that confused me.

We have

enter image description here

I thought that the dimensions of isomorphic free modules would be the same. But here it says that "since $\dim_k R = 4$, we have $\dim_k P = 4|I|$, and I found that a bit confusing...

Thanks in advance

1 Answers1

2

So you have that $P$ as an $R$-module would have dimension $|I|$, and the dimension of $R$ over $k$ is $4$, so the dimension of $P$ over $k$ is $4|I|$.

Is this what you were wondering?

Daniel Montealegre
  • 6,783
  • 1
    Make sure that you know the dimension over what field you are referring to. That might be the confusion. – Daniel Montealegre Sep 27 '13 at 01:59
  • 1
    Also, if you want another example of a projective module which is not free, let $R=\mathbb{Z}/6$, and let $P=\mathbb{Z}/2$. Then as $R$-modules you have that $R=P\oplus \mathbb{Z}/3$. Since $P$ is a direct summand of an $R$ module, it is projective, but it is not free since a non-empty free module over $\mathbb{Z}/6$ would have at least $6$ elements, and $\mathbb{Z}/2$ has only two elements. – Daniel Montealegre Sep 27 '13 at 02:02
  • There is a problem here pointing to a gap in the original proof, though. The computation relies on the reasoning that $dim_k(I)=dim_R(I)dim_k(R)$, but it is not immediate that you can reason with "dimensions" here like that. Indeed, there are examples of rings where $R\cong R^2$ as $R$ modules, so what is the "$R$ dimension" of that? To complete the proof, the author would need to somehow invoke the (true) fact that free modules of matrix rings over fields have unique rank. For example, he could just point out that $dim_k(R)<\infty$ precludes non-unique ranks. Maybe he did this earlier. – rschwieb Sep 27 '13 at 12:41