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If M is a free module over a commutative ring, then any 2 bases for M will have the same cardinality. I'm looking for an example of a module that doesn't have this property, as well as examples of the bases in question.

Zachary F
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  • A free module and a module with a basis are the same thing. So, if a module doesn't have the property you have written, then it simply doesn't have a basis at all. – wgrenard Nov 10 '17 at 00:46
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    @wgrenard no, I'm not looking for a free module without a basis, I'm looking for a free module with, for example, a basis A and a basis B where $\abs{A} ≠ \abs{B} $ – Zachary F Nov 10 '17 at 00:49
  • Oh I see what you mean now. Good question! – wgrenard Nov 10 '17 at 00:53
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    https://math.stackexchange.com/questions/335299/given-n-in-mathbb-n-is-there-a-free-module-with-a-basis-of-size-m-foral – Zach Teitler Nov 10 '17 at 01:19
  • Update: I've found one on my own, but I don't know how to type matrices in Latex, so I will post it later when I've learned – Zachary F Nov 10 '17 at 03:48
  • The duplicate I linked mentions the only two families of examples I know of: using the endomorphism ring of infinte dimensional vector spaces, and using Leavitt path algebras. – rschwieb Nov 10 '17 at 15:04
  • @rschwieb respectfully, I don't think this is a duplicate, as this one seems more concrete. Also that one is extremely difficult to understand (or even find) for someone like me who is just starting to learn about algebra. I've found a very simple example that is easy to understand that I was planning on posting here. Is there some way you could reopen this question? – Zachary F Nov 10 '17 at 15:24
  • @ZacharyF The Leavitt-path solution is a bit complex... but you should follow the second link in the post to read the simpler example of the linear transformations. I don't see how this is any more concrete than either of those posts. If you are having a specific difficulty with one of the answers you can ask at the post, or else formulate a new post about what is blocking you. I'm happy to help you here or there. – rschwieb Nov 10 '17 at 15:42
  • @ZacharyF Could you comment on your "extremely simple example"? I'm unaware of a simpler example than the ring of linear transformations of an infinite dimensional vector space. – rschwieb Nov 10 '17 at 15:44
  • @rschwieb sure. The example I found is the ring of 2x2 matrices taken as a module over itself. It has a basis of 1 element, namely the identity matrix. It also has a basis of 2 elements which is a little harder for me to describe without mathjax, namely the matrices with zeroes everywhere except for a 1 in the top corner (one has it in the left corner and the other has it in the right corner) – Zachary F Nov 10 '17 at 15:58
  • Let us continue this discussion in chat. – rschwieb Nov 10 '17 at 16:08

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