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Technically the question was

prove that if $R^m$ $\cong$ $R^n$, then m=n.

My proof was basically considering the case m>n, then consider an isomorphism $\phi$, then the image of the basis of $R^m$ under $\phi$ must still be linearly independent, but $R^n$ has a rank of n which is less than m, which will make the images of the basis of $R^m$ linearly dependent, contradiction.

My professor's comment of my solution was "long answer but you've shown nothing".

I do not understand what exactly is he looking for or the "proper" solution. Can anyone help explain?

edit: R is commutative

Ecotistician
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  • You should ask your professor what kind of solution was expected from you. It should depend on the course material. Is it the same course as this one? – Dietrich Burde Dec 21 '17 at 19:19
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    The question was closed assuming a context of commutative rings, where it has a definite answer. But we should also say, in case you are considering noncommutative rings, that this is false in general. – rschwieb Dec 21 '17 at 19:32
  • Sorry, R is indeed commutative in the problem statement. I am trying to read the proof in the link given above but I don't understand what is wrong with my argument. – Ecotistician Dec 21 '17 at 19:35
  • which will make the images of the basis of $R^m$ linearly dependent, contradiction. Well, you've sort of buried all the hard work in this vague sentence. How did you reason that it would "make the images of the basis linearly dependent"? As I've mentioned there do exist noncommutative rings with f.g. free modules with different finite basis cardinalities. So you'd have to say something more. – rschwieb Dec 21 '17 at 19:38
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    Actually, rschwieb has given a very good and elementary answer to your question at the duplicate. I suppose this is what you should read in detail. – Dietrich Burde Dec 21 '17 at 19:45
  • I am still trying to understand the proof in the link above. But if n is the rank of a module, and m>n, then wouldn't that implies m elements (the isomorphic image of the basis elements in R^m) must be linearly dependent? – Ecotistician Dec 21 '17 at 20:02

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