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A ring $R$ has Invariant Basis Number if, for every $n,m \ge 0$, $R^n \cong R^m$ (as right R-modules) implies $n=m$.

I want to show that that R has Invariant Basis Number if and only if $R^n \cong R^m$ (as left R-modules) implies $n=m$.

Is it true that $R^n \cong R^m$ (as right R-modules) if and only if $R^n \cong R^m$ (as left R-modules)?

effezeta
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1 Answers1

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Yes, having invariant basis number for left modules implies the same for right modules and vice versa.

To show this, show that having invariant basis number is equivalent to the following condition: Whenever $A, B$ are $m \times n$ and $n \times m$ matrices over $R$ such that $AB$ and $BA$ are identities then $m = n$.

You should be able to prove that equivalence for both the left and right versions of "invariant basis number", which means that those left and right versions are equivalent.

Jim
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  • I suppose that $R^n \cong R^m$ if and only if there exist $A$ and $B$, $n \times m$ and $m \times n$ matrices over R such that $AB=1_n$ and $BA=1_m$. But how can I show this? – effezeta Nov 12 '17 at 01:54
  • $A$ is the isomorphism between $R^n$ and $R^m$ and $B$ is its inverse. – Jim Nov 12 '17 at 04:22
  • Suppose that $R$ has IBN. If $AB=1_n$ and $BA=1_m$ then $A$ is invertibile, then there exists an isomorphism between $R^n$ and $R^m$, then $n=m$.

    Vice versa, suppose that $AB=1_n$ and $BA=1_m$ implies $n=m$. If $R^n \cong R^m$ then there exist an invertible matrix $A \in M_{n*m}(R)$, then $A^{-1}A=1_n$ and $AA^{-1}=1_m$, then $n=m$, that is, $R$ has IBN.

    Is it correct?

     – effezeta Nov 14 '17 at 00:44
  • Yes, that's all there is too it. – Jim Nov 14 '17 at 11:42