As the question says, I'm looking for a ring with all free left modules but some non-free right modules.

I had thought about looking for a ring not isomorphic to its opposite and try and use that a right $R$-module is a left $R^{op}$-module. But examples of such rings seem to be pretty complicated to construct. So I was wondering if there is an easier example.

Thank you


This is not possible. A ring whose left modules are free is a division ring, and so all of its right modules are also all free.

This is not hard to see: let $S$ be a simple left $R$ module. Since it's free, it is a direct sum of copies of $R$. Since it's simple, it cannot be a sum of more than one copy. This means that $R$ itself is a simple left $R$ module, and hence is a division ring. The right modules over a division ring are, for all intents and purposes, vector spaces.

You can find further information here:

Every $R$-module is free $\implies$ $R$ is a division ring

While the title visible here does not specify sides of the module, it is specified in the question there.

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  • $\begingroup$ Thank you very much, I was not aware of this result. $\endgroup$ – Chris Birkbeck Mar 31 '14 at 16:32

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