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Let $R \subseteq S$ be an extension of rings with unit. Suppose that $S$ is free as left $R$-module. I wonder what can said about the freeness of $S$ as right $R$-module. To be a little more precise let's consider the following questions:

  1. Does someone know an example such that $S$ is free as left $R$-module, but isn't free as right $R$-module.

  2. What are conditions that imply that $S$ is free as left module iff it's free as right module ?

  3. What are categorial examples such that $S$ is free as left as well as right $R$-module ?

An examples for 2. is

  • $R$ is a subring of the center of $S$

and examples for 3. are:

  • $R \subseteq S$ is a Frobenius extension

  • $S$ is a Hopf algebra over a field and $R$ a sub-Hopf algebra (freeness holds by the Nichols-Zoeller theorem)

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up vote 6 down vote accepted

Let $M$ be an $(R, R)$-bimodule and let $S$ be the square-zero extension $R \oplus M$ of $R$ with multiplication $$(r_1 \oplus m_1)(r_2 \oplus m_2) = r_1 r_2 \oplus (m_1 r_2 + r_1 m_2).$$

If $M$ is free as a left $R$-module then so is $S$, and if $S$ is free as a right $R$-module then $M$ is projective. So it suffices to find $M$ which is free as a left $R$-module but not projective as a right $R$-module.

So let $R = T \times T$ where $T$ is some nonzero ring and let $M = R$ with $R$ acting on the left by left multiplication but acting on the right as follows: the second factor of $T$ acts diagonally by right multiplication and the first factor of $T$ acts trivially. Then $M$ is free as a left $R$-module but torsion as a right $R$-module. Taking $T = \mathbb{F}_2$ gives a minimal counterexample in a fairly strong sense; we have $|R| = 4$ and if $|R| < 4$ then $R$ is a field.

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