What is an example of a ring $R$ and a subring $S \subseteq R$ such that $R$ is flat as a right module but not flat as a left module.

The following question is my motivation:

Faithful flatness for rings

But please note that I am looking for a more specific counterexample.


Take the associative algebra over a field $k$, with generators $x$ and $y$ subject to the relation $xy=0$. This admits a basis consisting of monomials of the form $y^a x^b$. It thus contains a subring $k[x]$, and is flat (even free) over this subring as a right module, on the basis $y^a$, $a\geq 0$. As a left module, it isn't flat, since it isn't even torsion free (for example, $y$ is annihilated by $(x)$).

  • $\begingroup$ Nice! Thanks a lot. $\endgroup$ Oct 13 at 14:50
  • $\begingroup$ BTW is it possible to produce examples that are torsion-free? $\endgroup$ Oct 13 at 15:03
  • 1
    $\begingroup$ Sure, it just gets annoying since your subring can't be a PID. I think something like the free associative algebra generated by $x, y, z$ subject to $xy=yx$ and $xz=y$ works (with subring $k[x, y]$) (but this is much larger than my original example and I haven't carefully checked it) $\endgroup$ Oct 13 at 17:52
  • $\begingroup$ Why can't it be a PID? $\endgroup$ Oct 14 at 9:44
  • 2
    $\begingroup$ Torsion-free modules over PIDs are flat $\endgroup$ Oct 14 at 11:19

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