Assume all algebras are finite dimensional quiver algebras over a field (no restriction of generaltiy if the field is algebraically closed).

Let A be a local Frobenius algebra. Is A isomorphic to its opposite algebra?

For non-Frobenius algebras this is false, see Do you know which is the minimal local ring that is not isomorphic to its opposite? (where the current question remained open, see the answer and comment).

Is there an easy example of a (not necessarily local) Frobenius algebra that is not isomorphic to its opposite algebra?

  • $\begingroup$ I think the example here is finite dimensional? It and the post linked to it were the examples I had in mind. $\endgroup$
    – rschwieb
    Feb 22 at 16:10
  • 2
    $\begingroup$ @rschwieb Thanks, I added that I assume that the algebras are quiver algebras. $\endgroup$
    – Mare
    Feb 22 at 16:11
  • $\begingroup$ Depending on how this goes, I might have to post the version for "finite quasi-Frobenius rings" $\endgroup$
    – rschwieb
    Feb 22 at 16:12

1 Answer 1


No. Consider, for example, the quantum complete intersection $A = k\langle X,Y\rangle/(X^2, Y^3, XY-qYX)$, $q\in k\setminus\{0\}$. This is a Frobenius local algebra (see Section 3 of arXiv:0709.3029), and the ideal $(X^2, Y^3, XY-qYX)$ in $k\langle X,Y\rangle$ is admissible.

Any isomorphism $f:A\to A^{\rm op}$ must satisfy $f(X) = aX+r$ and $f(Y) = bY + s$, where $a,b\in k\setminus\{0\}$ and $r,s\in\mathop{\rm rad}^2(A) = (XY, Y^2)$. Now $$f(XY) = f(Y)f(X) = (bY + s)(aX +r) = abq^{-1}XY + t,\quad t\in\mathop{\rm rad}\nolimits^3(A)$$ but also $$f(XY) = f(qYX) = qf(X)f(Y) = q(aX + r)(bY + s) = qabXY + u,\quad u\in\mathop{\rm rad}\nolimits^3(A)$$

Since $XY\notin\mathop{\rm rad}^3(A^{\rm op})$, this means that $abq^{-1}XY = qab XY$ and thus $q=q^{-1}$ (because $a,b\ne0$).

So unless $q = \pm1$, the algebras $A$ and $A^{\rm op}$ are not isomorphic.


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