The most popular examples are non-local rings and minimal has 16 elements. I am interested in knowing examples of local rings not isomorphic to their opposite.
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asked Nov 14, 2020 at 9:12
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I learned this example from MO-user Johannes Hahn: The algebra is $A=K<x,y>/(x^3,y*x,y^2,x^2*y)$ over a field $K$ with 2 elements.
Then $A$ as an $A$-module as 20 submodules, but $A^{op}$ as an $A^{op}$-module has 16 submodules. Thus $A$ and $A^{op}$ are not isomorphic. This also gives an example where $A$ and $A^{op}$ are not derived equivalent (since local algebras are derived equivalent iff they are isomorphic). Another argument (that works for any field $K$) is that $\Omega_A^{1}(I)$ has dimension 5 but $\Omega_{A^{op}}^{1}(I)$ has dimension 10 when $I$ is the indecomposable injective module.
One might wonder whether a finite local algebra over a finite field is isomorphic to its opposite algebra if and only if the number of submodules of the regular module coincide.
It might also be interesting to see a selfinjective local algebra not isomorphic to its opposite algebra.
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$\begingroup$ Is this example published in a paper? $\endgroup$ Nov 15, 2020 at 11:03
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1$\begingroup$ It might also be interesting to see a selfinjective local algebra not isomorphic to its opposite algebra. That does seem tricky, given that their left and right ideal lattices are certainly isomorphic in the Artinian case. $\endgroup$– rschwiebFeb 22, 2022 at 15:29
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$\begingroup$ @rschwieb I still do not know an example (or I dont remember), so I asked this here: mathoverflow.net/questions/416761/… $\endgroup$– MareFeb 22, 2022 at 15:57