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 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.

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.

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).

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.

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 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.