Given two finite dimensional algebra $A$ and $B$ such that $A$ is Gorenstein and $B$ is not. Can the trivial extension algebras of $A$ and $B$ be isomorphic? See <http://www.sciencedirect.com/science/article/pii/0022404984900586> 1.3. for the definition.
Gorenstein means here that the injective dimension of the regular module is finite as a left and as a right module.



Question on trivial extension algebras
If $S$ is a finite dimensional algebra, and $M$ a finite dimensional $S$-bimodule, and if we construct the algebras $A=S\ltimes M$ and $B=S\ltimes DM$, then the trivial extension algebras $T(A)=A\ltimes DA$ and $T(B)=B\ltimes DB$ are isomorphic. This is straightforward to check ($T(A)$ and $T(B)$ are naturally isomorphic as vector spaces, since both are $S\oplus DS\oplus M\oplus DM$, so you just need to check that this isomorphism preserves multiplication), and in fact it is proved in


*Wakamatsu, Takayoshi*, [**Note on trivial extensions of Artin algebras**](http://dx.doi.org/10.1080/00927878408822988), Commun. Algebra 12, 33-41 (1984). [ZBL0537.16008](https://zbmath.org/?q=an:0537.16008).


that if $T(A)\cong T(B)$ then there are $S$ and $M$ as above.


Take $S$ to be a non-Gorenstein algebra, and $M=S$. 


Then $A=S\ltimes S\cong S\otimes\_kk[x]/(x^2)$ is not Gorenstein, since a finite $A$-injective resolution of $A$ restricts to $S$ to give a finite $S$-injective resolution of $S\oplus S$, and similarly for a finite $A$-projective resolution of $DA$.


But $B=S\ltimes DS$ is symmetric, and therefore Gorenstein.


According to my calculations, the simplest example (taking $S=k[x,y]/(x^2,y^2,xy)$) produces:


$A=k[x,y,z]/(x^2,y^2,z^2,xy)$


and


$B=k[w,x,y,z]/(w^2,x^2,y^2,z^2,wy,wz,xy,xz,wx-yz).$


