Question on trivial extension algebras 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.
 A: 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, Commun. Algebra 12, 33-41 (1984). ZBL0537.16008.
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
$B=k[w,x,y,z]/(w^2,x^2,y^2,z^2,wy,wz,xy,xz,wx-yz).$
