Given a ring $S$ with a non-zero $S$-bimodule $M$, the trivial extension of $(S,M)$ is defined as the ring $R:=T_M(S)$ with $R= S \oplus M$ with multiplication $(s,m)(s',m')=(s s', sm' +m s')$. We then have that $M$ is an $T_S(M)$-ideal and $S \cong T_M(S)/M$. Thus we can recover $S$ from $T_S(M)$ as soon as we know $M$ as an $T_S(M)$-bimodule.
I am mostly interested in the following special case: Let $A$ be a finite dimensional algebra over a field $K$ and $D(A)=Hom_K(A,K)$. Let $T(A):=T_{D(A)}(A)$, which is a symmetric Frobenius algebra with twice the vector space dimension of $A$.
Question: Given $A$, is there a nice a way that when the trivial extension algebra $R=T(A)$ is given, one gets all two-sided ideals $I_i$ (as right modules and/or bimodules) of $R$ such that $R/I_i \cong A_i$ for an algebra $A_i$ such that $R \cong T(A_i)$? Especially: When is $T(A)$ unique as a trivial extension (meaning $T(A) \cong T(A')$ implies $A \cong A'$)?
For example for any field $K$, the ring $K[x]/(x^2)$ is unique as a trivial extension. More generally, I wonder whether the trival extensions of local finite dimensional algebra are unique as trivial extensions.
Note that all ideals $I_i$ would have the same vector space dimension. Are they related in a nice way, which could mean that in case you have one $I_i$ the other $I_j$ can be obtained from the one $I_i$ by a certain operation?
Is there a way to find all such ideals $I_i$ (as right $T(A)$-modules) as in question 2 with the GAP-package QPA for a given algebra $A$? Note that knowledge as right $T(A)$-modules would be enough since then one can recover the $A_i$ as $A_i \cong End_{T(A)}(T(A)/I_i)$.
