Given a ring $ S$ with an $ 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.

Call $ T_S(M)$ unique in case $ T_S(M) \cong T_{S’}(M’)$ implies $ S \cong S’$ and $ M \cong M’$ .

Question 1: Is there a nice criterion when a trivial extension $ T_S(M)$ is unique?

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 2: 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)$ .