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Let the bilinear trace form of a finite-dimensional associative algebra be defined as:

$(u,v) \mapsto Tr(L_u L_v)$

For $L_u$ the linear map given by multiplication on the left by $u$. In the literature, there seem to be good characterizations of algebras where this form is non-degenerate (semi-simple, special Frobenius), but what about the other extreme?

Is there a characterization for an associative algebra $A$ whose bilinear trace form, considered as a linear map $V \to V^*$ is rank 1?

In particular, when $A$ is unital and commutative, there seems to be only one possibility for the trace form, if its rank 1:

$(u,v)\mapsto \frac{1}{dimV} Tr(L_u)Tr(L_v)$

This seems like quite the coincidence. Could anyone shed some light on this?

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Note that $Tr(L_uL_v)=Tr(L_{uv})$. Assume characteristic $0$. The kernel of trace has codimension $1$, except in the case of the algebra $0$. This kernel contains the ideal $I$ consisting of all those $u$ such that for all $v$ we have $Tr(L_{uv})=0$, so if that ideal has codimension $1$ then it's equal to $ker(Tr)$. Then $I$, considered as a (nonunital) algebra in its own right, is such that the trace of multiplication by any element is zero. Conversely, given any nonunital algebra with that property, adjoin a $1$ to make it unital and you've got an example of what you're asking about. In such an example, $Tr(L_{uv})$ is proportional to $Tr(L_u)Tr(L_v)$ even without commutativity, but note that it's not equal; the trace of the identity is the vector space dimension.

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