First let us recall some terminologies. Let $A$ be a finite dimensional algebra over a field $k$. A linear map $t: A\to k$ is called symmetric (resp. non-degenerate) if the bilinear map $Q_t: A\times A\to k$ given by $(x,y)\mapsto t(xy)$ is symmetric (resp. non-degenerate).

The trace form of $A$ is the linear map $T_A: A\to k$ given by $a\mapsto tr(\lambda_a)$, where $\lambda_a: A\to A$ is the left multiplication of $a$ on $A$ and $tr$ is the trace of linear maps. Note that $T_A$ is always symmetric. However, it is not necessary non-degenerate even in the case that $A$ is separable.

Now suppose $A$ is separable. One may choose an element $\sum u_i\otimes v_i \in A\otimes A$ such that $\sum au_i\otimes v_i = u_i\otimes v_ia$ for all $a\in A$ and $\sum u_iv_i=1$. It follows that $\sum u_ixv_i$ lies in the center $C$ of $A$ for all $x\in A$. Note that $T_C:C\to k$ is symmetric and non-degenerate. My question is that is the map $t: A\to k$ given by $x\mapsto T_C(\sum u_ixv_i)$ symmetric and non-degenerate?

Any comments are welcome. Thanks a lot.

  • $\begingroup$ you might want some assumptions on the characteristic. For example, if $char(k)\mid n,$ then the trace map $T_A$ can be the zero map (this will be the case if $A$ is central simple, for example) $\endgroup$ – GreginGre Mar 23 '18 at 11:48
  • $\begingroup$ Yes, it is. But what I ask is about the map $t:A\to k$ defined through an separable idempotent and the trace map $T_C$ of the center $C$. $\endgroup$ – G.-S. Zhou Mar 23 '18 at 13:15
  • $\begingroup$ Non degeneracy of a bilinear form, and separability for algebras are preserved after extending scalars to a separable closure, so you may reduce to the case where $A=M_{n_1}(k)\times\cdots\times M_{n_r}(k)$. Have you worked out this example ? $\endgroup$ – GreginGre Mar 25 '18 at 9:11
  • $\begingroup$ Also, you might want to check first that the answer to your question is independent from the choice of your separability idempotent. $\endgroup$ – GreginGre Mar 25 '18 at 9:12

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