In the sequel, "$k$-algebra" means "associative unital finite dimensional $k$-algebra.
Apologies for the very long exposition.
If $A$ is a $k$-algebra and $s:A\to k$ is $k$-linear, we say that $s$ is $\mathit{nondegenerate}$ if for all $x\in A$, $s(xa)=0$ for all $a\in A\Rightarrow x=0$ (maybe I need to assume right nondegeneracy as well, I don't know)
Let us fix a base field $k$. I would like to understand the existence and "uniqueness" of nondegenerate linear maps functorially associated to algebras in the sense that the assignment $A\mapsto s_A$ needs to satisfy the following conditions:
(1) for all field extensions $K/k$ and for all morphisms $\varphi:A\to B$ of $K$-algebras, we have $s_A=s_B\circ\varphi$;
(2) for any morphism of field extensions of $k$ $f:K\to L$, and for all $K$-algebras $A$, and all $a\in A$, we have $s_{A\otimes_K L}(a\otimes 1)=f(s_A(a))$, where the tensor product is taken with respect to $f$;
(3) for all field extensions $K/k$ and for all for all $K$-algebras $A$, the linear map $s_A$ is nondegenerate.
Of course, I do not expect such an assignment to exist for all $k$-algebras (even if the dimension is fixed), but before precising more the context, notice that there is always an obvious assignement satisfying (1) and (2), namely the trace map: if $A$ is a $K$-algebra, we denote by $t_A$ the linear map $t_A:a\in A\mapsto tr(\ell_a)$, where $\ell_a$ is the endomorphism of left multiplication by $a$.
The problem is that $t_A$ is not always non degenerate. It is even the zero map if $E$ is an inseparable field extension of $k$, for example. The following theorem is known (I found it browsing the web, but forgot to write down the reference. I'll find it later for those who are interested)
Thm If $char(k)=0$, i$t_A$ is nondegenerate if and only if $A$ is a separable $k$-algebra.
If $char(k)$ divides the dimension of $A$, then $t_A$ is the zero map. However, if $A$ is separable, there is a notion of reduced trace, which satisfies (1),(2) and (3), so for separable algebras, assignments satisfying (1),(2),(3) exist, in every characteristic.
From now on, I will fix a $k$-algebra $B$, and only consider $K$-algebras $A$ such that $A\otimes_K K_{alg}\simeq B\otimes_K K_{alg}$. Such algebras will be called $B$-forms.
If $B$ is a direct product of matrices, we then recover separable algebras.
Now for the questions:
Q1 Fixing $B$ and restricting to $B$-forms, does there exist assignments satisfying (1),(2),(3) ?
Q1' If the answer to Q1 is no, do you know examples of $B$, other than a direct product of matrices, such that Q1 has a positive answer ?
Q2 Assume we are in the case where Q1 has a positive answer (e.g. the case of separable algebras), is such assignment unique up to multiplication by an element of $k$ or perhaps more reasonably by an element of $Z(B)$?
My thoughts I think that unfortunately, answer to Q1 is NO for arbitrary $B$. I have not finished my computations yet, but I think that if $B=K[X]/(X^2)$ any assignment satisfying (1) and (2) is identically zero. Can anyone confirm ?
However, I am pretty much convinced that answer to Q2 is YES, but I cannot figure a way to prove it yet. Note that (2) implies that $s_A$ is invariant under any $k$-algebra automorphism of $A$.
If $A$ is a $K$-algebra ($K/k$ is a field extension) and $s_A$ and $s'_A$ are non degenerate assignments, then there exists $\alpha\in A$ such that $s'_A(a)=s_A(\alpha a)$ for all $a\in A$ by non degeneracy. A non degeneracy argument then shows that $\varphi(\alpha)=\alpha$ for all automorphisms $\varphi$ of the $k$-algebra $A$. In particular, applying this to inner automorphisms show that
(*) $\alpha a=a\alpha$ for all $a\in A^\times$.
Side question. If $\alpha$ satisfies $(*)$, does it implies that $\alpha\in Z(A)$ ? i'll be happy to assume $K$ infinite if necessary.
If this is true, I think that the fact that $A$ is a $B$-form must imply that in fact $\alpha\in B_0$ , and so we should have $\alpha\in Z(B)$. Can someone confirm ?
I have absolutely no clue regarding Q1', but I would really really be interested in an example of $B$ such that Q1 has a positive answer, for a $B$ which is not a direct product of matrices. Unfortunately, I have the intuition that such $B$ do not exist...
Thanks for reading this long message.
Greg
