Let $A$ be a finite dimensional , associative, unital $F$-algebra, where $F$ is a field.

Let $s_A:A\to F$ be an $F$-linear map. Now consider an arbitrary field extension $K/F$, and define $s_{A\otimes_FK}$ as the unique $K$-linear map such that $s_{A\otimes_F K}(a\otimes 1)=s_A(a) $ for all $a\in A$.

Question. Assume that for all automorphism $f:A\to A$ of $F$-algebras, we have $s_A\circ f=s_A$.

Do we have the same property for $A\otimes_FK$, that is that for all automorphism $g:A\otimes_FK\to A\otimes_FK$ of $K$-algebras, $s_{A\otimes_F K}\circ g=s_{A\otimes_FK}$ ?

If the answer is no , can we find necessary and sufficient conditions on $s_A$ for this to be true ?

The answer is positive for all examples I know (separable algebras with the reduced traces, plus few more non separable examples), but I have no clue whether it is true or not , and how to proceed to prove it if it is indeed true.

I will be happy to assume that $F$ is infinite if this is necessary.

I am also interested in the same question when $s_A$ is replaced by a bilinear form over $A$, but I suspect that the answer would be the same.

  • 2
    $\begingroup$ Probably there's a counterexample. It should be possible to arrange something for which $\mathrm{Aut}(A)_F\to \mathrm{Aut}(A)_K$ is not surjective on connected components. $\endgroup$
    – YCor
    Jun 25, 2019 at 9:46

1 Answer 1


Here's an example where the answer is no ($d+1$-dimensional for $d\ge 2$). Let $L$ be a separable extension of $F$ of finite degree $d\ge 2$ (we assume this exist, i.e. $F$ is not separably closed), and assume that $L$ splits over $K$ (e.g., assume that $K$ is separably closed). Say, $F=\mathbf{R}$ and $L=K=\mathbf{C}$ if this language is not familiar.

Consider $A=F\times L$ as $A$-algebra, and $s_F$ is the projection from $A$ to $F$ with kernel $L$ ($s_F$ is a more accurate notation than $s_A$ since I can write $s_K=s_F\otimes_F K$). Since $\mathrm{Aut}(A)_F$ consists of the $$F\times L\ni (t,u)\mapsto (t,\alpha(u)), \quad \alpha\in\mathrm{Aut}_F(L),$$ we indeed have $s_F\circ f=s_F$ for every $f\in\mathrm{Aut}(A)_F$.

But on the "complexification" $A\otimes_FK$ is isomorphic to $K^{d+1}$ as $K$-algebra, and in particular all the symmetric group $\mathfrak{S}_{d+1}$ acts by permuting, and the set of linear forms commuting with all automorphism reduced to the "sum of coordinates" map, which vanishes on no nonzero idempotent. Since $s_F$ vanished on the nonzero idempotent $1_L$, we deduce that $s_K$ does not commute with all automorphisms.

  • $\begingroup$ Thanks for the nice answer ! do you have any ideas about necessary and sufficient conditiosn for this to hold ? or at least a family of examples for which it does hold (apart from separable algebras) ? For example, what if $\ker(s_A)$ does not contain nonzero left ideal of $A$, that is the bilinear map $(x,y)\in A\times A\mapsto s_A(xy)\in F$ is non degenerate ? $\endgroup$
    – GreginGre
    Jun 25, 2019 at 11:05

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