# Linear maps preserved by algebra automorphisms

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.

• 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. – YCor Jun 25 '19 at 9:46

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.
• 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 ? – GreginGre Jun 25 '19 at 11:05