The trace form is the unique non-deg. symmetric bilinear form such that ... Let $L \vert K$ be a field extension. Recall the various definitions of the trace map $Tr_{L\vert K}:L\mapsto K$ (via matrices, minimal polynomials, field embeddings).
With that, one gets the trace form $B:L\times L \mapsto K$, $B(x,y):=Tr_{L\vert K}(xy)$, a symmetric bilinear form on the $K$-vector space $L$. It is well-known to be non-degenerate as soon as the field extension is separable, which we assume from now on. In fact, if it helps, we can assume $char(K)=0$.
The trace form is known to give "conceptual" definitions of the different and the discriminant of (e.g. number) field extensions.
I was wondering if there are, in turn, "conceptual" definitions of the trace form as suggested in the title: as the unique (or unique up to scaling) non-degenerate symmetric BLF which satisfies properties x,y,z ...
My motivation for that comes partly from Lie algebras, where on an absolutely simple Lie algebra the Killing form is characterized, up to scaling, as the unique non-degenerate symmetric BLF which is invariant under $ad$.
Candidates for such properties:
Being invariant under field embeddings to an algebraic closure (in case of a Galois extension, under Galois) in either / both components.
Satisfying $B(x,x) = [L:K] x^2$ for $x\in K$ (this would take care of scaling).
 A: In the case that $L$ is a Galois extension of $K$ with $[L:K]$ not divisible by the characteristic, there is the following characterization.  The trace form is the unique bilinear form $B:L\times L\to K$ which satisfies the following three properties:

*

*$B(ab,c)=B(b,ac)$ for all $a,b,c\in L$.  (Multiplication by elements of $L$ is self-adjoint)

*$B(g(a),g(b))=B(a,b)$ for all $a,b\in L$ and $g\in Gal(L/K)$. (Galois-invariance)

*$B(1,1)=[L:K]$. (Normalization)
The proof is easy.  Property (1) implies that $B(a,b)=B(1,ab)$, so $B$ is uniquely determined by the linear functional $T:L\to K$ given by $T(x)=B(1,x)$.  Property (2) then says that $T$ is Galois-invariant, which implies $[L:K]T(x)$ is equal to the sum of the values of $T$ on the Galois-conjugates of $x$.  But by linearity that sum is just $T(Tr(x))$, and then property (3) implies $T(Tr(x))=[L:K]Tr(x)$ so $T(x)=Tr(x)$ since we are assuming $[L:K]$ is not divisible by the characteristic.
Let me remark that I would expect something like property (1) to be crucial for any sort of characterization.  In particular, you need some property that relates $B$ to the multiplication of $L$ (and not merely its structure as a module over the Galois group as in property (2)).  The following example may be instructive.  Suppose the characteristic is different from $2$ and $L=K(a)$ where $a^2\in L$ and $a\not\in L$.  To define a bilinear form on $L$ we need to pick values for $B(1,1)$, $B(1,a)$, $B(a,1)$ and $B(a,a)$.  Normalization will fix the value of $B(1,1)$, and Galois-invariance will force $B(1,a)$ and $B(a,1)$ to be $0$ since $a$ is conjugate to $-a$.  But any value at all for $B(a,a)$ will still satisfy Galois-invariance, and any nonzero value will make $B$ symmetric and nondegenerate.  To force $B(a,a)$ to be $2a^2$, you need to use some property like property (1) that knows what the value of $a\cdot a$ is using the multiplication of $L$.
