Algebraic Number Theory – Unique Non-Degenerate Symmetric Bilinear Form and Trace Form

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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).

Best Answer

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:

  1. $B(ab,c)=B(b,ac)$ for all $a,b,c\in L$. (Multiplication by elements of $L$ is self-adjoint)
  2. $B(g(a),g(b))=B(a,b)$ for all $a,b\in L$ and $g\in Gal(L/K)$. (Galois-invariance)
  3. $B(1,1)=[L:K]$. (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 $L$.

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