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Associated to a finite, separable field extension $L/K$, there is a natural nondegenerate bilinear form, the trace form, defined by $$\langle x,y \rangle := \mathrm{Tr}_{L/K}(xy)$$

Now, given a finite dimensional $K$-vector space $V$ with a nondegenerate bilinear form $\langle,\rangle$ what are some interesting/useful necessary or sufficient conditions for $\langle, \rangle$ to be equal to the bilinear form associated to a finite separable field extension $L/K$? Or more generally to some algebra $A/K$?

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  • $\begingroup$ Yes, it is. You cannot see anything general about the number of hyperbolic planes contained in a trace form. For example, if $L/K$ is a Galois field extension of odd degree, its trace form is isomorphic to the unit form, which is anisotropic as soon as $K$ is an ordered field. However, if $L=K(\alpha)$, where $\alpha^n+aX+b=0$ then the trace form contains many hypoerbolic planes. $\endgroup$ – GreginGre Jan 29 '18 at 20:59
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    $\begingroup$ Some useful references on trace forms are: Conner, Perlis - A survey of trace forms of algebraic number fields, and Mantilla-Soler - On the arithmetic determination of the trace. $\endgroup$ – Somatic Custard Apr 8 '18 at 19:14
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In general, this is a difficult question.

The answer is completely known if $K$ is a number field. I'm going to rephrase the result in terms of quadratic forms, but it is the same, really. Before that, i'm gonna give some necessary conditions, valid over an arbitrary field.

Recall first that, if $q$ is a non degenerate quadratic form over an arbitrary field $K$ of characteristic different from $2$, it is isomorphic to a diagonal form $\langle a_1,\ldots,a_n\rangle$, $a_i\in K^\times$.

If $K$ is an ordered field, the signature of $q$ with respect to this ordering is the number of positive $a_i's$ minus the number of negative ones. This is an integer which does not depend on the choice of the diagonalization (and not a pair of natural integers).

We say that a quadratic form is positive if all its signatures (w.r.t. the orderings of $K$) are non-negative.

Of course, if $K$ has no ordering, any quadratic form is positive.

One may show that the trace form of a separable extension $E/K$ (or more generally of an etale $K$-algebra), that i will denote by $q_E$, satisfies the following conditions (here $K$ is an arbitrary field of characteristic different from $2$):

  1. the quadratic form $q_E$ positive.

  2. If $n=\dim_K(E)$, then $q_E\simeq r_n\perp q'$, where $r_n$ is defined as follows:

    Write $n=\displaystyle\sum_{i=1}^h 2^{m_i},0\leq m_1<\cdots<m_h$.

Then $r_n=\langle 1,1,\ldots,1\rangle$ if $\displaystyle\sum_i m_i$ is even, and $r_n=\langle 2,1,\ldots,1\rangle$ if $\displaystyle \sum_i m_i$ is odd, where $r_n$ has rank $h$.

The following result of Epkenhans says Condition 1. characterizes the trace forms of field extensions of number fields.

Thm. Let $K$ be a number field, and let $q$ be a non degenerate quadratic form of rank $n$ over $K.$ Then, $q$ is isomorphic to the trace form of a separable field extension $E/K$ if and only if $q$ is a positive quadratic form and one of the following conditions hold:

1) $n=1$ and $q\simeq \langle 1\rangle$

2) $n=2$ and $q\simeq \langle 2,2D\rangle$ for some $D\in K^\times$

3) $n=3$ and $q\simeq \langle 1, 2,2D\rangle$ for some $D\in K^\times$

4) $n\geq 4$.

For general fields of characteristic different from $2$, Serre characterized the trace forms of étale algebras of dimension $\leq 7.$ You can find his results in "Cohomological invariants in Galois cohomology", Garibaldi, Merkurjev,Serre, AMS University Lecture Series Vol.28, 2003 (p.75-80)

I also kinda remember that there is a theorem (of Mestre ? or maybe Serre...)saying that a quadratic form is isomorphic to the trace form of an étale algebra if and only if it is the trace form of a separable field extension.

You also have and article of Drees, Epkenhans and Kruskemper, where they compute and characterize trace forms of various Galois extensions:

On the computation of the trace forms of some Galois field extensions, Journal of Algebra Volume 192, Issue 1, 1 June 1997, Pages 209-234

https://www.sciencedirect.com/science/article/pii/S0021869396969392?via%3Dihub

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    $\begingroup$ The reference is Epkenhans, Martin, On trace forms of algebraic number fields. Arch. Math. (Basel) 60 (1993), no. 6, 527–529. Searching on MathSciNet for Epkenhans + trace gives some articles. $\endgroup$ – user19475 Jan 29 '18 at 10:53
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    $\begingroup$ I seems that Hilbert's irreducibility theorem is used. $\endgroup$ – user19475 Jan 29 '18 at 12:57
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I think the starting point should be the algebra of selfadjoint operators, i.e. the subalgebra $A\subset End_K(V)$ which consists of operators $f$ such that $$\langle fx,y\rangle=\langle x, fy\rangle.$$ If it is indeed a field extension, then there is an injective map $\phi:V\to A$ (multiplication by a constant) such that $$\phi(x)y=\phi(y)x.$$ I am not sure if these conditions are sufficient, but they must be close.

By the way, $\phi$ is not necessarily surjective (one obvious example is the complex conjugation).

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    $\begingroup$ GreginGre gave a much better answer. In short, it is very difficult to say something reasonable when $K$ is a general field. $\endgroup$ – Alex Gavrilov Jan 29 '18 at 12:21

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