# Field trace

For other uses, see Trace

In mathematics, the field trace is a particular function defined with respect to a finite field extension L/K, which is a K-linear map from L to K.

## Definition

Let K be a field and L a finite extension (and hence an algebraic extension) of K. L can be viewed as a vector space over K. Multiplication by α, an element of L,

$m_\alpha:L\to L \text{ given by } m_\alpha (x) = \alpha x$,

is a K-linear transformation of this vector space into itself. The trace, TrL/K(α), is defined as the (linear algebra) trace of this linear transformation.[1]

For α in L, let σ1(α), ..., σn(α) be the roots (counted with multiplicity) of the minimal polynomial of α over K (in some extension field of L), then

$\operatorname{Tr}_{L/K}(\alpha)=[L:K(\alpha)]\sum_{j=1}^n\sigma_j(\alpha)$.

If L/K is separable then each root appears only once and the coefficient above is one.[2]

More particularly, if L/K is a Galois extension and α is in L, then the trace of α is the sum of all the Galois conjugates of α, i.e.

$\operatorname{Tr}_{L/K}(\alpha)=\sum_{g\in\operatorname{Gal}(L/K)}g(\alpha)$,

where Gal(L/K) denotes the Galois group of L/K.

## Example

Let $L = \mathbb{Q}(\sqrt{d})$ be a quadratic extension of $\mathbb{Q}$. Then a basis of $L/\mathbb{Q} \text{ is }1, \sqrt{d}.$ If $\alpha = a + b\sqrt{d}$ then the matrix of $m_{\alpha}$ is:

$\left [ \begin{matrix} a & bd \\ b & a \end{matrix} \right ]$,

and so, $\operatorname{Tr}_{L/\mathbb{Q}}(\alpha) = 2a$.[1] The minimal polynomial of α is X2 - 2a X + a2 - d b2.

## Properties of the trace

Several properties of the trace function hold for any finite extension.[3]

The trace TrL/K : LK is a K-linear map (a K-linear functional), that is

$\operatorname{Tr}_{L/K}(\alpha a + \beta b) = \alpha \operatorname{Tr}_{L/K}(a)+ \beta \operatorname{Tr}_{L/K}(b) \text{ for all }\alpha, \beta \in K$.

If α ∈ K then $\operatorname{Tr}_{L/K}(\alpha) = [L:K] \alpha.$

Additionally, trace behaves well in towers of fields: if M is a finite extension of L, then the trace from M to K is just the composition of the trace from M to L with the trace from L to K, i.e.

$\operatorname{Tr}_{M/K}=\operatorname{Tr}_{L/K}\circ\operatorname{Tr}_{M/L}$.

## Finite fields

Let L = GF(qn) be a finite extension of a finite field K = GF(q). Since L/K is a Galois extension, if α is in L, then the trace of α is the sum of all the Galois conjugates of α, i.e.[4]

$\operatorname{Tr}_{L/K}(\alpha)=\alpha + \alpha^q + \cdots + \alpha^{q^{n-1}}$.

In this setting we have the additional properties,[5]

• $\operatorname{Tr}_{L/K}(a^q) = \operatorname{Tr}_{L/K}(a) \text{ for } a \in L$
• $\text{for any }\alpha \in K, \text{ we have }|\{b \in L \colon \operatorname{Tr}_{L/K}(b) = \alpha \}| = q^{n-1}$

And,[6]

Theorem. For bL, let Fb be the map $a \mapsto \operatorname{Tr}_{L/K}(ba).$ Then FbFc if bc. Moreover the K-linear transformations from L to K are exactly the maps of the form Fb as b varies over the field L.

When K is the prime subfield of L, the trace is called the absolute trace and otherwise it is a relative trace.[4]

### Application

A quadratic equation, $ax^2 + bx + c = 0, \text{ with }a \ne 0,$ and coefficients in the finite field $\operatorname{GF}(q) = \mathbb{F}_q$ has either 0, 1 or 2 roots in GF(q) (and two roots, counted with multiplicity, in the quadratic extension GF(q2)). If the characteristic of GF(q) is odd, the discriminant, Δ = b2 - 4ac indicates the number of roots in GF(q) and the classical quadratic formula gives the roots. However, when GF(q) has even characteristic (i.e., q = 2h for some positive integer h), these formulas are no longer applicable.

Consider the quadratic equation ax2 + bx + c = 0 with coefficients in the finite field GF(2h).[7] If b = 0 then this equation has the unique solution $x = \sqrt{\frac{c}{a}}$ in GF(q). If b ≠ 0 then the substitution y = ax/b converts the quadratic equation to the form:

$y^2 + y + \delta = 0, \text { where } \delta = \frac{ac}{b^2}$.

This equation has two solutions in GF(q) if and only if the absolute trace $\operatorname{Tr}_{GF(q)/GF(2)}(\delta) = 0.$ In this case, if y = s is one of the solutions, then y = s + 1 is the other. Let k be any element of GF(q) with $\operatorname{Tr}_{GF(q)/GF(2)}(k) = 1.$ Then a solution to the equation is given by:

$y = s = k \delta^2 + (k + k^2)\delta^4 + \ldots + (k + k^2 + \ldots + k^{2^{h-2}})\delta^{2^{h-1}}$.

When h = 2m + 1, a solution is given by the simpler expression:

$y = s = \delta + \delta^{2^2} + \delta^{2^4} + \ldots + \delta^{2^{2m}}$.

## Trace form

When L/K is separable, the trace provides a duality theory via the trace form: the map from L × L to K sending (xy) to TrL/K(xy) is a nondegenerate, symmetric, bilinear form called the trace form. An example of where this is used is in algebraic number theory in the theory of the different ideal.

The trace form for a finite degree field extension L/K has non-negative signature for any field ordering of K.[8] The converse, that every Witt equivalence class with non-negative signature contains a trace form, is true for algebraic number fields K.[8]

If L/K is an inseparable extension, then the trace form is identically 0.[9]

## Notes

1. ^ a b Rotman 2002, p. 940
2. ^ Rotman 2002, p. 941
3. ^ Roman 1995, p. 151 (1st ed.)
4. ^ a b Lidl & Niederreiter 1997, p.54
5. ^ Mullen & Panario 2013, p. 21
6. ^ Lidl & Niederreiter 1997, p.56
7. ^ Hirschfeld 1979, pp. 3-4
8. ^ a b Lorenz (2008) p.38
9. ^ Isaacs 1994, p. 369 as footnoted in Rotman 2002, p. 943