# Field 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 onto 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,

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

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

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

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.

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

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

## Example

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

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

and so, ${\displaystyle \operatorname {Tr} _{L/\mathbb {Q} }(\alpha )=2a}$.[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

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

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

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.

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

## 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]

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

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

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

Theorem.[6] For bL, let Fb be the map ${\displaystyle a\mapsto \operatorname {Tr} _{L/K}(ba).}$ 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, ${\displaystyle ax^{2}+bx+c=0,{\text{ with }}a\neq 0,}$ and coefficients in the finite field ${\displaystyle \operatorname {GF} (q)=\mathbb {F} _{q}}$ 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 ${\displaystyle x={\sqrt {\frac {c}{a}}}}$ in GF(q). If b ≠ 0 then the substitution y = ax/b converts the quadratic equation to the form:

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

This equation has two solutions in GF(q) if and only if the absolute trace ${\displaystyle \operatorname {Tr} _{GF(q)/GF(2)}(\delta )=0.}$ 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 ${\displaystyle \operatorname {Tr} _{GF(q)/GF(2)}(k)=1.}$ Then a solution to the equation is given by:

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

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

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

## 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