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,
,
is a K-
linear transformation of this vector space into itself. The trace, TrL/K(α), is defined as the
trace (in the
linear algebra sense) 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 K). Then
If L/K is
separable then each root appears only once[2] (however this does not mean the coefficient above is one; for example if α is the identity element 1 of K then the trace is [L:K ] times 1).
and so, .[1] The minimal polynomial of α is X2 − 2aX + (a2 − db2).
Properties of the trace
Several properties of the trace function hold for any finite extension.[3]
The trace TrL/K : L → K is a K-
linear map (a K-linear functional), that is
.
If α ∈ K then
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.
In this setting we have the additional properties:[5]
.
For any , there are exactly elements with .
Theorem.[6] For b ∈ L, let Fb be the map Then Fb ≠ Fc if b ≠ c. 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, ax2 + bx + c = 0 with a ≠ 0, and coefficients in the finite field 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
discriminantΔ = 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
integerh), these formulas are no longer applicable.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 in GF(q). If b ≠ 0 then the substitution y = ax/b converts the quadratic equation to the form:
This equation has two solutions in GF(q)
if and only if the absolute trace 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 Then a solution to the equation is given by:
When h = 2m' + 1, a solution is given by the simpler expression:
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 (x, y) to TrL/K(xy) is a
nondegenerate,
symmetric bilinear form called the trace form. If L/K is a Galois extension, the trace form is invariant with respect to the Galois group.
Lorenz, Falko (2008). Algebra. Volume II: Fields with Structure, Algebras and Advanced Topics. Springer.
ISBN978-0-387-72487-4.
Zbl1130.12001.
Mullen, Gary L.; Panario, Daniel (2013), Handbook of Finite Fields, CRC Press,
ISBN978-1-4398-7378-6
Roman, Steven (2006), Field theory, Graduate Texts in Mathematics, vol. 158 (Second ed.), Springer, Chapter 8,
ISBN978-0-387-27677-9,
Zbl1172.12001
Rotman, Joseph J. (2002), Advanced Modern Algebra, Prentice Hall,
ISBN978-0-13-087868-7
Further reading
Conner, P.E.; Perlis, R. (1984). A Survey of Trace Forms of Algebraic Number Fields. Series in Pure Mathematics. Vol. 2. World Scientific.
ISBN9971-966-05-0.
Zbl0551.10017.