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2.1 Norms and Traces
2.1.1 Definitions and Comments
If E/F is a field extension of finite degree n, then in particular, E is a finite-dimensional
vector space over F, and the machinery of basic linear algebra becomes available. If x is
any element of E, we can study the F-linear transformation m(x)giv en by multiplication
by x, that is, m(x)y = xy. We define the norm and the trace of x, relative to the extension
E/F, as
NE/F (x)= detm(x)and TE/F (x)= trace m(x).
We will write N(x)and T(x)if E/F is understood. If the matrix A(x) = [aij(x)] represents
m(x)with respect to some basis for E over F, then the norm of x is the determinant
of A(x)and the trace of x is the trace of A(x), that is, the sum of the main diagonal
entries. The characteristic polynomial of x is defined as the characteristic polynomial of
the matrix A(x), that is,
charE/F (x)= det[XI − A(x)]
where I is an n by n identity matrix. It follows from the definitions that the norm, the
trace and the coefficients of the characteristic polynomial are elements belonging to the
base field F.
2.1.2 Example
Let E = C and F = R. A basis for C over R is {1, i} and, with x = a + bi, we have
(a + bi)(1) = a(1)+ b(i)and (a + bi)(i) = −b(1)+ a(i).
Thus
A(a + bi) =
a −b
b a
The norm, trace and characteristic polynomial of a + bi are
N(a + bi) = a2 + b2, T(a + bi) = 2a, char(a + bi) = X2 − 2aX + a2 + b2.
The computation is exactly the same if E = Q(i)and F = Q.
2.1.3 Some Basic Properties
Notice that in (2.1.2), the coefficient of the second highest power of X in the characteristic
polynomial is minus the trace, and the constant term is the norm. In general, it follows
from the definition of characteristic polynomial that
char(x) = Xn − T(x)Xn−1 + · · · + (−1)nN(x). (1)
[The only terms multiplying Xn−1 in the expansion of the determinant defining the characteristic
polynomial are −aii(x), i = 1, . . . , n. Set X = 0 to show that the constant term
of char(x)is (−1)n detA(x).]
If x, y ∈ E and a, b ∈ F, then
T(ax + by) = aT(x) + bT (y)and N(xy) = N(x)N(y). (2)
[This holds because m(ax + by) = am(x) + bm(y)and m(xy) = m(x) ◦ m(y).]
If a ∈ F, then
N(a) = an, T(a) = na, and char(a) = (X − a)n. (3)
[Note that the matrix representing multiplication by the element a in F is aI.]
It is natural to look for a connection between the characteristic polynomial of x and
the minimal polynomial min(x, F)of x over F.
2.1.4 Proposition
charE/F (x)= [min(x, F)]r, where r = [E : F(x)].
Proof. First assume that r = 1, so that E = F(x). By the Cayley-Hamilton theorem,
the linear transformation m(x)satisfies char(x). Since m(x)is multiplication by x, it
follows that x itself is a root of char(x). Thus min(x, F)divides char(x), and since both
polynomials are monic of degree n, the result follows. In the general case, let y1, . . . , ys
be a basis for F(x)ove r F, and let z1, . . . , zr be a basis for E over F(x). Then the yizj
form a basis for E over F. Let A = A(x)b e the matrix representing multiplication by x
in the extension F(x)/F, so that xyi =
k akiyk and x(yizj) =
k aki(ykzj ). Order the
2.1. NORMS AND TRACES 3
basis for E/F as y1z1, y2z1, . . . , ysz1; y1z2, y2z2 . . . , ysz2; · · · ; y1zr, y2zr, . . . , yszr. Then
m(x)is represented in E/F as
A 0 · · · 0
0 A · · · 0
...
...
...
0 0 · · · A
with r blocks, each consisting of the s by s matrix A. Thus charE/F (x)= [det(XI −A)]r,
which by the r = 1 case coincides with [min(x, F)]r. ♣
2.1.5Corollary
Let [E : F] = n and [F(x) : F] = d. Let x1, . . . , xd be the roots of min(x, F), counting
multiplicity, in a splitting field. Then
N(x) = (
d
i=1
xi)n/d, T(x) = n
d
d
i=1
xi, char(x) = [
d
i=1
(X − xi)]n/d.
Proof. The formula for the characteristic polynomial follows from (2.1.4). By (2.1.3),
the norm is (−1)n times the constant term of char(x). Evaluating the characteristic
polynomial at X = 0 produces another factor of (−1)n, which yields the desired expression
for the norm. Finally, if min(x, F) = Xd+ad−1Xd−1+· · ·+a1X+a0, then the coefficient
of Xn−1 in [min(x, F)]n/d is (n/d)ad−1 = −(n/d)
d
i=1 xi. Since the trace is the negative
of this coefficient [see (2.1.3)], the result follows. ♣
If E is a separable extension of F, there are very useful alternative expressions for the
trace, norm and characteristic polynomial.
2.1.6 Proposition
Let E/F be a separable extension of degree n, and let σ1, . . . , σn be the distinct Fembeddings
(that is, F-monomorphisms)of E into an algebraic closure of E, or equally
well into a normal extension L of F containing E. Then
NE/F (x) =
n
i=1
σi(x), TE/F (x) =
n
i=1
σi(x), charE/F (x) =
n
i=1
(X − σi(x)).
Proof. Each of the d distinct F-embeddings τi of F(x)in to L takes x into a unique
conjugate xi, and extends to exactly n/d = [E : F(x)] F-embeddings of E into L, all
of which also take x to xi. Thus the list of elements {σ1(x), . . . , σn(x)} consists of the
τi(x) = xi, i = 1, . . . , d, each appearing n/d times. The result follows from (2.1.5). ♣
We may now prove a basic transitivity property.
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