5.9. Norms and traces.

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The trace of a square matrix is the sum of its diagonal elements, Tr(aij) =

_

i aii. Since

Tr(UAU−1) = Tr(A), we can define the trace of an endomorphism α of a finite-dimensional

vector space V to be the trace of the matrix of α with respect to any basis of V .

Similarly, we can define the determinant and characteristic polynomial of α to be the

determinant and characteristic polynomial of the matrix of α with respect to any basis of V .

In a little more detail, a direct computation shows that Tr(AB) = Tr(BA), which shows

that Tr(UAU−1) = Tr(A) and hence Tr(α) is well-defined. The characteristic polynomial of

α can be defined to be

cα(X) = Xn + c1Xn−1 + ・ ・ ・ + cn, ci = (−1)i Tr(α|ΛiV ), n= dimV ;

in particular, c1 = −Tr(A) and cn = (−1)n detA. If A is the matrix of α with respect to

some basis for V , then cα(X) = det(XIn − A).

For α and β endomorphisms of a finite-dimensional F-vector space V , we have

Tr(α) ∈ F;Tr (α + β) =Tr(α) + Tr(β);

det(α) ∈ F;de t(αβ) =det(α) det(β).

Now let E be a finite extension of F of degree n, and regard E as an F-vector space. Then

α ∈ E defines an F-linear map αL : E → E, x → αx.

Define:

TrE/F(α) =Tr(αL);T r is a homomorphism (E, +) → (F, +);

NmE/F(α) =det(αL);N m is homomorphism (E×,×) → (F×,×);

cα(X) =cαL(X).

50 J.S. MILNE

Note that α → αL is an injective ring F-homomorphism from E into the ring of endomorphisms

of E as a vector space over F, and so the minimum polynomial of α (in the sense of

Section 1.8) is the same as the minimum polynomial of αL (in the sense of linear algebra).

Example 5.32. (a) Consider the field extension C ⊃ R;the matrix of αL, α = a + bi,

relative to the basis 1, i is

_

a −b

b a

_

, and so

TrC/R(α) = 2"(α), NmC/R(α) = |α|2.

(b) For α ∈ F, Tr(α) = rα, Nm(α) = αr , r = [E : F].

(c) Let E = Q[α, i] be the splitting field of X8 − 2. What are the norm and the trace of

α? The definition requires us to compute a 16 × 16 matrix. We shall see a quicker way of

computing them presently.

Proposition 5.33. Consider a finite field extension E/F, and let f(X) be the minimum

polynomial of α ∈ E (in the sense of Section 1.8). Then

cα(X) = f(X)[E:F[α]].

Proof. Suppose first that E = F[α]. In this case, we have to show that cα(X) = f(X).

But f(X)|cα(X) because cα(αL) = 0 (Cayley-Hamilton theorem), and the injectivity of

E → EndF-linear(E) then implies that cα(α) = 0. Since the polynomials are monic of the

same degree, they must be equal.

For the general case, write V for E regarded as an F-vector space. The endomorphism

αL of V defines an action of F[X] on V (see Math 593), and this action factors through

F[X]/(f(X)) = F[α]. Because F[α] is a field, V is a free F[α]-module, and in fact, V ≈

F[α]m with m = [E : F[α]] (count dimensions over F). Hence the characteristic polynomial

of α acting on V is the mth power of its characteristic polynomial acting on F[α], which,

according to case already proved, is f(X).

Alternatively, we can be more explicit. Let β1, ..., βn be a basis for F[α] over F, and let

γ1, ..., γm be a basis for E over F[α]. As we saw in the proof of (1.10), {βiγk} is a basis

for E over F. Write αβi =

_

ajiβj;the n A = (aij) has characteristic polynomial f(X)

according to the first case proved. Note that αβiγk =

_

ajiβjγk. Therefore the matrix of

αL in End(E) breaks up into n × n blocks with A’s down the diagonal and zero matrices

elsewhere. Therefore its characteristic polynomial is f(X)m.

Corollary 5.34. Suppose that the roots of the minimum polynomial of α are α1, . . . , αn

(in some splitting field containing E), and that [E : F[α]] = m. Then

Tr(α) = m

_n

i=1

αi, NmE/F α =

_

_n

i=1

αi

_m

.

Proof. Write the minimum polynomial of α as

f(X) = Xn + a1Xn−1 + ・ ・ ・ + an =

_

(X − αi).

Then

cα(X) = (f(X))m = Xmn + ma1Xmn−1 + ・ ・ ・ + amn

,

and so

TrE/F (α) = −ma1 = m

_

αi,

FIELDS AND GALOIS THEORY 51

and

NmE/F(α) = (−1)mnamn

= (

_

αi)m.

Example 5.35. (a) Consider the extension C ⊃ R. If α ∈ C \ R, then

cα(X) = f(X) = X2 − 2"(α)X + |α|2.

If α ∈ R, then cα(X) = (X − a)2.

(b) Let E = Q[α, i] be the splitting field of X8 − 2 (see Exercise 16). The minimum

polynomial of α = 8

√

2 is X8 − 2, and so

TrQ[α]/Q α = 0;T rE/Q α = 0.

NmQ[α]/Q α = −2;N mE/Q α = 4.

Remark 5.36. Assume E is separable over F, and let Ω be an algebraic closure of F;let

σ1, ..., σr be the distinct embeddings of E into Ω. Then

TrE/F α =

_

σiα

NmE/F α =

_

σiα.

When E = F[α], this follows from the observation (cf. 2.1b) that the σiα are the roots of

the minimum polynomial f(X) of α over F. In the general case, σ1α, ..., σrα are still roots

of f(X) in Ω, but now each root of f(X) occurs [E : F[α]] times (cf. the proof of 2.7).

For example, if E is Galois over F with Galois group G, then

TrE/F α =

_

σ∈G

σα

NmE/F α =

_

σ∈G

σα.

Proposition 5.37. For finite extensions E ⊃ M ⊃ F, we have

TrE/M ◦ TrM/F = TrE/F ,

NmE/M ◦NmM/F = NmE/F .

Proof. If E is separable over F, then this can be proved fairly easily using the descriptions

in the above remark. We omit the proof in the general case.

Proposition 5.38. Let f(X) ∈ F[X] factor as f(X) =

_m

i=1(X − αi) in some splitting

field, and let α = α1. Then, with f_ = df

dX , we have

disc f(X) = (−1)m(m−1)/2 NmF[α]/F f_(α).

Proof. Compute that

disc f(X) df =

_

i<j

(αi − αj)2

= (−1)m(m−1)/2 ・

_

i

(

_

j_=i

(αi − αj))

= (−1)m(m−1)/2 ・

_

f_(αj)

= (−1)m(m−1)/2 NmF[α]/F (f

_

(α)).

52 J.S. MILNE

Example 5.39. We compute the discriminant of

f(X) = Xn + aX + b, a, b ∈ F,

assumed to be irreducible and separable. Let α be a root of f(X), and let γ = f_(α) =

nαn−1 + a. We compute its norm. On multiplying the equation

αn + aα + b = 0

by nα−1 and rearranging, we obtain the equation

nαn−1 = −na − nbα

−1.

Hence

γ = nαn−1 + a = −(n − 1)a − nbα−1.

Solving for α gives

α =

−nb

γ + (n − 1)a

,

from which it is clear that F[α] = F[γ], and so the minimum polynomial of γ over F has

degree n also. If we write

f(

−nb

X + (n − 1)a

) =

P(X)

Q(X)

,

then P(γ) = f(α) = 0. Since

P(X) = (X + (n − 1)a)n − na(X + (n − 1)a)n−1 + (−1)nnnbn−1

is monic of degree n, it must be the minimum polynomial of γ. Therefore Nmγ is (−1)n

times the constant term of this polynomial, and so we find that

Nmγ = nnbn−1 + (−1)n−1(n − 1)n−1an.

Finally we obtain the formula,

disc(Xn + aX + b) = (−1)n(n−1)/2(nnbn−1 + (−1)n−1(n − 1)n−1an),

which is something Maple doesn’t know (because it doesn’t understand symbols as exponents).

For example,

disc(X5 + aX + b) = 55b4 + 44a5.

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