as a linear combination over K of the d cosets

1 + (p(X)), X + (p(X)), X 2 + (p(X)), . . . , X d’1 + (p(X)),

where d = deg p(X). Via the isomorphism µu under which µu (X k + (p(X))) = uk , we see that

the elements 1, u, . . . , ud’1 form a basis for K(u) over K.

√√ √√

Example 2.10. For the extension Q( 2, 3)/Q we have [Q( 2, 3) : Q] = 4.

√

Proof. By Example 2.4 we know that [Q( 2) : Q] = 2. We have the following tower of

extensions.

√√

Q( 2, 3)

√√ √

[Q( 2, 3):Q( 2)]

√ √√ √√ √

Q( 2) [Q( 2, 3):Q]=2[Q( 2, 3):Q( 2)]

2

Q

√√ √

We will show that [Q( 2, √3) √Q( 2)] = 2. √

: √ √ √

Notice that if u ∈ Q( 2, 3) = Q( 2)( 3) then u = a + b 3 for some a, b ∈ Q(√2),

√ √√ √ √

so 1, 3 span Q( 2, 3) over Q( 2). But if these are linearly dependent then 3 ∈ Q( 2).

Writing √ √

3=v+w 2

with v, w ∈ Q, we ¬nd that √

v 2 + 2w2 + 2vw 2 = 3 ∈ Q,

√

and hence 2vw 2 ∈√ The possibilities v = 0 or w = 0 are easily ruled out, while v, w √ 0

Q. =

√

would implies that 2 ∈ Q which is false. So 1, 3 are linearly√

√ independent over Q( 2)

√√ √

and√ √

therefore form a basis of Q( 2, 3). This shows that [Q( 2, 3) : Q( 2)] = 2 and so

[Q( 2, 3) : Q] = 4.

√√

Remark 2.11. There are some other sub¬elds of Q( 2, 3) which are conveniently displayed

in the following diagram.

√√

Q( 2, 3)

vvv

r

2 rrrr vvv

2

vvv

rr 2

rrr v

√ √ √

Q( 2) Q( 3) Q( 6)

ww r

rrr

ww

ww rr

ww 2

rrr 2

ww

2

rrr

w

Q

One idea in the veri¬cation of Example 2.10 can be extended to provide a useful general

result whose proof is left as an exercise.

Proposition 2.12. Let p1 , . . . , pn be a sequence of distinct primes pi > 0. Then

√ √ √

pn ∈ Q( p1 , . . . , pn’1 ).

/

√ √ √ √ √

√

Hence [Q( p1 , . . . , pn ) : Q( p1 , . . . , pn’1 )] = 2 and [Q( p1 , . . . , pn ) : Q] = 2n .

√ √

Example 2.13. For the extension Q( 2, i)/Q we have [Q( 2, i) : Q] = 4.

2.2. SIMPLE AND FINITELY GENERATED EXTENSIONS 27

√ √ √

Proof. We know that [Q( 2) : Q] = 2. Also, i ∈ Q( 2) since i is not real and Q( 2) R.

/√

√ √ √

Since i2 + 1 = 0, we have Q( 2, i) = Q( 2)(i) and [Q( 2, i) : Q( 2)] = 2. Using the formula

√ √ √ √

[Q( 2, i) : Q] = [Q( 2, i) : Q( 2)] [Q( 2) : Q],

√

we obtain [Q( 2, i) : Q] = 4.

√ √

This example also has several other sub¬elds, with only Q( 2) = Q( 2, i) © R being a

sub¬eld of R.

C

2

R ∞

√

Q( 2, i)

∞

uuu

t

tt uuu

2 2

tt uuu

tt 2

u

tt

√ √

Q(i)

Q( 2) Q( 2 i)

uuu

rr

uuu

rrr

u rr 2

2

2 uuu

rrr

u r

Q

1, let En = Q(21/n ) R, where 21/n ∈ R denotes the positive real

Example 2.14. For n

n-th root of 2.

(i) Show that [En : Q] = n.

(ii) If m 1 with m | n, show that Em En and determine [En : Em ].

(iii) If m, n are coprime, show that Emn = Q(21/m , 21/n ).

Solution. (i) Consider the evaluation homomorphism µ21/n : Q[X] ’’ En . Applying the

Eisenstein Test 1.37 using the prime 2 to the polynomial X n ’ 2 ∈ Z[X], we ¬nd that

ker µ21/n = (X n ’ 2) Q[X],

and the induced homomorphism µ21/n : Q[X]/(X n ’ 2) ’’ En is an isomorphism. Hence

[En : Q] = n.

(ii) Since n/m is an integer,

21/m = (21/n )n/m ∈ En ,

so

Em = Q(21/m ) ⊆ En .

By Theorem 2.6 we have

n = [En : Q] = [En : Em ] [Em : Q] = m[En : Em ],

whence [En : Em ] = n/m.

(iii) By (ii) we have Em Emn and En Emn , hence Q(21/m , 21/n ) Emn . As gcd(m, n) = 1,

there are integers r, s for which rm + sn = 1 and so

1 rm + sn r s

= =+.

mn mn nm

This shows that

21/mn = (21/n )r (21/m )s ∈ Q(21/m , 21/n ),

Q(21/m , 21/n ). Combining these inclusions we obtain Emn = Q(21/m , 21/n ).

whence Emn

28 2. FIELDS AND THEIR EXTENSIONS

Exercises on Chapter 2

√ √

2.1. Let p ∈ N be an prime. Show that the extension Q( p)/Q has [Q( p) : Q] = 2.

√√ √

2.2. Let p, q > 0 be distinct primes. Show that [Q( p, q) : Q( p)] = 2.

2.3. Prove Proposition 2.12 by induction on n.

2.4. Let K a ¬eld with char K = 2 and suppose that L/K is an extension. If a, b ∈ K are

distinct, suppose that u, v ∈ L satisfy u2 = a and v 2 = b. Show that K(u, v) = K(u + v).

[Hint: ¬rst show that u±v = 0 and deduce that u’v ∈ K(u+v); then show that u, v ∈ K(u+v).]

2.5. Show that [Q(i) : Q] = 2.

√ √

2.6. Show that [Q( 3, i) : Q] = 4. Find the three sub¬elds L Q( 3, i) with [L : Q] = 2 and

display their relationship in a diagram, indicating which ones are sub¬elds of R.

2.7. Let ζ5 = e2πi/5 ∈ C.

(a) Explain why [Q(ζ5 ) : Q] = 4.

(b) Show that cos(2π/5), sin(2π/5) i ∈ Q(ζ5 ).

(c) Show that for t ∈ R,

cos 5t = 16 cos5 t ’ 20 cos3 t + 5 cos t.

(d) Show that the numbers cos(2kπ/5) with k = 0, 1, 2, 3, 4 are roots of the polynomial

f (X) = 16X 5 ’ 20X 3 + 5X ’ 1 = (X ’ 1)(4X 2 + 2X ’ 1)2

and deduce that [Q(cos(2π/5)) : Q] = 2.

(e) Display the relationship between the ¬elds Q, Q(cos(2π/5)), and Q(ζ5 ) in a suitable

diagram.

2.8. This question is for those who like lots of calculation or using Maple. Let ζ7 = e2πi/7 ∈ C.

(a) Explain why [Q(ζ7 ) : Q] = 6.

(b) Show that cos(2π/7), sin(2π/7) i ∈ Q(ζ7 ).

(c) Show

cos 7t = 64 cos7 t ’ 112 cos5 t + 56 cos3 t ’ 7 cos t.

Show that the numbers cos(2kπ/7) with k = 0, 1, . . . , 6 are roots of the polynomial

f (X) = 64X 7 ’ 112X 5 + 56X 3 ’ 7X ’ 1 = (X ’ 1)(8X 3 + 4X 2 ’ 4X ’ 1)2

and deduce that [Q(cos(2π/7)) : Q] = 3.

(d) Show that sin(2π/7) i is a root of

g(X) = 64X 7 + 112X 5 + 56X 3 + 7X = X(64X 6 + 112X 4 + 56X 2 + 7)

and that 64X 6 + 112X 4 + 56X 2 + 7 ∈ Q[X] is irreducible. What is [Q(sin(2π/7) i) : Q]?

(e) Display the relationship between the ¬elds Q, Q(cos(2π/7)), Q(sin(2π/7) i) and Q(ζ7 )

in a diagram.

(f) Is i ∈ Q(ζ7 )?

2.9. In this question we continue to consider the situation described in Example 2.14.

(a) Show that

{id} if n is odd,

AutQ (En ) =

{id, „n } ∼ Z/2 if n is even,

=

where „n has composition order 2.

En R. Show that AutQ (E) = {id}.

(b) Let E =

n>1

(c) Display the 6 sub¬elds of E12 in a diagram.

(d) Which of the sub¬elds in part (c) contain the element 21/2 + 21/3 ?

CHAPTER 3

Algebraic extensions of ¬elds

3.1. Algebraic extensions

Let L/K be an extension of ¬elds. From Theorem 2.9(ii), recall the following notion.

Definition 3.1. An element t ∈ L is algebraic over K if there is a non-zero polynomial

p(X) ∈ K[X] for which p(t) = 0.

Notice in particular that for an element t ∈ K, the polynomial p(X) = X ’ t ∈ K[X]

satis¬es p(t) = 0, so such a t is algebraic over K.

Theorem 2.9 allows us to characterize algebraic elements in other ways.

Proposition 3.2. Let t ∈ L. Then the following conditions are equivalent.

(i) t is algebraic over K.

(ii) The evaluation homomorphism µt : K[X] ’’ L has non-trivial kernel.

(iii) The extension K(t)/K is ¬nite dimensional.

Definition 3.3. If t ∈ L is algebraic over K then by Proposition 3.2,

ker µt = (minpolyK,t (X)) = (0),

where minpolyK,t (X) ∈ K[X] is an irreducible monic polynomial called the minimal polynomial

of t over K. The degree of minpolyK,t (X) is called the degree of t over K and denoted by

degK t.

Proposition 3.4. If t ∈ L is algebraic over K then

[K(t) : K] = deg minpolyK,t (X) = degK t.

Proof. This follows from Theorem 2.9(ii).

Remark 3.5. Suppose that t ∈ L is algebraic over K and p(X) ∈ ker µt with deg p(X) =

deg minpolyK,t (X). Then minpolyK,t (X) | p(X) and so

p(X) = u minpolyK,t (X)

for some u ∈ K. In particular, when p(X) is monic,

p(X) = minpolyK,t (X).

We will often use this without further comment.

√

Example 3.6. Consider C/Q. The minimal polynomial of 2 ∈ C over Q is

minpolyQ,√2 (X) = X 2 ’ 2.

√

Proof. Clearly X 2 ’ 2 ∈ ker µ√2 since ( 2)2 ’ 2 = 0. By Example 2.4,

√

deg minpolyQ,√2 (X) = [Q( 2) : Q] = 2,

hence

minpolyQ,√2 (X) = X 2 ’ 2.

Example 3.7. Consider C/Q. The minimal polynomial of i ∈ C over Q is X 2 + 1.

Proof. Clearly X 2 + 1 ∈ ker µi since i2 + 1 = 0. As [Q(i) : Q] = 2, we have

minpolyQ,i (X) = X 2 + 1.

29

30 3. ALGEBRAIC EXTENSIONS OF FIELDS

Example 3.8. Consider C/Q. Find the minimal polynomial of the primitive 6-th root of

unity, ζ6 ∈ C over Q.

Solution. Recall from Example 1.43 that ζ6 is a root of the irreducible cyclotomic poly-

nomial

¦6 (X) = X 2 ’ X + 1.

Then ¦6 (X) ∈ ker µζ6 so minpolyQ,ζ6 (X) | ¦6 (X). Since ¦6 (X) is irreducible and monic, we

must have

minpolyQ,ζ6 (X) = ¦6 (X)

and so degQ ζ6 = 2.

√ √

Example 3.9. Consider C/Q. Find the minimal polynomial of 2 + 3 over Q.

Solution. Notice that

√ √√ √

√ √ √ √

( 3 ’ 2)( 3 + 2) 1

√ √ =√ √ ∈ Q( 2 + 3).

3’ 2=

( 3 + 2) 2+ 3

So we have

√ 1√ √ √ √ √ √

2= ( 2 + 3) ’ ( 3 ’ 2) ∈ Q( 2 + 3),

2

√ 1√ √ √ √ √ √

3= ( 2 + 3) + ( 3 ’ 2) ∈ Q( 2 + 3),