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where deg f (X) < deg p(X). Hence every element of K[X]/(p(X)) can be uniquely expressed
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),

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