ak X k = a0 + a1 X + · · · + ak X k + · · · ,

k=0

with ak ∈ k.

(a) Show that this can be made into an integral domain containing k[X] as a subring by

de¬ning addition and multiplication in the obvious way.

(b) Show that ∞ ak X k ∈ k[[X]] is a unit if and only if a0 = 0.

k=0

(c) Show that Fr(k[[X]]) consists of all ¬nite tailed Laurent series

∞

ak X k = a X + a +1

+ · · · + ak X k + · · ·

+1 X

k=

∈ Z and a| ∈ k.

for some

1.10. Taking k = Q, ¬nd the quotient and remainder when performing long division of f (X) =

6X 4 ’ 6X 3 + 3X 2 ’ 3X ’ 2 by d(X) = 2X 3 + X + 3.

1.11. Taking k = F3 , ¬nd the quotient and remainder when performing long division of

f (X) = 2X 3 + 2X 2 + X + 1 by d(X) = 2X 3 + 2X.

1.12. Let p > 0 be a prime. Suppose that f (X) = a0 + a1 X + · · · + an X n ∈ Z[X] with

p an and that f (X) ∈ Fp [X] denotes the polynomial obtained by reducing the coe¬cients of

f (X) modulo p. If f (X) is irreducible, show that f (X) is irreducible. Which of the following

polynomials in Z[X] is irreducible?

X 3 ’ X + 1, X 3 + 2X + 1, X 3 + X ’ 1, X 5 ’ X + 1, X 5 + X ’ 1, 5X 3 ’ 10X + X 2 ’ 2.

1.13. Find generators for each of the following ideals:

√

I1 = {f (X) ∈ Q[X] : f (i) = 0} Q[X], I2 = {f (X) ∈ Q[X] : f ( 2 i) = 0} Q[X],

√ √

I3 = {f (X) ∈ Q[X] : f ( 2) = 0} Q[X], I4 = {f (X) ∈ R[X] : f ( 2) = 0} R[X],

√

I5 = {f (X) ∈ R[X] : f ( 2 i) = 0} R[X], I6 = {f (X) ∈ R[X] : f (ζ3 ) = 0} R[X].

1.14. Consider the inclusion inc : Q ’’ C and its extension to µ√2 : Q[X] ’’ C. Determine

the image µ√2 Q[X] ⊆ C. What is µ’√2 Q[X] ⊆ C? Find ker µ√2 Q[X] and ker µ’√2 Q[X];

are these maximal ideals?

√

1.15. Let ω = (’1 + 3i)/2 ∈ C. Consider the inclusion inc : Q ’’ C and its extension

to µω : Q[X] ’’ C. Determine the image µω Q[X] ⊆ C. Determine ker µω Q[X] and decide

whether it is maximal. Find another evaluation homomorphism with the same kernel and image.

1.16. Consider the inclusion inc : Q ’’ C and its extension to µ± : Q[X] ’’ C where ± is

one of the 4 complex roots of the polynomial f (X) = X 4 ’ 2 ∈ Q[X]. Determine the image

µ± Q[X] ⊆ C and the ideal ker µ± Q[X]; is the latter ideal maximal? What happens if ± is

replaced by one of the other roots of f (X)?

Repeat this problem starting with the inclusion of the real numbers into the complex num-

bers inc : R ’’ C and µ± : R[X] ’’ C.

1.17. Use Cardan™s method to ¬nd the complex roots of the polynomial

f (X) = X 3 ’ 9X 2 + 21X ’ 5.

1.18. Consider the real numbers

√ √ 2 7 2 7

3 3 3 3

±= 10 + 108 + 10 ’ 108, β= 1+ + 1’ .

3 3 3 3

22 1. INTEGRAL DOMAINS, FIELDS AND POLYNOMIAL RINGS

Find rational cubic polynomials f (X) and g(X) for which f (±) = 0 = g(β). Hence determine

these real numbers.

1.19. Prove the ¬nal part of Example 1.58 by showing that there is an isomorphism of groups

A¬ 1 (k) ∼ Aut| (k[X]).

=

1.20. Let k be any ¬eld. Consider the 6 automorphisms ±j : k(X) ’’ k(X) (j = 1, . . . , 6)

de¬ned by

±1 (f (X)) = f (X), ±2 (f (X)) = f (1 ’ X), ±3 (f (X)) = f (1/X),

±4 (f (X)) = f ((X ’ 1)/X), ±5 (f (X)) = f (1/(1 ’ X)), ±6 (f (X)) = f (X/(X ’ 1)).

Show that the set consisting of these elements is a subgroup “| Aut| (k(X)) isomorphic to

the symmetric group S3 . When k = F2 , show that “| ∼ GL2 (k).

=

1.21. Determine the cyclotomic polynomial ¦20 (X).

1.22. Let p > 0 be a prime.

(a) Show that for k 1, the cyclotomic polynomial ¦pk (X) satis¬es

k’1

¦pk (X) = ¦p (X p )

and has as its complex roots the primitive pk -th roots of 1.

(b) Show that ¦pk (X) ∈ Q[X] is irreducible.

(c) Generalize part (a) to show that if n = pr1 · · · prk is the prime power factorization of n

1 k

with the pi being distinct primes and ri > 0, then

r ’1

r1 ’1

···pkk

¦n (X) = ¦p1 ···pk (X p1 ).

1.23. For n 2, show that

X •(n) ¦n (X ’1 ) = ¦n (X).

’1

1.24. Show that for n 1, ζn + ζn = 2 cos(2π/n).

’1 ’2

2

Find expressions for ζ5 +ζ5 and ζ5 +ζ5 in terms of cos(2π/5). Hence ¬nd a rational polynomial

which has cos(2π/5) as a root.

1.25. Let p > 0 be a prime and K be a ¬eld with char K = p.

(a) Show that if ζ ∈ K is a p-th root of 1 then ζ = 1. Deduce that if m, n > 0 and p n,

then every npm -th root of 1 in K is an n-th root of 1.

(b) If a ∈ K, show that the polynomial X p ’ a ∈ K[X] has either no roots or exactly one

root in K.

CHAPTER 2

Fields and their extensions

2.1. Fields and sub¬elds

Definition 2.1. Let K and L be ¬elds and suppose that K ⊆ L is a subring. Then we say

that K is a sub¬eld of L; L is also said to be an extension (¬eld ) of K. We write K L to

indicate this and K < L if K is a proper sub¬eld of L, i.e., if K = L. We will also sometimes

write L/K when discussing the extension of K to L.

An important fact about an extension of ¬elds L/K is that L is a K-vector space whose

addition is the addition in the ¬eld L while scalar multiplication is de¬ned by

u · x = ux (u ∈ K, x ∈ L).

Definition 2.2. We will call dimK L the degree or index of the extension L/K and use the

notation [L : K] = dimK L. An extension of ¬elds L/K is ¬nite if [L : K] < ∞, otherwise it is

in¬nite.

Example 2.3. Show that the extension C/R is ¬nite, while R/Q and C/Q are in¬nite.

Solution. We have

C = {x + yi : x, y ∈ R},

so 1, i span C as a vector space over R. Since i ∈ R, these elements are also linearly independent

/

over R and therefore they form a basis, whence [C : R] = 2. The in¬niteness of R/Q and C/Q are

consequences of the fact that any ¬nite dimensional vector space over Q is countable, however

R and C are uncountable. A basis for the Q-vector space R is known as a Hamel basis.

√

Example 2.4. Consider the extension Q( 2)/Q where

√ √

Q( 2) = {x + y 2 : x, y ∈ Q}.

√

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

√ √

Solution. The elements 1, 2 clearly span the Q-vector space Q( 2). √ Now recall that

√ √

2 ∈ Q. If the elements 1, 2 were linearly dependent we would have u + v 2 = 0 for some

/

u, v ∈ Q not both zero; in fact it is easy to see that we would then also have u, v both non-zero.

Thus we would have √ u

2 = ’ ∈ Q,

v

√ √

which we know √ be false. Hence 1, 2 are linearly independent and so form a basis for Q( 2)

to

over Q and [Q( 2) : Q] = 2.

If we have two extensions L/K and M/L then it is a straightforward to verify that K M

and so we have another extension M/K.

Definition 2.5. Given two extensions L/K and M/L, we say that L/K is a subextension

of M/K and sometimes write L/K M/L.

Theorem 2.6. Let L/K be a subextension of M/K.

(i) If one or both of the dimensions [L : K] or [M : L] is in¬nite then so is [M : K].

(ii) If the dimensions [L : K] and [M : L] are both ¬nite then so is [M : K] and

[M : K] = [M : L] [L : K].

23

24 2. FIELDS AND THEIR EXTENSIONS

Proof. (i) If [M : K] is ¬nite, choose a basis m1 , . . . , mr of M over K. Now any element

u ∈ M can be expressed as

u = t1 m 1 + · · · + tr m r ,

where t1 , . . . , tr ∈ K; but since K ⊆ L, this means that m1 , . . . , mr spans M over L and so

[M : L] < ∞. Also L is a K-vector subspace of the ¬nite dimensional K-vector space M , hence

[L : K] < ∞.

(ii) Setting r = [L : K] and s = [M : L], choose a basis 1 , . . . , r of L over K and a basis

m1 , . . . , ms of M over L.

Now let v ∈ M . Then there are elements y1 , . . . , ys ∈ L for which

v = y1 m1 + · · · + ys ms .

But each yj can be expressed in the form

yj = x1j + · · · + xrj

1 r

for suitable xij ∈ K. Hence,

s r s r

v= xij mj = xij ( i mj ),

i

j=1 i=1 j=1 i=1

where each coe¬cient xij is in K. Thus the elements i mj (i = 1, . . . , r, j = 1, . . . , s) span the

K-vector space M .

Now suppose that for some tij ∈ K we have

s r

tij ( i mj ) = 0.

j=1 i=1

On collecting terms we obtain

s r

tij mj = 0,

i

j=1 i=1

r

where each coe¬cient i=1 tij i is in L. By the linear independence of the mj over L, this

means that for each j,

r

tij = 0.

i

i=1

By the linear independence of the i over K, each tij = 0.

Hence the i mj form a basis of M over K and so

[M : K] = rs = [M : L] [L : K].

We will often indicate subextensions in diagrammatic form where larger ¬elds always go

above smaller ones and the information on the lines indicates dimensions

MF

A

[M :L]

6

1 [M :K]=[M :L] [L:K]

L

&

!

[L:K]

K

We often suppress ˜composite™ lines such as the dashed one. Such towers of extensions are our

main objects of study. We can build up sequences of extensions and form towers of arbitrary

2.2. SIMPLE AND FINITELY GENERATED EXTENSIONS 25

length. Thus, if L1 /K, L2 /L1 , . . . , Lk /Lk’1 is a such a sequence of extensions, there is a

diagram

Lk

Lk’1

L1

K

2.2. Simple and ¬nitely generated extensions

Definition 2.7. Let F be a ¬eld and K F . Given elements u1 , . . . , ur ∈ F we set

K(u1 , . . . , ur ) = L

K6L6F

u1 ,...,ur ∈L

which is the smallest sub¬eld in F that contains K and the elements u1 , . . . , ur . The ex-

tension K(u1 , . . . , ur )/K is said to be generated by the elements u1 , . . . , ur ; we also say that

K(u1 , . . . , ur )/K is a ¬nitely generated extension of K. An extension of the form K(u)/K is

called a simple extension of K with generator u.

We can extend this to the case of an in¬nite sequence u1 , . . . , ur , . . . in F and denote by

K(u1 , . . . , ur , . . .) F the smallest extension ¬eld of K containing all the elements ur .

It is not di¬cult to show that

(2.1) K(u1 , . . . , ur ) =

f (u1 , . . . , ur )

∈ F : f (X1 , . . . , Xr ), g(X1 , . . . , Xr ) ∈ K[X1 , . . . , Xr ], g(u1 , . . . , ur ) = 0 .

g(u1 , . . . , ur )

Reordering the ui does not change K(u1 , . . . , un ).

Proposition 2.8. Let K(u)/K and K(u, v)/K(u) be simple extensions. Then

K(u, v) = K(u)(v) = K(v)(u).

More generally,

K(u1 , . . . , un ) = K(u1 , . . . , un’1 )(un )

and this is independent of the order of the sequence u1 , . . . , un .

Theorem 2.9. For a simple extension K(u)/K exactly one of the following conditions holds.

(i) The evaluation at u homomorphism µu : K[X] ’’ K(u) is a monomorphism and on

passing to the fraction ¬eld gives an isomorphism (µu )— : K(X) ’’ K(u). In this case,

K(u)/K is in¬nite and u is said to be transcendental over K.

(ii) The evaluation at u homomorphism µu : K[X] ’’ K(u) has a non-trivial kernel

ker µu = (p(X)) where p(X) ∈ K[X] is an irreducible monic polynomial of positive de-

gree and the quotient homomorphism µu : K[X]/(p(X)) ’’ K(u) is an isomorphism.

In this case K(u)/K is ¬nite with [K(u) : K] = deg p(X) and u is said to be algebraic

over K.

Proof. (i) If ker µu = (0), all that needs checking is that (µu )— is an epimorphism; but as

u is in the image of (µu )— this is obvious.

(ii) When ker µu = (0), Theorem 1.30(iv) implies that the image of µu is a sub¬eld of K(u) and

26 2. FIELDS AND THEIR EXTENSIONS

since it contains u it must equal K(u). Hence µu is an isomorphism. Using Long Division, we

¬nd that every element of K[X]/(p(X)) can be uniquely expressed as a coset of the form

f (X) + (p(X)),