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6.4.2 Corollary
If E is a п¬Ѓnite п¬Ѓeld of characteristic p, then E/Fp is a Galois extension. The Galois group
is cyclic and is generated by the Frobenius automorphism Пѓ(x) = xp , x в€€ E.

Proof. E is a splitting п¬Ѓeld for a separable polynomial over Fp , so E/Fp is Galois; see
(6.3.1). Since xp = x for each x в€€ Fp , Fp is contained in the п¬Ѓxed п¬Ѓeld F( Пѓ ). But
each element of the п¬Ѓxed п¬Ѓeld is a root of X p в€’ X, so F( Пѓ ) has at most p elements.
Consequently, F( Пѓ ) = Fp . Now Fp = F(Gal(E/Fp ) by (6.1.2), so by the fundamental
theorem, Gal(E/Fp ) = Пѓ . в™Ј

6.4.3 Corollary
Let E/F be a п¬Ѓnite extension of a п¬Ѓnite п¬Ѓeld, with |E| = pn , |F | = pm . Then E/F is a
Galois extension. Moreover, m divides n, and Gal(E/F ) is cyclic and is generated by the
m
automorphism П„ (x) = xp , x в€€ E. Furthermore, F is the only subп¬Ѓeld of E of size pm .

Proof. If the degree of E/F is d, then as in (6.4.1), (pm )d = pn , so d = n/m and m | n.
m
We may then reproduce the proof of (6.4.2) with Fp replaced by F , Пѓ by П„ , xp by xp ,
m
and X p by X p . Uniqueness of F as a subп¬Ѓeld of E with pm elements follows because
m
there is only one splitting п¬Ѓeld over Fp for X p в€’ X inside E; see (3.2.1). в™Ј

How do we know that п¬Ѓnite п¬Ѓelds (other than the Fp ) exist? There is no problem.
Given any prime p and positive integer n, we can construct E = GF (pn ) as a splitting
n
п¬Ѓeld for X p в€’ X over Fp . We have just seen that if E contains a subп¬Ѓeld F of size pm ,
then m is a divisor of n. The converse is also true, as a consequence of the following basic
result.

6.4.4 Theorem
The multiplicative group of a п¬Ѓnite п¬Ѓeld is cyclic. More generally, if G is a п¬Ѓnite subgroup
of the multiplicative group of an arbitrary п¬Ѓeld, then G is cyclic.

Proof. G is a п¬Ѓnite abelian group, hence contains an element g whose order r is the
exponent of G, that is, the least common multiple of the orders of all elements of G; see
Section 1.1, Problem 9. Thus if x в€€ G then the order of x divides r, so xr = 1. Therefore
each element of G is a root of X r в€’ 1, so |G| в‰¤ r. But |G| is a multiple of the order of
every element, so |G| is at least as big as the least common multiple, so |G| в‰Ґ r. We
conclude that the order and the exponent are the same. But then g has order |G|, so
G = g and G is cyclic. в™Ј

6.4.5 Proposition
GF (pm ) is a subп¬Ѓeld of E = GF (pn ) if and only if m is a divisor of n.

Proof. The вЂњonly ifвЂќ part follows from (6.4.3), so assume that m divides n. If t is any
positive integer greater than 1, then m | n iп¬Ђ (tm в€’ 1) | (tn в€’ 1). (A formal proof is not
diп¬ѓcult, but I prefer to do an ordinary long division of tn в€’ 1 by tm в€’ 1. The successive
12 CHAPTER 6. GALOIS THEORY

quotients are tnв€’m , tnв€’2m , tnв€’3m , . . . , so the division will be successful iп¬Ђ n в€’ rm = 0 for
some positive integer r.) Taking t = p, we see that pm в€’ 1 divides |E в€— |, so by (6.4.4)
and (1.1.4), E в€— has a subgroup H of order pm в€’1. By LagrangeвЂ™s theorem, each x в€€ Hв€Є{0}
m
satisп¬Ѓes xp = x. As in the proof of (6.4.1), H в€Є {0} coincides with the set of roots of
m m
X p в€’ X. Thus we may construct entirely inside GF (pn ) a splitting п¬Ѓeld for X p в€’ X
over Fp . But this splitting п¬Ѓeld is a copy of GF (pm ). в™Ј

In practice, п¬Ѓnite п¬Ѓelds are constructed by adjoining roots of carefully selected irre-
ducible polynomials over Fp . The following result is very helpful.

6.4.6 Theorem
n
Let p be a prime and n a positive integer. Then X p в€’ X is the product of all monic
irreducible polynomials over Fp whose degree divides n.
n
Proof. Let us do all calculations inside E = GF (pn ) = the set of roots of f (X) = X p в€’X.
If g(X) is any monic irreducible factor of f (X), and deg g = m, then all roots of g lie
in E. If О± is any root of g, then Fp (О±) is a п¬Ѓnite п¬Ѓeld with pm elements, so m divides n by
(6.4.5) or (6.4.3). Conversely, let g(X) be a monic irreducible polynomial over Fp whose
degree m is a divisor of n. Then by (6.4.5), E contains a subп¬Ѓeld with pm elements,
and this subп¬Ѓeld must be isomorphic to Fp (О±). If ОІ в€€ E corresponds to О± under this
isomorphism, then g(ОІ) = 0 (because g(О±) = 0) and f (ОІ) = 0 (because ОІ в€€ E). Since g is
the minimal polynomial of ОІ over Fp , it follows that g(X) divides f (X). By (6.4.1), the
roots of f are distinct, so no irreducible factor can appear more than once. The theorem
is proved. в™Ј

6.4.7 The Explicit Construction of a Finite Field
By (6.4.4), the multiplicative group E в€— of a п¬Ѓnite п¬Ѓeld E = GF (pn ) is cyclic, so E в€— can
be generated by a single element О±. Thus E = Fp (О±) = Fp [О±], so that О± is a primitive
element of E. The minimal polynomial of О± over Fp is called a primitive polynomial. The
key point is that the nonzero elements of E are not simply the nonzero polynomials of
degree at most n в€’ 1 in О±, they are the powers of О±. This is signiп¬Ѓcant in applications to
coding theory. LetвЂ™s do an example over F2 .
The polynomial g(X) = X 4 + X + 1 is irreducible over F2 . One way to verify this is
to factor X 16 в€’ X = X 16 + X over F2 ; the factors are the (necessarily monic) irreducible
polynomials of degrees 1,2 and 4. To show that g is primitive, we compute powers of О±:
О±0 = 1, О±1 = О±, О±2 = О±2 , О±3 = О±3 , О±4 = 1 + О± (since g(О±) = 0),
О±5 = О± + О±2 , О±6 = О±2 + О±3 , О±7 = О±3 + О±4 = 1 + О± + О±3 , О±8 = О± + О±2 + О±4 = 1 + О±2
(since 1+1=0 in F2 ),
О±9 = О±+О±3 , О±10 = 1+О±+О±2 , О±11 = О±+О±2 +О±3 , О±12 = 1+О±+О±2 +О±3 , О±13 = 1+О±2 +О±3 ,
О±14 = 1 + О±3 ,
and at this point we have all 24 в€’ 1 = 15 nonzero elements of GF (16). The pattern now
repeats, beginning with О±15 = О± + О±4 = 1.
For an example of a non-primitive polynomial, see Problem 1.
6.5. CYCLOTOMIC FIELDS 13

Problems For Section 6.4
1. Verify that the irreducible polynomial X 4 + X 3 + X 2 + X + 1 в€€ F2 [X] is not primitive.
2. Let F be a п¬Ѓnite п¬Ѓeld and d a positive integer. Show that there exists an irreducible
polynomial of degree d in F [X].
3. In (6.4.5) we showed that m | n iп¬Ђ (tm в€’ 1) | (tn в€’ 1) (t = 2, 3, . . . ). Show that an
equivalent condition is (X m в€’ 1) divides (X n в€’ 1).
If E is a п¬Ѓnite extension of a п¬Ѓnite п¬Ѓeld, or more generally a п¬Ѓnite separable extension
of a п¬Ѓeld F , then by the theorem of the primitive element, E = F (О±) for some О± в€€ E.
We now develop a condition equivalent to the existence of a primitive element.
4. Let E/F be a п¬Ѓnite extension, with E = F (О±) and F в‰¤ L в‰¤ E. Suppose that the min-
rв€’1
imal polynomial of О± over L is g(X) = i=0 bi X i + X r , and let K = F (b0 , . . . , brв€’1 ).
If h is the minimal polynomial of О± over K, show that g = h, and conclude that
L = K.
5. Continuing Problem 4, show that there are only п¬Ѓnitely many intermediate п¬Ѓelds L
between E and F .
6. Conversely, let E = F (О±1 , . . . , О±n ) be a п¬Ѓnite extension with only п¬Ѓnitely many inter-
mediate п¬Ѓelds between E and F . We are going to show by induction that E/F has a
primitive element. If n = 1 there is nothing to prove, so assume the result holds for
all integers less than n. If L = F (О±1 , . . . , О±nв€’1 ), show that E = F (ОІ, О±n ) for some
ОІ в€€ L.
7. Now assume (without loss of generality) that F is inп¬Ѓnite. Show that there are distinct
elements c, d в€€ F such that F (cОІ + О±n ) = F (dОІ + О±n ).
8. Continuing Problem 7, show that E = F (cОІ + О±n ). Thus a п¬Ѓnite extension has a
primitive element iп¬Ђ there are only п¬Ѓnitely many intermediate п¬Ѓelds.
9. Let О± be an element of the п¬Ѓnite п¬Ѓeld GF (pn ). Show that О± and О±p have the same
minimal polynomial over Fp .
10. Suppose that О± is an element of order 13 in the multiplicative group of nonzero
elements in GF (3n ). Partition the integers {0, 1, . . . , 12} into disjoint subsets such
that if i and j belong to the same subset, then О±i and О±j have the same minimal
polynomial. Repeat for О± an element of order 15 in GF (2n ). [Note that elements of
the speciп¬Ѓed orders exist, because 13 divides 26 = 33 в€’ 1 and 15 = 24 в€’ 1.]

6.5 Cyclotomic Fields
6.5.1 Deп¬Ѓnitions and Comments
Cyclotomic extensions of a п¬Ѓeld F are formed by adjoining nth roots of unity. Formally, a
cyclotomic extension of F is a splitting п¬Ѓeld E for f (X) = X n в€’ 1 over F . The roots of f
are called nth roots of unity, and they form a multiplicative subgroup of the group E в€— of
nonzero elements of E. This subgroup must be cyclic by (6.4.4). A primitive nth root of
unity is one whose order in E в€— is n.
14 CHAPTER 6. GALOIS THEORY

It is tempting to say вЂњobviously, primitive nth roots of unity must exist, just take a
generator of the cyclic subgroupвЂќ. But suppose that F has characteristic p and p divides n,
say n = mp. If П‰ is an nth root of unity, then

0 = П‰ n в€’ 1 = (П‰ m в€’ 1)p

so the order of П‰ must be less than n. To avoid this diп¬ѓculty, we assume that the
characteristic of F does not divide n. Then f (X) = nX nв€’1 = 0, so the greatest common
divisor of f and f is constant. By (3.4.2), f is separable, and consequently E/F is Galois.
Since there are n distinct nth roots of unity, there must be a primitive nth root of unity П‰,
and for any such П‰, we have E = F (П‰).
If Пѓ is any automorphism in the Galois group Gal(E/F ), then Пѓ must take a primitive
root of unity П‰ to another primitive root of unity П‰ r , where r and n are relatively prime.
(See (1.1.5).) We can identify Пѓ with r, and this shows that Gal(E/F ) is isomorphic to a
subgroup of Un , the group of units mod n. Consequently, the Galois group is abelian.
Finally, by the fundamental theorem (or (3.5.9)), [E : F ] = | Gal(E/F )|, which is a
divisor of |Un | = П•(n).
Cyclotomic п¬Ѓelds are of greatest interest when the underlying п¬Ѓeld F is Q, the rational
numbers, and from now on we specialize to that case. The primitive nth roots of unity
are ei2ПЂr/n where r and n are relatively prime. Thus there are П•(n) primitive nth roots
of unity. Finding the minimal polynomial of a primitive nth root of unity requires some
rather formidable equipment.

6.5.2 Deп¬Ѓnition
The nth cyclotomic polynomial is deп¬Ѓned by

(X в€’ П‰i )
ОЁn (X) =
i

where the П‰i are the primitive nth roots of unity in the п¬Ѓeld C of complex numbers. Thus
the degree of ОЁn (X) is П•(n).
ВїFrom the deп¬Ѓnition, we have ОЁ1 (X) = X в€’ 1 and ОЁ2 (X) = X + 1. In general, the
cyclotomic polynomials can be calculated by the following recursion formula, in which d
runs through all positive divisors of n.

6.5.3 Proposition

Xn в€’ 1 = ОЁd (X).
d|n

In particular, if p is prime, then

Xp в€’ 1
= X pв€’1 + X pв€’2 + В· В· В· + X + 1.
ОЁp (X) =
X в€’1
6.5. CYCLOTOMIC FIELDS 15

Proof. If П‰ is an nth root of unity, then its order in Cв€— is a divisor d of n, and in this
case, П‰ is a primitive dth root of unity, hence a root of ОЁd (X). Conversely, if d | n, then
any root of ОЁd (X) is a dth , hence an nth , root of unity. в™Ј
ВїFrom (6.5.3) we have
ОЁ3 (X) = X 2 + X + 1,
ОЁ4 (X) = X 2 + 1, ОЁ5 (X) = X 4 + X 3 + X 2 + X + 1,
X 6 в€’1 X 6 в€’1 3
ОЁ6 (X) = (Xв€’1)(X+1)(X 2 +X+1) = (X 3 в€’1)(X+1) = X +1 = X 2 в€’ X + 1.
X+1
It is a natural conjecture that all coeп¬ѓcients of the cyclotomic polynomials are integers,
and this turns out to be correct.

6.5.4 Proposition
ОЁn (X) в€€ Z[X].
Proof. By (6.5.3), we have

Xn в€’ 1 = [ ОЁd (X)]ОЁn (X).
d|n,d<n

By deп¬Ѓnition, the cyclotomic polynomials are monic, and by induction hypothesis, the
expression in brackets is a monic polynomial in Z[X]. Thus ОЁn (X) is the quotient of two
monic polynomials with integer coeп¬ѓcients. At this point, all we know for sure is that
the coeп¬ѓcients of ОЁn (X) are complex numbers. But if we apply ordinary long division,
even in C, we know that the process will terminate, and this forces the quotient ОЁn (X)
to be in Z[X]. в™Ј
We now show that the nth cyclotomic polynomial is the minimal polynomial of each
primitive nth root of unity.

6.5.5 Theorem
ОЁn (X) is irreducible over Q.
Proof. Let П‰ be a primitive nth root of unity, with minimal polynomial f over Q. Since
П‰ is a root of X n в€’ 1, we have X n в€’ 1 = f (X)g(X) for some g в€€ Q[X]. Now it follows
from (2.9.2) that if a monic polynomial over Z is the product of two monic polynomials f
and g over Q, then in fact the coeп¬ѓcients of f and g are integers.
If p is a prime that does not divide n, we will show that П‰ p is a root of f . If not,
then it is a root of g. But g(П‰ p ) = 0 implies that П‰ is a root of g(X p ), so f (X) divides
g(X p ), say g(X p ) = f (X)h(X). As above, h в€€ Z[X]. But by the binomial expansion
modulo p, g(X)p в‰Ў g(X p ) = f (X)h(X) mod p. Reducing the coeп¬ѓcients of a polynomial
k(X) mod p is equivalent to viewing it as an element k в€€ Fp [X], so we may write g(X)p =
f (X)h(X). Then any irreducible factor of f must divide g, so f and g have a common
factor. But then X n в€’ 1 has a multiple root, contradicting (3.4.2). [This is where we use
the fact that p does not divide n.]
Now we claim that every primitive nth root of unity is a root of f , so that deg f в‰Ґ
П•(n) =deg ОЁn , and therefore f = ОЁn by minimality of f . The best way to visualize this
16 CHAPTER 6. GALOIS THEORY

is via a concrete example with all the features of the general case. If П‰ is a primitive nth
root of unity where n = 175, then П‰ 72 is a primitive nth root of unity because 72 and 175
are relatively prime. Moreover, since 72 = 23 Г— 32 , we have

П‰ 72 = (((((П‰)2 )2 )2 )3 )3

and the result follows. в™Ј

6.5.6 Corollary
The Galois group G of the nth cyclotomic extension Q(П‰)/Q is isomorphic to the group Un
of units mod n.

Proof. By the fundamental theorem, |G| = [Q(П‰) : Q] = deg ОЁn = П•(n) = |Un |. Thus the
monomorphism of G and a subgroup of Un (see (6.5.1)) is surjective. в™Ј

Problems For Section 6.5
1. If p is prime and p divides n, show that ОЁpn (X) = ОЁn (X p ). (This formula is sometimes
useful in computing the cyclotomic polynomials.)
2. Show that the group of automorphisms of a cyclic group of order n is isomorphic to
the group Un of units mod n. (This can be done directly, but it is easier to make use
of the results of this section.)
We now do a detailed analysis of subgroups and intermediate п¬Ѓelds associated with the
cyclotomic extension Q7 = Q(П‰)/Q where П‰ = ei2ПЂ/7 is a primitive 7th root of unity.
The Galois group G consists of automorphisms Пѓi , i = 1, 2, 3, 4, 5, 6, where Пѓi (П‰) = П‰ i .
3. Show that Пѓ3 generates the cyclic group G.
4. Show that the subgroups of G are 1 (order 1), Пѓ6 (order 2), Пѓ2 (order 3), and
G = Пѓ3 (order 6).
5. The п¬Ѓxed п¬Ѓeld of 1 is Q7 and the п¬Ѓxed п¬Ѓeld of G is Q. Let K be the п¬Ѓxed п¬Ѓeld
of Пѓ6 . Show that П‰ + П‰ в€’1 в€€ K, and deduce that K = Q(П‰ + П‰ в€’1 ) = Q(cos 2ПЂ/7).
6. Let L be the п¬Ѓxed п¬Ѓeld of Пѓ2 . Show that П‰ + П‰ 2 + П‰ 4 belongs to L but not to Q.
7. Show that L = Q(П‰ + П‰ 2 + П‰ 4 ).
8. If q = pr , p prime, r > 0, show that

ОЁq (X) = tpв€’1 + tpв€’2 + В· В· В· + 1
rв€’1
where t = X p .
9. Assuming that the п¬Ѓrst 6 cyclotomic polynomials are available [see after (6.5.3)], cal-
culate ОЁ18 (X) in an eп¬Ђortless manner.
6.6. THE GALOIS GROUP OF A CUBIC 17

6.6 The Galois Group of a Cubic
Let f be a polynomial over F , with distinct roots x1 , . . . , xn in a splitting п¬Ѓeld E over F .
The Galois group G of f permutes the xi , but which permutations belong to G? When f
is a quadratic, the analysis is straightforward, and is considered in Section 6.3, Problem 4.
In this section we look at cubics (and some other manageable cases), and the appendix
to Chapter 6 deals with the quartic.

6.6.1 Deп¬Ѓnitions and Comments
Let f be a polynomial with roots x1 , . . . , xn in a splitting п¬Ѓeld. Deп¬Ѓne

(xi в€’ xj ).
в€†(f ) =
i<j

The discriminant of f is deп¬Ѓned by

(xi в€’ xj )2 .
D(f ) = в€†2 =
i<j
в€љ
LetвЂ™s look at a quadratic polynomial f (X) = X 2 + bX + c, with roots 1 (в€’b В± b2 в€’ 4c).
2
In order to divide by 2, we had better assume that the characteristic of F is not 2, and
this assumption is usually made before deп¬Ѓning the discriminant. In this case we have
(x1 в€’x2 )2 = b2 в€’4c, a familiar formula. Here are some basic properties of the discriminant.

6.6.2 Proposition
Let E be a splitting п¬Ѓeld of the separable polynomial f over F , so that E/F is Galois.
(a) D(f ) belongs to the base п¬Ѓeld F .
(b) Let Пѓ be an automorphism in the Galois group G of f . Then Пѓ is an even permutation
(of the roots of f ) iп¬Ђ Пѓ(в€†) = в€†, and Пѓ is odd iп¬Ђ Пѓ(в€†) = в€’в€†.
(c) G вЉ† An , that is, G consists entirely of even permutations, iп¬Ђ D(f ) is the square of
an element of F (for short, D в€€ F 2 ).
Proof. Let us examine the eп¬Ђect of a transposition Пѓ = (i, j) on в€†. Once again it is
useful to consider a concrete example with all the features of the general case. Say
n = 15, i = 7, j = 10. Then

x3 в€’ x7 в†’ x3 в€’ x10 , x3 в€’ x10 в†’ x3 в€’ x7
x10 в€’ x12 в†’ x7 в€’ x12 , x7 в€’ x12 в†’ x10 в€’ x12
x7 в€’ x8 в†’ x10 в€’ x8 , x8 в€’ x10 в†’ x8 в€’ x7
x7 в€’ x10 в†’ x10 в€’ x7 .

The point of the computation is that the net eп¬Ђect of (i, j) on в€† is to take xi в€’ xj to
its negative. Thus Пѓ(в€†) = в€’в€† when Пѓ is a transposition. Thus if Пѓ is any permutation,
we have Пѓ(в€†) = в€† if в€† is even, and Пѓ(в€†) = в€’в€† if Пѓ is odd. Consequently, Пѓ(в€†2 ) =
18 CHAPTER 6. GALOIS THEORY

(Пѓ(в€†))2 = в€†2 , so D belongs to the п¬Ѓxed п¬Ѓeld of G, which is F . This proves (a), and (b)
follows because в€† = в€’в€† (remember that the characteristic of F is not 2). Finally G вЉ† An
iп¬Ђ Пѓ(в€†) = в€† for every Пѓ в€€ G iп¬Ђ в€† в€€ F(G) = F . в™Ј

6.6.3 The Galois Group of a Cubic
In the appendix to Chapter 6, it is shown that the discriminant of the abbreviated cubic
X 3 + pX + q is в€’4p3 в€’ 27q 2 , and the discriminant of the general cubic X 3 + aX 2 + bX + c
is

a2 (b2 в€’ 4ac) в€’ 4b3 в€’ 27c2 + 18abc.

Alternatively, the change of variable Y = X + a eliminates the quadratic term without
3
changing the discriminant.
We now assume that the cubic polynomial f is irreducible as well as separable. Then
the Galois group G is isomorphic to a transitive subgroup of S3 (see Section 6.3, Prob-
lem 3). By direct enumeration, G must be A3 or S3 , and by (6.6.2(c)), G = A3 iп¬Ђ the
discriminant D is a square in F .
If G = A3 , which is cyclic of order 3, there are no proper subgroups except {1}, so
there are no intermediate п¬Ѓelds strictly between E and F . However, if G = S3 , then the
proper subgroups are

{1, (2, 3)}, {1, (1, 3)}, {1, (1, 2)}, A3 = {1, (1, 2, 3), (1, 3, 2)}.

If the roots of f are О±1 , О±2 and О±3 , then the corresponding п¬Ѓxed п¬Ѓelds are

F (О±1 ), F (О±2 ), F (О±3 ), F (в€†)

where A3 corresponds to F (в€†) because only even permutations п¬Ѓx в€†.

6.6.4 Example
Let f (X) = X 3 в€’ 31X + 62 over Q. An application of the rational root test (Section 2.9,
Problem 1) shows that f is irreducible. The discriminant is в€’4(в€’31)3 в€’27(62)2 = 119164в€’
103788 = 15376 = (124)2 , which is a square in Q. Thus the Galois group of f is A3 .
We now develop a result that can be applied to certain cubics, but which has wider
applicability as well. The preliminary steps are also of interest.

6.6.5 Some Generating Sets of Sn
(i) Sn is generated by the transpositions (1, 2), (1, 3), . . . , (1, n).
[An arbitrary transposition (i, j) can be written as (1, i)(1, j)(1, i).]
(ii) Sn is generated by transpositions of adjacent digits, i.e., (1, 2), (2, 3), . . . , (nв€’1, n).
[Since (1, j в€’ 1)(j в€’ 1, j)(1, j в€’ 1) = (1, j), we have

(1, 2)(2, 3)(1, 2) = (1, 3), (1, 3)(3, 4)(1, 3) = (1, 4), etc.,

and the result follows from (i).]
6.6. THE GALOIS GROUP OF A CUBIC 19

(iii) Sn is generated by the two permutations Пѓ1 = (1, 2) and П„ = (1, 2, . . . , n).
[If Пѓ2 = П„ Пѓ1 П„ в€’1 , then Пѓ2 is obtained by applying П„ to the symbols of Пѓ1 (see Section 5.2,
Problem 1). Thus Пѓ2 = (2, 3). Similarly,

Пѓ3 = П„ Пѓ2 П„ в€’1 = (3, 4), . . . , Пѓnв€’1 = П„ Пѓnв€’2 П„ в€’1 = (n в€’ 1, n),

and the result follows from (ii).]
(iv) Sn is generated by (1, 2) and (2, 3, . . . , n).
[(1, 2)(2, 3, . . . , n) = (1, 2, 3, . . . , n), and (iii) applies.]

6.6.6 Lemma
If f is an irreducible separable polynomial over F of degree n, and G is the Galois group
of f , then n divides |G|. If n is a prime number p, then G contains a p-cycle.

Proof. If О± is any root of f , then [F (О±) : F ] = n, so by the fundamental theorem, G
contains a subgroup whose index is n. By LagrangeвЂ™s theorem, n divides |G|. If n = p,
then by CauchyвЂ™s theorem, G contains an element Пѓ of order p. We can express Пѓ as a
product of disjoint cycles, and the length of each cycle must divide the order of Пѓ. Since
p is prime, Пѓ must consist of disjoint p-cycles. But a single p-cycle already uses up all the
symbols to be permuted, so Пѓ is a p-cycle. в™Ј

6.6.7 Proposition
If f is irreducible over Q and of prime degree p, and f has exactly two nonreal roots in
the complex п¬Ѓeld C, then the Galois group G of f is Sp .

Proof. By (6.6.6), G contains a p-cycle Пѓ. Now one of the elements of G must be complex
conjugation П„ , which is an automorphism of C that п¬Ѓxes R (hence Q). Thus П„ permutes
the two nonreal roots and leaves the p в€’ 2 real roots п¬Ѓxed, so П„ is a transposition. Since
p is prime, Пѓ k is a p-cycle for k = 1, . . . , p в€’ 1. It follows that by renumbering symbols if
necessary, we can assume that (1, 2) and (1, 2, . . . , p) belong to G. By (6.6.5) part (iii),
G = Sp . в™Ј

Problems For Section 6.6
In Problems 1вЂ“4, all polynomials are over the rational п¬Ѓeld Q, and in each case, you are
asked to п¬Ѓnd the Galois group G.

1. f (X) = X 3 в€’ 2 (do it two ways)
2. f (X) = X 3 в€’ 3X + 1
3. f (X) = X 5 в€’ 10X 4 + 2
4. f (X) = X 3 + 3X 2 в€’ 2X + 1 (calculate the discriminant in two ways)
5. If f is a separable cubic, not necessarily irreducible, then there are other possibilities
for the Galois group G of f besides S3 and A3 . What are they?
20 CHAPTER 6. GALOIS THEORY

6. Let f be an irreducible cubic over Q with exactly one real root. Show that D(f ) < 0,
and conclude that the Galois group of f is S3 .
7. Let f be an irreducible cubic over Q with 3 distinct real roots. Show that D(f ) > 0,
в€љ в€љ
so that the Galois group is A3 or S3 according as D в€€ Q or D в€€ Q /

6.7 Cyclic and Kummer Extensions
The problem of solving a polynomial equation by radicals is thousands of years old, but
it can be given a modern п¬‚avor. We are looking for roots of f в€€ F [X], and we are only
allowed to use algorithms that do ordinary arithmetic plus the extraction of nth roots.
The idea is to identify those polynomials whose roots can be found in this way. Now if
в€љ
a в€€ F and our algorithm computes Оё = n a in some extension п¬Ѓeld of F , then Оё is a root
of X n в€’ a, so it is natural to study splitting п¬Ѓelds of X n в€’ a.

6.7.1 Assumptions, Comments and a Deп¬Ѓnition
Assume
(i) E is a splitting п¬Ѓeld for f (X) = X n в€’ a over F , where a = 0.
(ii) F contains a primitive nth root of unity П‰.
These are natural assumption if we want to allow the computation of nth roots. If Оё is
any root of f in E, then the roots of f are Оё, П‰Оё, . . . , П‰ nв€’1 Оё. (The roots must be distinct
because a, hence Оё, is nonzero.) Therefore E = F (Оё). Since f is separable, the extension
E/F is Galois (see (6.3.1)). If G = Gal(E/F ), then |G| = [E : F ] by the fundamental
theorem (or by (3.5.9)).
In general, a cyclic extension is a Galois extension whose Galois group is cyclic.

6.7.2 Theorem
Under the assumptions of (6.7.1), E/F is a cyclic extension and the order of the Galois
group G is a divisor of n. We have |G| = n if and only if f (X) is irreducible over F .
Proof. Let Пѓ в€€ G; since Пѓ permutes the roots of f by (3.5.1), we have Пѓ(Оё) = П‰ u(Пѓ) Оё.
[Note that Пѓ п¬Ѓxes П‰ by (ii).] We identify integers u(Пѓ) with the same residue mod n. If
Пѓi (Оё) = П‰ u(Пѓi ) Оё, i = 1, 2, then

Пѓ1 (Пѓ2 (Оё)) = П‰ u(Пѓ1 )+u(Пѓ2 ) Оё,

so

u(Пѓ1 Пѓ2 ) = u(Пѓ1 ) + u(Пѓ2 )

and u is a group homomorphism from G to Zn . If u(Пѓ) is 0 mod n, then Пѓ(Оё) = Оё, so Пѓ is
the identity and the homomorphism is injective. Thus G is isomorphic to a subgroup of
Zn , so G is cyclic and |G| divides n.
If f is irreducible over F , then |G| = [E : F ] = [F (Оё) : F ] = deg f = n. If f is not
irreducible over F , let g be a proper irreducible factor. If ОІ is a root of g in E, then ОІ is
also a root of f , so E = F (ОІ) and |G| = [E : F ] = [F (ОІ) : F ] = deg g < n. в™Ј
6.7. CYCLIC AND KUMMER EXTENSIONS 21

Thus splitting п¬Ѓelds of X n в€’ a give rise to cyclic extensions. Conversely, we can prove
that a cyclic extension comes from such a splitting п¬Ѓeld.

6.7.3 Theorem
Let E/F be a cyclic extension of degree n, where F contains a primitive nth root of
unity П‰. Then for some nonzero a в€€ F , f (X) = X n в€’ a is irreducible over F and E is a
splitting п¬Ѓeld for f over F .
Proof. Let Пѓ be a generator of the Galois group of the extension. By DedekindвЂ™s lemma
(6.1.6), the distinct automorphisms 1, Пѓ, Пѓ 2 , . . . , Пѓ nв€’1 are linearly independent over E.
Thus 1 + П‰Пѓ + П‰ 2 Пѓ 2 + В· В· В· + П‰ nв€’1 Пѓ nв€’1 is not identically 0, so for some ОІ в€€ E we have

Оё = ОІ + П‰Пѓ(ОІ) + В· В· В· + П‰ nв€’1 Пѓ nв€’1 (ОІ) = 0.

Now

Пѓ(Оё) = Пѓ(ОІ) + П‰Пѓ 2 (ОІ) + В· В· В· + П‰ nв€’2 Пѓ nв€’1 (ОІ) + П‰ nв€’1 Пѓ n (ОІ) = П‰ в€’1 Оё

since Пѓ n (ОІ) = ОІ. We take a = Оёn . To prove that a в€€ F , note that

Пѓ(Оёn ) = (Пѓ(Оё))n = (П‰ в€’1 Оё)n = Оёn

and therefore Пѓ п¬Ѓxes Оёn . Since Пѓ generates G, all other members of G п¬Ѓx Оёn , hence a
belongs to the п¬Ѓxed п¬Ѓeld of Gal(E/F ), which is F .
Now by deп¬Ѓnition of a, Оё is a root of f (X) = X n в€’ a, so the roots of X n в€’ a
are Оё, П‰Оё, . . . , П‰ nв€’1 Оё. Therefore F (Оё) is a splitting п¬Ѓeld for f over F . Since Пѓ(Оё) = П‰ в€’1 Оё,
the distinct automorphisms 1, Пѓ, . . . , Пѓ nв€’1 can be restricted to distinct automorhisms
of F (Оё). Consequently,

n в‰¤ | Gal(F (Оё)/F )| = [F (Оё) : F ] в‰¤ deg f = n

so [F (Оё) : F ] = n. It follows that E = F (Оё) and (since f must be the minimal polynomial
of Оё over F ) f is irreducible over F . в™Ј
A п¬Ѓnite abelian group is a direct product of cyclic groups (or direct sum, in additive
notation; see (4.6.4)). It is reasonable to expect that our analysis of cyclic Galois groups
will help us to understand abelian Galois groups.

6.7.4 Deп¬Ѓnition
A Kummer extension is a п¬Ѓnite Galois extension with an abelian Galois group.

6.7.5 Theorem
Let E/F be a п¬Ѓnite extension, and assume that F contains a primitive nth root of unity П‰.
Then E/F is a Kummer extension whose Galois group G has an exponent dividing n if
and only if there are nonzero elements a1 , . . . , ar в€€ в€љ such that E is a splitting п¬Ѓeld of
F в€љ
(X в€’ a1 ) В· В· В· (X в€’ ar ) over F . [For short, E = F ( a1 , . . . , n ar ).]
n n n
22 CHAPTER 6. GALOIS THEORY

Proof. We do the вЂњifвЂќ part п¬Ѓrst. As in (6.7.1), we have E = F (Оё1 , . . . , Оёr ) where Оёi is a
root of X n в€’ ai . If Пѓ в€€ Gal(E/F ), then Пѓ maps Оёi to another root of X n в€’ ai , so

Пѓ(Оёi ) = П‰ ui (Пѓ) Оёi .

Thus if Пѓ and П„ are any two automorphisms in the Galois group G, then ПѓП„ = П„ Пѓ and G
is abelian. [The ui are integers, so ui (Пѓ) + ui (П„ ) = ui (П„ ) + ui (Пѓ).] Now restrict attention
to the extension F (Оёi ). By (6.7.2), the Galois group of F (Оёi )/F has order dividing n, so
Пѓ n (Оёi ) = Оёi for all i = 1, . . . , r. Thus Пѓ n is the identity, and the exponent of G is a divisor
of n.
For the вЂњonly ifвЂќ part, observe that since G is a п¬Ѓnite abelian group, it is a direct
product of cyclic groups C1 , . . . , Cr . For each i = 1, . . . , r, let Hi be the product of the
Cj for j = i; by (1.5.3), Hi G. We have G/Hi в€ј Ci by the п¬Ѓrst isomorphism theorem.
=
(Consider the projection mapping x1 В· В· В· xr в†’ xi в€€ Ci .) Let Ki be the п¬Ѓxed п¬Ѓeld of Hi . By
the fundamental theorem, Ki /F is a Galois extension and its Galois group is isomorphic
to G/Hi , hence isomorphic to Ci . Thus Ki /F is a cyclic extension of degree di = |Ci |,
and di is a divisor of n. (Since G is the direct product of the Ci , some element of G has
order di , so di divides the exponent of G and therefore divides n.) We want to apply
(6.7.3) with n replaced by di , and this is possible because F contains a primitive dth root i
d
of unity, namely П‰ n/di . We conclude that Ki = F (Оёi ), where Оёi i is a nonzero element
d (n/di ) n/d
bi в€€ F . But Оёi = Оёi i = bi i = ai в€€ F .
n

Finally, in the Galois correspondence, the intersection of the Hi is paired with the
r
composite of the Ki , which is F (Оё1 , . . . , Оёr ); see Section 6.3, Problem 7. But i=1 Hi = 1,
so E = F (Оё1 , . . . , Оёr ), and the result follows. в™Ј

Problems For Section 6.7
в€љв€љв€љв€љ
1. Find the Galois group of the extension Q( 2, 3, 5, 7) [the splitting п¬Ѓeld of (X 2 в€’
2)(X 2 в€’ 3)(X 2 в€’ 5)(X 2 в€’ 7)] over Q.
2. Suppose that E is a splitting п¬Ѓeld for f (X) = X n в€’ a over F , a = 0, but we drop
the second assumption in (6.7.1) that F contains a primitive nth root of unity. Is it
possible for the Galois group of E/F to be cyclic?
3. Let E be a splitting п¬Ѓeld for X n в€’ a over F , where a = 0, and assume that the
characteristic of F does not divide n. Show that E contains a primitive nth root of
unity.

We now assume that E is a splitting п¬Ѓeld for f (X) = X p в€’ c over F , where c = 0, p is
prime and the characteristic of F is not p. Let П‰ be a primitive pth root of unity in E (see
Problem 3). Assume that f is not irreducible over F , and let g be an irreducible factor
of f of degree d, where 1 в‰¤ d < p. Let Оё be a root of g in E.
4. Let g0 be the product of the roots of g. (Since g0 is В± the constant term of g, g0 в€€ F .)
p
Show that g0 = Оёdp = cd .
5. Since d and p are relatively prime, there are integers a and b such that ad + bp = 1.
Use this to show that if X p в€’ c is not irreducible over F , then it must have a root
in F .
6.8. SOLVABILITY BY RADICALS 23

6. Continuing Problem 5, show that if X p в€’ c is not irreducible over F , then E = F (П‰).
7. Continuing Problem 6, show that if X p в€’ c is not irreducible over F , then X p в€’ c
splits over F if and only if F contains a primitive pth root of unity.

Let E/F be a cyclic Galois extension of prime degree p, where p is the characteristic of F .
Let Пѓ be a generator of G = Gal(E/F ). It is a consequence of HilbertвЂ™s Theorem 90 (see
the Problems for Section 7.3) that there is an element Оё в€€ E such that Пѓ(Оё) = Оё + 1.
Prove the Artin-Schreier theorem:

8. E = F (Оё).
9. Оё is a root of f (X) = X p в€’ X в€’ a for some a в€€ F .
10. f is irreducible over F (hence a = 0).

Conversely, Let F be a п¬Ѓeld of prime characteristic p, and let E be a splitting п¬Ѓeld for
f (X) = X p в€’ X в€’ a, where a is a nonzero element of F .

11. If Оё is any root of f in E, show that E = F (Оё) and that f is separable.
12. Show that every irreducible factor of f has the same degree d, where d = 1 or p. Thus
if d = 1, then E = F , and if d = p, then f is irreducible over F .
13. If f is irreducible over F , show that the Galois group of f is cyclic of order p.

6.8 Solvability By Radicals
6.8.1 Deп¬Ѓnitions and Comments
We wish to solve the polynomial equation f (X) = 0, f в€€ F [X], under the restriction that
we are only allowed to perform ordinary arithmetic operations (addition, subtraction,
multiplication and division) on the coeп¬ѓcients, along with extraction of nth roots (for
any n = 2, 3, . . . ). A sequence of operations of this type gives rise to a sequence of
extensions

F в‰¤ F (О±1 ) в‰¤ F (О±1 , О±2 ) в‰¤ В· В· В· в‰¤ F (О±1 , . . . , О±r ) = E

where О±1 1 в€€ F and О±i i в€€ F (О±1 , . . . , О±iв€’1 ), i = 2, . . . , r. Equivalently, we have
n n

F = F0 в‰¤ F1 в‰¤ В· В· В· в‰¤ Fr = E

where Fi = Fiв€’1 (О±i ) and О±i i в€€ Fiв€’1 , i = 1, . . . , r. We say that E is a radical extension
n

of F . It is convenient (and legal) to assume that n1 = В· В· В· = nr = n. (Replace each ni
by the product of all the ni . To justify this, observe that if О±j belongs to a п¬Ѓeld L, then
О±mj в€€ L, m = 2, 3, . . . .) Unless otherwise speciп¬Ѓed, we will make this assumption in all
hypotheses, conclusions and proofs.
We have already seen three explicit classes of radical extensions: cyclotomic, cyclic
and Kummer. (In the latter two cases, we assume that the base п¬Ѓeld contains a primitive
nth root of unity.)
24 CHAPTER 6. GALOIS THEORY

We say that the polynomial f в€€ F [X] is solvable by radicals if the roots of f lie in
some radical extension of F , in other words, there is a radical extension E of F such
that f splits over E.
Since radical extensions are formed by successively adjoining nth roots, it follows that
the transitivity property holds: If E is a radical extension of F and L is a radical extension
of E, then L is a radical extension of F .
A radical extension is always п¬Ѓnite, but it need not be normal or separable. We
will soon specialize to characteristic 0, which will force separability, and we can achieve
normality by taking the normal closure (see (3.5.11)).

6.8.2 Proposition
Let E/F be a radical extension, and let N be the normal closure of E over F . Then N/F
is also a radical extension.
Proof. E is obtained from F by successively adjoining О±1 , . . . , О±r , where О±i is the nth
root of an element in Fiв€’1 . On the other hand, N is obtained from F by adjoining
not only the О±i , but their conjugates О±i1 , . . . , О±im(i) . For any п¬Ѓxed i and j, there is an
automorphism Пѓ в€€ Gal(N/F ) such that Пѓ(О±i ) = О±ij (see (3.2.3), (3.5.5) and (3.5.6)).
Thus
О±ij = Пѓ(О±i )n = Пѓ(О±i )
n n

n n
and since О±i belongs to F (О±1 , . . . , О±iв€’1 ), it follows from (3.5.1) that Пѓ(О±i ) belongs to
iв€’1
the splitting п¬Ѓeld Ki of j=1 min(О±j , F ) over F . [Take K1 = F , and note that since
О±1 = b1 в€€ F , we have Пѓ(О±1 ) = Пѓ(b1 ) = b1 в€€ F. Alternatively, observe that by (3.5.1), Пѓ
n n

must take a root of X n в€’ b1 to another root of this polynomial.] Thus we can display N
as a radical extension of F by successively adjoining
О±11 , . . . , О±1m(1) , . . . , О±r1 , . . . , О±rm(r) . в™Ј

6.8.3 Preparation for the Main Theorem
If F has characteristic 0, then a primitive nth root of unity П‰ can be adjoined to F to
reach an extension F (П‰); see (6.5.1). If E is a radical extension of F and F = F0 в‰¤
F1 в‰¤ В· В· В· в‰¤ Fr = E, we can replace Fi by Fi (П‰), i = 1, . . . , r, and E(П‰) will be a radical
extension of F . By (6.8.2), we can pass from E(П‰) to its normal closure over F . Here is
the statement we are driving at:
Let f в€€ F [X], where F has characteristic 0. If f is solvable by radicals, then there is
a Galois radical extension N = Fr в‰Ґ В· В· В· в‰Ґ F1 в‰Ґ F0 = F containing a splitting п¬Ѓeld K for
f over F , such that each intermediate п¬Ѓeld Fi , i = 1, . . . , r, contains a primitive nth root
of unity П‰. We can assume that F1 = F (П‰) and for i > 1, Fi is a splitting п¬Ѓeld for X n в€’ bi
over Fiв€’1 . [(Look at the end of the proof of (6.8.2).] By (6.5.1), F1 /F is a cyclotomic
(Galois) extension, and by (6.7.2), each Fi /Fiв€’1 , i = 2, . . . , r is a cyclic (Galois) extension.
We now do some further preparation. Suppose that K is a splitting п¬Ѓeld for f over F ,
and that the Galois group of K/F is solvable, with
В·В·В·
Gal(K/F ) = H0 H1 Hr = 1
6.8. SOLVABILITY BY RADICALS 25

with each Hiв€’1 /Hi abelian. By the fundamental theorem, we have the corresponding
sequence of п¬Ѓxed п¬Ѓelds

F = K0 в‰¤ K1 в‰¤ В· В· В· в‰¤ Kr = K

with Ki /Kiв€’1 Galois and Gal(Ki /Kiв€’1 ) isomorphic to Hiв€’1 /Hi . Let us adjoin a primitive
nth root of unity П‰ to each Ki , so that we have п¬Ѓelds Fi = Ki (П‰) with

F в‰¤ F0 в‰¤ F 1 в‰¤ В· В· В· в‰¤ F r .

We take n = | Gal(K/F )|. Since Fi can be obtained from Fiв€’1 by adjoining everything
in Ki \ Kiв€’1 , we have

Fi = Fiв€’1 Ki = Ki Fiв€’1

the composite of Fiв€’1 and Ki , i = 1, . . . , r. We may now apply Theorem 6.2.2. In the
diamond diagram of Figure 6.2.1, at the top of the diamond we have Fi , on the left Ki ,
on the right Fiв€’1 , and on the bottom Ki в€© Fiв€’1 вЉ‡ Kiв€’1 (see Figure 6.8.1). We conclude
that Fi /Fiв€’1 is Galois, with a Galois group isomorphic to a subgroup of Gal(Ki /Kiв€’1 ).
Since Gal(Ki /Kiв€’1 ) в€ј Hiв€’1 /Hi , it follows that Gal(Fi /Fiв€’1 ) is abelian. Moreover, the
=
exponent of this Galois group divides the order of H0 , which coincides with the size of
Gal(K/F ). (This explains our choice of n.)

u i uuuu
F
uu uu
uu uu
uu uu
uu
u
Fiв€’1
Ki s
ss ss
ss ss
ss ss
ss s
s ss

Kiв€’1

Figure 6.8.1

6.8.4 GaloisвЂ™ Solvability Theorem
Let K be a splitting п¬Ѓeld for f over F , where F has characteristic 0. Then f is solvable
by radicals if and only if the Galois group of K/F is solvable.

Proof. If f is solvable by radicals, then as in (6.8.3), we have

F = F 0 в‰¤ F 1 в‰¤ В· В· В· в‰¤ Fr = N
26 CHAPTER 6. GALOIS THEORY

where N/F is Galois, N contains a splitting п¬Ѓeld K for f over F , and each Fi /Fiв€’1 is
Galois with an abelian Galois group. By the fundamental theorem, the corresponding
sequence of subgroups is

В·В·В·
1 = Hr Hrв€’1 H0 = G = Gal(N/F )

with each Hiв€’1 /Hi abelian. Thus G is solvable, and since

Gal(K/F ) в€ј Gal(N/F )Gal(N/K)
=

[map Gal(N/F ) в†’ Gal(K/F ) by restriction; the kernel is Gal(N/K)], Gal(K/F ) is solv-
able by (5.7.4).
Conversely, assume that Gal(K/F ) is solvable. Again as in (6.8.3), we have

F в‰¤ F 0 в‰¤ F1 в‰¤ В· В· В· в‰¤ Fr

where K в‰¤ Fr , each Fi contains a primitive nth root of unity, with n = | Gal(K/F )|,
and Gal(Fi /Fiв€’1 ) is abelian with exponent dividing n for all i = 1, . . . , r. Thus each
Fi /Fiв€’1 is a Kummer extension whose Galois group has an exponent dividing n. By
(6.7.5) (or (6.5.1) for the case i = 1), each Fi /Fiв€’1 is a radical extension. By transitivity
(see (6.8.1)), Fr is a radical extension of F . Since K вЉ† Fr , f is solvable by radicals. в™Ј

6.8.5 Example
Let f (X) = X 5 в€’ 10X 4 + 2 over the rationals. The Galois group of f is S5 , which is not
solvable. (See Section 6.6, Problem 3 and Section 5.7, Problem 5.) Thus f is not solvable
There is a fundamental idea that needs to be emphasized. The signiп¬Ѓcance of GaloisвЂ™
solvability theorem is not simply that there are some examples of bad polynomials. The
key point is there is no general method for solving a polynomial equation over the rationals
by radicals, if the degree of the polynomial is 5 or more. If there were such a method,
then in particular it would work on Example (6.8.5), a contradiction.

Problems For Section 6.8
In the exercises, we will sketch another classical problem, that of constructions with ruler
and compass. In Euclidean geometry, we start with two points (0, 0) and (1, 0), and we
are allowed the following constructions.
(i) Given two points P and Q, we can draw a line joining them;
(ii) Given a point P and a line L, we can draw a line through P parallel to L;
(iii) Given a point P and a line L, we can draw a line through P perpendicular to L;
(iv) Given two points P and Q, we can draw a circle with center at P passing through Q;
(v) Let A, and similarly B, be a line or a circle. We can generate new points, called
constructible points, by forming the intersection of A and B. If (c, 0) (equivalently
(0, c)) is a constructible point, we call c a constructible number. It follows from (ii)
and (iii) that (a, b) is a constructible point iп¬Ђ a and b are constructible numbers. It
6.8. SOLVABILITY BY RADICALS 27

can be shown that every rational number is constructible, and that the constructible
numbers form a п¬Ѓeld. Now in (v), the intersection of A and B can be found by
ordinary arithmetic plus at worst the extraction of a square root. Conversely, the
square roof of any nonnegative constructible number can be constructed. Therefore
c is constructible iп¬Ђ there are real п¬Ѓelds Q = F0 в‰¤ F1 В· В· В· в‰¤ Fr such that c в€€ Fr and
each [Fi : Fiв€’1 ] is 1 or 2. Thus if c is constructible, then c is algebraic over Q and
[Q(c) : Q] is a power of 2.
1. (Trisecting the angle) If it is possible to trisect any angle with ruler and compass, then
in particular a 60 degree angle can be trisected, so that О± = cos 20в—¦ is constructible.
Using the identity
ei3Оё = cos 3Оё + i sin 3Оё = (cos Оё + i sin Оё)3 ,
2. (Duplicating the cube) Show that it is impossible to construct, with ruler and compass,
в€љ
a cube whose volume is exactly 2. (The side of such a cube would be 3 2.)
3. (Squaring the circle) Show that if it were possible to construct a square with area ПЂ,
then ПЂ would be algebraic over Q. (It is known that ПЂ is transcendental over Q.)
To construct a regular n-gon, that is, a regular polygon with n sides, n в‰Ґ 3,we must
be able to construct an angle of 2ПЂ/n; equivalently, cos 2ПЂ/n must be a constructible
number. Let П‰ = ei2ПЂ/n , a primitive nth root of unity.
Show that [Q(П‰) : Q(cos 2ПЂ/n)] = 2.
4.
5. Show that if a regular n-gon is constructible, then the Euler phi function П•(n) is a
power of 2.
Conversely, assume that П•(n) is a power of 2.
6. Show that Gal(Q(cos 2ПЂ/n)/Q) is a 2-group, that is, a p-group with p = 2.
7. By Section 5.7, Problem 7, every nontrivial п¬Ѓnite p-group has a subnormal series in
which every factor has order p. Use this (with p = 2) to show that a regular n-gon is
constructible.
8. ВїFrom the preceding, a regular n-gon is constructible if and only if П•(n) is a power
of 2. Show that an equivalent condition is that n = 2s q1 В· В· В· qt , s, t = 0, 1, . . . , where
the qi are distinct Fermat primes, that is, primes of the form 2m + 1 for some positive
integer m.
Show that if 2m + 1 is prime, then m must be a power of 2. The only known Fermat
9.
primes have m = 2a , where a = 0, 1, 2, 3, 4 (232 + 1 is divisible by 641). [The key
point is that if a is odd, then X + 1 divides X a + 1 in Z[X]; the quotient is X aв€’1 в€’
X aв€’2 + В· В· В· в€’ X + 1 (since a в€’ 1 is even).]
Let F be the п¬Ѓeld of rational functions in n variables e1 , . . . , en over a п¬Ѓeld K with
characteristic 0, and let f (X) = X n в€’ e1 X nв€’1 + e2 X nв€’2 в€’ В· В· В· + (в€’1)n en в€€ F [X]. If
О±1 , . . . , О±n are the roots of f in a splitting п¬Ѓeld over F , then the ei are the elementary
symmetric functions of the О±i . Let E = F (О±1 , . . . , О±n ), so that E/F is a Galois
extension and G = Gal(E/F ) is the Galois group of f .
Show that G в€ј Sn .
10. =
11. What can you conclude from Problem 10 about solvability of equations?
28 CHAPTER 6. GALOIS THEORY

6.9 Transcendental Extensions
6.9.1 Deп¬Ѓnitions and Comments
An extension E/F such that at least one О± в€€ E is not algebraic over F is said to be
transcendental. An idea analogous to that of a basis of an arbitrary vector space V turns
out to be proп¬Ѓtable in studying transcendental extensions. A basis for V is a subset of V
that is linearly independent and spans V . A key result, whose proof involves the Steinitz
exchange, is that if {x1 , . . . , xm } spans V and S is a linearly independent subset of V ,
then |S| в‰¤ m. We are going to replace linear independence by algebraic independence
and spanning by algebraic spanning. We will п¬Ѓnd that every transcendental extension has
a transcendence basis, and that any two transcendence bases for a given extension have
the same cardinality. All these terms will be deп¬Ѓned shortly. The presentation in the
text will be quite informal; I believe that this style best highlights the strong connection
between linear and algebraic independence. An indication of how to formalize the devel-
opment is given in a sequence of exercises. See also Morandi, вЂњFields and Galois TheoryвЂќ,
pp. 173вЂ“182.
Let E/F be an extension. The elements t1 , . . . , tn в€€ E are algebraically dependent
over F (or the set {t1 , . . . , tn } is algebraically dependent over F ) if there is a nonzero
polynomial f в€€ F [X1 , . . . , Xn ] such that f (t1 , . . . , tn ) = 0; otherwise the ti are alge-
braically independent over F . Algebraic independence of an inп¬Ѓnite set means algebraic
independence of every п¬Ѓnite subset.
Now if a set T spans a vector space V , then each x in V is a linear combination
of elements of T , so that x depends on T in a linear fashion. Replacing вЂњlinearвЂќ by
вЂњalgebraicвЂќ, we say that the element t в€€ E depends algebraically on T over F if t is
algebraic over F (T ), the п¬Ѓeld generated by T over F (see Section 3.1, Problem 1). We
say that T spans E algebraically over F if each t in E depends algebraically on T over F ,
that is, E is an algebraic extension of F (T ). A transcendence basis for E/F is a subset
of E that is algebraically independent over F and spans E algebraically over F . (From
now on, we will frequently regard F as п¬Ѓxed and drop the phrase вЂњover F вЂќ.)

6.9.2 Lemma
If S is a subset of E, the following conditions are equivalent.

(i) S is a transcendence basis for E/F ;
(ii) S is a maximal algebraically independent set;
(iii) S is a minimal algebraically spanning set.

Thus by (ii), S is a transcendence basis for E/F iп¬Ђ S is algebraically independent and E
is algebraic over F (S).

Proof. (i) implies (ii): If S вЉ‚ T where T is algebraically independent, let u в€€ T \ S.
Then u cannot depend on S algebraically (by algebraic independence of T ), so S cannot
span E algebraically.
6.9. TRANSCENDENTAL EXTENSIONS 29

(ii) implies (i): If S does not span E algebraically, then there exists u в€€ E such
that u does not depend algebraically on S. But then S в€Є {u} is algebraically independent,
contradicting maximality of S.
(i) implies (iii): If T вЉ‚ S and T spans E algebraically, let u в€€ S \ T . Then u depends
algebraically on T , so T в€Є {u}, hence S, is algebraically dependent, a contradiction.
(iii) implies (i): If S is algebraically dependent, then some u в€€ S depends algebraically
on T = S \ {u}. But then T spans E algebraically, a contradiction. в™Ј

6.9.3 Proposition
Every transcendental extension has a transcendence basis.

Proof. The standard argument via ZornвЂ™s lemma that an arbitrary vector space has a
maximal linearly independent set (hence a basis) shows that an arbitrary transcendental
extension has a maximal algebraically independent set, which is a transcendence basis
by (6.9.2). в™Ј

For completeness, if E/F is an algebraic extension, we can regard в€… as a transcendence
basis.

6.9.4 The Steinitz Exchange
If {x1 , . . . , xm } spans E algebraically and S вЉ† E is algebraically independent, then
|S| в‰¤ m.

Proof. Suppose that S has at least m + 1 elements y1 , . . . , ym+1 . Since the xi span E
algebraically, y1 depends algebraically on x1 , . . . , xm . The algebraic dependence relation
must involve at least one xi , say x1 . (Otherwise, S would be algebraically dependent.)
Then x1 depends algebraically on y1 , x2 , . . . , xm , so {y1 , x2 , . . . , xm } spans E algebraically.
We claim that for every i = 1, . . . , m, {y1 , . . . , yi , xi+1 , . . . , xm } spans E algebraically. We
have just proved the case i = 1. If the result holds for i, then yi+1 depends algebraically on
{y1 , . . . , yi , xi+1 , . . . , xm }, and the dependence relation must involve at least one xj , say
xi+1 for convenience. (Otherwise, S would be algebraically dependent.) Then xi+1 de-
pends algebraically on y1 , . . . , yi+1 , xi+2 , . . . , xm , so {y1 , . . . , yi+1 , xi+2 , . . . , xm } spans E
algebraically, completing the induction.
Since there are more yвЂ™s than xвЂ™s, eventually the xвЂ™s disappear, and y1 , . . . , ym span E
algebraically. But then ym+1 depends algebraically on y1 , . . . , ym , contradicting the alge-
braic independence of S. в™Ј

6.9.5 Corollary
Let S and T be transcendence bases of E. Then either S and T are both п¬Ѓnite or they
are both inп¬Ѓnite; in the former case, |S| = |T |.

Proof. Assume that one of the transcendence bases, say T , is п¬Ѓnite. By (6.9.4), |S| в‰¤ |T |,
so S is п¬Ѓnite also. By a symmetrical argument, |T | в‰¤ |S|, so |S| = |T |. в™Ј
30 CHAPTER 6. GALOIS THEORY

6.9.6 Proposition
If S and T are arbitrary transcendence bases for E, then |S| = |T |. [The common value
is called the transcendence degree of E/F .]
Proof. By (6.9.5), we may assume that S and T are both inп¬Ѓnite. Let T = {yi : i в€€ I}.
If x в€€ S, then x depends algebraically on п¬Ѓnitely many elements yi1 , . . . , yir in T . Deп¬Ѓne
I(x) to be the set of indices {i1 , . . . , ir }. It follows that I = в€Є{I(x) : x в€€ S}. For if j
belongs to none of the I(x), then we can remove yj from T and the resulting set will still
span E algebraically, contradicting (6.9.2) part (iii). Now an element of в€Є{I(x) : x в€€ S}
is determined by selecting an element x в€€ S and then choosing an index in I(x). Since
I(x) is п¬Ѓnite, we have |I(x)| в‰¤ в„µ0 . Thus

|I| = | {I(x) : x в€€ S}| в‰¤ |S|в„µ0 = |S|

since S is inп¬Ѓnite. Thus |T | в‰¤ |S|. By symmetry, |S| = |T |. в™Ј

6.9.7 Example
Let E = F (X1 , . . . , Xn ) be the п¬Ѓeld of rational functions in the variables X1 , . . . , Xn
with coeп¬ѓcients in F . If f (X1 , . . . , Xn ) = 0, then f is the zero polynomial, so S =
{X1 , . . . , Xn } is an algebraically independent set. Since E = F (S), E is algebraic over
F (S) and therefore S spans E algebraically. Thus S is a transcendence basis.
Now let T = {X1 1 , . . . , Xn n }, where u1 , . . . , un are arbitrary positive integers. We
u u

claim that T is also a transcendence basis. As above, T is algebraically independent.
Moreover, each Xi is algebraic over F (T ). To see what is going on, look at a concrete
example, say T = {X1 , X2 , X3 }. If f (Z) = Z 3 в€’ X2 в€€ F (T )[Z], then X2 is a root of f , so
5 3 4 3

X2 , and similarly each Xi , is algebraic over F (T ). By (3.3.3), E is algebraic over F (T ),
so T is a transcendence basis.

Problems For Section 6.9
1. If S is an algebraically independent subset of E over F , T spans E algebraically over F ,
and S вЉ† T , show that there is a transcendence basis B such that S вЉ† B вЉ† T .
2. Show that every algebraically independent set can be extended to a transcendence
basis, and that every algebraically spanning set contains a transcendence basis.
3. Prove carefully, for an extension E/F and a subset T = {t1 , . . . , tn } вЉ† E, that the
following conditions are equivalent.
(i) T is algebraically independent over F ;
(ii) For every i = 1, . . . , n, ti is transcendental over F (T \ {ti });
(iii) For every i = 1, . . . , n, ti is transcendental over F (t1 , . . . , tiв€’1 ) (where the state-
ment for i = 1 is that t1 is transcendental over F ).
4. Let S be a subset of E that is algebraically independent over F . Show that if t в€€ E \ S,
then t is transcendental over F (S) if and only if S в€Є {t} is algebraically independent
over F .
6.9. TRANSCENDENTAL EXTENSIONS 31

[Problems 3 and 4 suggest the reasoning that is involved in formalizing the results of this
section.]

5. Let F в‰¤ K в‰¤ E, with S a subset of K that is algebraically independent over F , and T
a subset of E that is algebraically independent over K. Show that S в€ЄT is algebraically
independent over F , and S в€© T = в€….
6. Let F в‰¤ K в‰¤ E, with S a transcendence basis for K/F and T a transcendence basis
for E/K. Show that S в€ЄT is a transcendence basis for E/F . Thus if tr deg abbreviates
transcendence degree, then by Problem 5,

tr deg(E/F ) = tr deg(K/F ) + tr deg(E/K).

7. Let E be an extension of F , and T = {t1 , . . . , tn } a п¬Ѓnite subset of E. Show that
F (T ) is F -isomorphic to the rational function п¬Ѓeld F (X1 , . . . , Xn ) if and only if T is
algebraically independent over F .
8. An algebraic function п¬Ѓeld F in one variable over K is a п¬Ѓeld F/K such that there
exists x в€€ F transcendental over K with [F : K(x)] < в€ћ. If z в€€ F , show that z is
transcendental over K iп¬Ђ [F : K(z)] < в€ћ.
9. Find the transcendence degree of the complex п¬Ѓeld over the rationals.

Appendix To Chapter 6
We will develop a method for calculating the discriminant of a polynomial and apply the
result to a cubic. We then calculate the Galois group of an arbitrary quartic.

A6.1 Deп¬Ѓnition
If x1 , . . . , xn (n в‰Ґ 2) are arbitrary elements of a п¬Ѓeld, the Vandermonde determinant of
the xi is

В·В·В·
1 1 1
В·В·В·
x1 x2 xn
det V = .
.
.
В·В·В·
xnв€’1 xnв€’1 xnв€’1
n
1 2

A6.2 Proposition

(xj в€’ xi ).
det V =
i<j

Proof. det V is a polynomial h of degree 1 + 2 + В· В· В· + (n в€’ 1) = ( n ) in the variables
2
x1 , . . . , xn , as is g = i<j (xj в€’ xi ). If xi = xj for i < j, then the determinant is 0, so
by the remainder theorem (2.5.2), each factor of g, hence g itself, divides h. Since h and
g have the same degree, h = cg for some constant c. Now look at the leading terms of h
32 CHAPTER 6. GALOIS THEORY

and g, i.e., those terms in which xn appears to as high a power as possible, and subject
to this constraint, xnв€’1 appears to as high a power as possible, etc. In both cases, the
leading term is x2 x2 В· В· В· xnв€’1 , and therefore c must be 1. (For this step it is proп¬Ѓtable to
3 n
regard the xi as abstract variables in a polynomial ring. Then monomials xr1 В· В· В· xrn with
n
1
diп¬Ђerent sequences (r1 , . . . , rn ) of exponents are linearly independent.) в™Ј

A6.3 Corollary
If f is a polynomial in F [X] with roots x1 , . . . , xn in some splitting п¬Ѓeld over F , then the
discriminant of f is (det V )2 .

Proof. By deп¬Ѓnition of the discriminant D of f (see 6.6.1), we have D = в€†2 where
в€† = В± det V . в™Ј

A6.4 Computation of the Discriminant
The square of the determinant of V is det(V V t ), which is the determinant of
пЈ® пЈ№пЈ® пЈ№
В·В·В· В·В·В· xnв€’1
1 1 1 1 x1 1
пЈЇ x1 пЈє пЈЇ1 xnв€’1 пЈє
В·В·В· В·В·В·
x2 xn пЈє пЈЇ x2
пЈЇ пЈє
2
пЈЇ пЈєпЈЇ пЈє
. .
пЈ° пЈ»пЈ° пЈ»
. .
. .
В·В·В·
xnв€’1 xnв€’1 xnв€’1 xnв€’1
1 xn ...
n
1 2 n

and this in turn is

В·В·В·
t0 t1 tnв€’1
В·В·В·
t1 t2 tn
.
.
.
В·В·В·
tnв€’1 tn t2nв€’2

where the power sums tr are given by
n
xr , r в‰Ґ 1.
t0 = n, tr = i
i=1

We must express the power sums in terms of the coeп¬ѓcients of the polynomial f . This
will involve, improbably, an exercise in diп¬Ђerential calculus. We have
n n
(1 в€’ xi z) = ci z i with c0 = 1;
F (z) =
i=1 i=0

the variable z ranges over real numbers. Take the logarithmic derivative of F to obtain
в€ћ в€ћ
n n
в€’xi
F (z) d
xj+1 z j = в€’
=в€’ tj+1 z j .
= log F (z) =
1 в€’ xi z i
F (z) dz i=1 i=1 j=0 j=0
6.9. TRANSCENDENTAL EXTENSIONS 33

Thus
в€ћ
tj+1 z j = 0,
F (z) + F (z)
j=0

that is,
в€ћ
n n
iв€’1 i
tj z jв€’1 = 0.
ici z + ci z
i=1 i=0 j=1

Equating powers of z rв€’1 , we have, assuming that n в‰Ґ r,

rcr + c0 tr + c1 trв€’1 + В· В· В· + crв€’1 t1 = 0; (1)

if r > n, the п¬Ѓrst summation does not contribute, and we get

tr + c1 trв€’1 + В· В· В· + cn trв€’n = 0. (2)

Our situation is a bit awkward here because the roots of F (z) are the reciprocals of the xi .
n
The xi are the roots of i=0 ai z i where ai = cnв€’i (so that an = c0 = 1). The results can
be expressed as follows.

A6.5 NewtonвЂ™s Identities
n
If f (X) = i=0 ai X i (with an = 1) is a polynomial with roots x1 , . . . , xn , then the power
sums ti satisfy

tr + anв€’1 trв€’1 + В· В· В· + anв€’r+1 t1 + ranв€’r = 0, r в‰¤ n (3)

and

tr + anв€’1 trв€’1 + В· В· В· + a0 trв€’n = 0, r > n. (4)

A6.6 The Discriminant of a Cubic
First consider the case where the X 2 term is missing, so that f (X) = X 3 + pX + q. Then
n = t0 = 3, a0 = q, a1 = p, a2 = 0 (a3 = 1). NewtonвЂ™s identities yield
t1 + a2 = 0, t1 = 0; t2 + a2 t1 + 2a1 = 0, t2 = в€’2p;
t3 + a2 t2 + a1 t1 + 3a0 = 0, t3 = в€’3a0 = в€’3q;
t4 + a2 t3 + a1 t2 + a0 t1 = 0, t4 = в€’p(в€’2p) = 2p2

в€’2p
3 0
в€’2p в€’3q = в€’4p3 в€’ 27q 2 .
D= 0
в€’2p в€’3q 2p2

We now go to the general case f (X) = X 3 + aX 2 + bX + c. The quadratic term can be
eliminated by the substitution Y = X + a . Then
3
a a a
f (X) = g(Y ) = (Y в€’ )3 + a(Y в€’ )2 + b(Y в€’ ) + c
3 3 3
34 CHAPTER 6. GALOIS THEORY

a2 2a3 ba
p = b в€’ ,q = в€’
3
= Y + pY + q where + c.
3 27 3
Since the roots of f are translations of the roots of g by the same constant, the two
polynomials have the same discriminant. Thus D = в€’4p3 в€’ 27q 2 , which simpliп¬Ѓes to

D = a2 (b2 в€’ 4ac) в€’ 4b3 в€’ 27c2 + 18abc.

We now consider the Galois group of a quartic X 4 + aX 3 + bX 2 + cX + d, assumed
irreducible and separable over a п¬Ѓeld F . As above, the translation Y = X + a eliminates
4
the cubic term without changing the Galois group, so we may assume that f (X) =
X 4 + qX 2 + rX + s. Let the roots of f be x1 , x2 , x3 , x4 (distinct by separability), and
let V be the four group, realized as the subgroup of S4 containing the permutations
(1, 2)(3, 4), (1, 3)(2, 4) and (1, 4)(2, 3), along with the identity. By direct veriп¬Ѓcation (i.e.,
brute force), V S4 . If G is the Galois group of f (regarded as a group of permutations
of the roots), then V в€© G G by the second isomorphism theorem.

A6.7 Lemma
F(V в€© G) = F (u, v, w), where

u = (x1 + x2 )(x3 + x4 ), v = (x1 + x3 )(x2 + x4 ), w = (x1 + x4 )(x2 + x3 ).

Proof. Any permutation in V п¬Ѓxes u, v and w, so GF (u, v, w) вЉ‡ V в€© G. If Пѓ в€€ G
but Пѓ в€€ V в€© G then (again by direct veriп¬Ѓcation) Пѓ moves at least one of u, v, w. For
/
example, (1,2,3) sends u to w, and (1,2) sends v to w. Thus Пѓ в€€ GF (u, v, w). Therefore
/
GF (u, v, w) = V в€© G, and an application of the п¬Ѓxed п¬Ѓeld operator F completes the
proof. в™Ј

A6.8 Deп¬Ѓnition
The resolvent cubic of f (X) = X 4 + qX 2 + rX + s is g(X) = (X в€’ u)(X в€’ v)(X в€’ w).
To compute g, we must express its coeп¬ѓcients in terms of q, r and s. First note that
u в€’ v = в€’(x1 в€’ x4 )(x2 в€’ x3 ), u в€’ w = в€’(x1 в€’ x3 )(x2 в€’ x4 ), v в€’ w = в€’(x1 в€’ x2 )(x3 в€’ x4 ).
Thus f and g have the same discriminant. Now

X 4 + qX 2 + rX + s = (X 2 + kX + l)(X 2 в€’ kX + m)

where the appearance of k and в€’k is explained by the missing cubic term. Equating
coeп¬ѓcients gives l + m в€’ k 2 = q, k(m в€’ l) = r, lm = s. Solving the п¬Ѓrst two equations for
m and adding, we have 2m = k 2 + q + r/k, and solving the п¬Ѓrst two equations for l and
adding, we get 2l = k 2 + q в€’ r/k. Multiply the last two equations and use lm = s to get
a cubic in k 2 , namely

k 6 + 2qk 4 + (q 2 в€’ 4s)k 2 в€’ r2 = 0.

(This gives a method for actually п¬Ѓnding the roots of a quartic.) To summarize,

f (X) = (X 2 + kX + l)(X 2 в€’ kX + m)
6.9. TRANSCENDENTAL EXTENSIONS 35

where k 2 is a root of

h(X) = X 3 + 2qX 2 + (q 2 в€’ 4s)X в€’ r2 .

We claim that the roots of h are simply в€’u, в€’v, в€’w. For if we arrange the roots of f so
that x1 and x2 are the roots of X 2 + kX + l, and x3 and x4 are the roots of X 2 в€’ kX + m,
then k = в€’(x1 + x2 ), в€’k = в€’(x3 + x4 ), so в€’u = k 2 . The argument for в€’v and в€’w
is similar. Therefore to get g from h, we simply change the sign of the quadratic and
constant terms, and leave the linear term alone.

A6.9 An Explicit Formula For The Resolvent Cubic:

g(X) = X 3 в€’ 2qX 2 + (q 2 в€’ 4s)X + r2 .

We need some results concerning subgroups of Sn , n в‰Ґ 3.

A6.10 Lemma
(i) An is generated by 3-cycles, and every 3-cycle is a commutator.
(ii) The only subgroup of Sn with index 2 is An .

Proof. For the п¬Ѓrst assertion of (i), see Section 5.6, Problem 4. For the second assertion
of (i), note that

(a, b)(a, c)(a, b)в€’1 (a, c)в€’1 = (a, b)(a, c)(a, b)(a, c) = (a, b, c).

To prove (ii), let H be a subgroup of Sn with index 2; H is normal by Section 1.3,
Problem 6. Thus Sn /H has order 2, hence is abelian. But then by (5.7.2), part 5,
Sn в‰¤ H, and since An also has index 2, the same argument gives Sn в‰¤ An . By (i),
An в‰¤ Sn , so An = Sn в‰¤ H. Since An and H have the same п¬Ѓnite number of elements
n!/2, it follows that H = An . в™Ј

A6.11 Proposition
Let G be a subgroup of S4 whose order is a multiple of 4, and let V be the four group
(see the discussion preceding A6.7). Let m be the order of the quotient group G/(G в€© V ).
Then

(a) If m = 6, then G = S4 ;
(b) If m = 3, then G = A4 ;
(c) If m = 1, then G = V ;
(d) If m = 2, then G = D8 or Z4 or V ;
(e) If G acts transitively on {1, 2, 3, 4}, then the case G = V is excluded in (d). [In all
cases, equality is up to isomorphism.]
36 CHAPTER 6. GALOIS THEORY

Proof. If m = 6 or 3, then since |G| = m|G в€© V |, 3 is a divisor of |G|. By hypothesis, 4 is
also a divisor, so |G| is a multiple of 12. By A6.10 part (ii), G must be S4 or A4 . But

|S4 /(S4 в€© V )| = |S4 /V | = 24/4 = 6

and

|A4 /(A4 в€© V )| = |A4 /V | = 12/4 = 3

proving both (a) and (b). If m = 1, then G = G в€© V , so G в‰¤ V , and since |G| is a multiple
of 4 and |V | = 4, we have G = V , proving (c).
If m = 2, then |G| = 2|G в€© V |, and since |V | = 4, |G в€© V | is 1, 2 or 4. If it is 1,
then |G| = 2 Г— 1 = 2, contradicting the hypothesis. If it is 2, then |G| = 2 Г— 2 = 4, and
G = Z4 or V (the only groups of order 4). Finally, assume |G в€© V | = 4, so |G| = 8. But a
subgroup of S4 of order 8 is a Sylow 2-subgroup, and all such subgroups are conjugate and
therefore isomorphic. One of these subgroups is D8 , since the dihedral group of order 8
is a group of permutations of the 4 vertices of a square. This proves (d).
If m = 2, G acts transitively on {1, 2, 3, 4} and |G| = 4, then by the orbit-stabilizer
theorem, each stabilizer subgroup G(x) is trivial (since there is only one orbit, and its size
is 4). Thus every permutation in G except the identity moves every integer 1, 2, 3, 4. Since
|G в€© V | = 2, G consists of the identity, one other element of V , and two elements not in V ,
which must be 4-cycles. But a 4-cycle has order 4, so G must be cyclic, proving (e). в™Ј

A6.12 Theorem
Let f be an irreducible separable quartic, with Galois group G. Let m be the order of
the Galois group of the resolvent cubic. Then:

(a) If m = 6, then G = S4 ;
(b) If m = 3, then G = A4 ;
(c) If m = 1, then G = V ;
(d) If m = 2 and f is irreducible over L = F (u, v, w), where u, v and w are the roots of
the resolvent cubic, then G = D8 ;
(e) If m = 2 and f is reducible over L, then G = Z4 .

Proof. By A6.7 and the fundamental theorem, [G : G в€© V ] = [L : F ]. Now the roots of
the resolvent cubic g are distinct, since f and g have the same discriminant. Thus L is
a splitting п¬Ѓeld of a separable polynomial, so L/F is Galois. Consequently, [L : F ] = m
by (3.5.9). To apply (A6.11), we must verify that |G| is a multiple of 4. But this follows
from the orbit-stabilizer theorem: since G acts transitively on the roots of f , there is only
one orbit, of size 4 = |G|/|G(x)|. Now (A6.11) yields (a), (b) and (c), and if m = 2, then
G = D8 or Z4 .
To complete the proof, assume that m = 2 and G = D8 . Thinking of D8 as the
group of symmetries of a square with vertices 1,2,3,4, we can take D8 to be generated by
(1, 2, 3, 4) and (2, 4), with V = {1, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}. The elements of V
are symmetries of the square, hence belong to D8 ; thus V = Gв€©V = Gal(E/L) by (A6.7).
6.9. TRANSCENDENTAL EXTENSIONS 37

[E is a splitting п¬Ѓeld for f over F .] Since V is transitive, for each i, j = 1, 2, 3, 4, i = j,
there is an L-automorphism П„ of E such that П„ (xi ) = xj . Applying П„ to the equation
h(xi ) = 0, where h is the minimal polynomial of xi over L, we see that each xj is a root
of h, and therefore f | h. But h | f by minimality of h, so h = f , proving that f is
irreducible over L.
Finally, assume m = 2 and G = Z4 , which we take as {1, (1, 2, 3, 4), (1, 3)(2, 4),
(1, 4, 3, 2)}. Then G в€© V = {1, (1, 3)(2, 4)}, which is not transitive. Thus for some i = j,
xi and xj are not roots of the same irreducible polynomial over L. In particular, f is
reducible over L. в™Ј

A6.13 Example
Let f (X) = X 4 + 3X 2 + 2X + 1 over Q, with q = 3, r = 2, s = 1. The resolvent cubic is,
by (A6.9), g(X) = X 3 в€’ 6X 2 + 5X + 4. To calculate the discriminant of g, we can use the
general formula in (A6.6), or compute g(X + 2) = (X + 2)3 в€’ 6(X + 2)2 + 5(X + 2) + 4 =
X 3 в€’ 7X в€’ 2. [The rational root test gives irreducibility of g and restricts a factorization
of f to (X 2 + aX В± 1)(X 2 в€’ aX В± 1), a в€€ Z, which is impossible. Thus f is irreducible
as well.] We have D(g) = в€’4(в€’7)3 в€’ 27(в€’2)2 = 1264, which is not a square in Q. Thus
m = 6, so the Galois group of f is S4 .
Chapter 7

Introducing Algebraic Number
Theory

(Commutative Algebra 1)

The general theory of commutative rings is known as commutative algebra. The main
applications of this discipline are to algebraic number theory, to be discussed in this
chapter, and algebraic geometry, to be introduced in Chapter 8.
Techniques of abstract algebra have been applied to problems in number theory for
a long time, notably in the eп¬Ђort to prove FermatвЂ™s Last Theorem. As an introductory
example, we will sketch a problem for which an algebraic approach works very well. If p
is an odd prime and p в‰Ў 1 mod 4, we will prove that p is the sum of two squares, that is,
p can be expressed as x2 + y 2 where x and y are integers. Since pв€’1 is even, it follows
2
that -1 is a quadratic residue (that is, a square) mod p. [Pair each of the numbers 2,3,
. . . ,p в€’ 2 with its inverse mod p and pair 1 with p в€’ 1 в‰Ў в€’1 mod p. The product of the
numbers 1 through p в€’ 1 is, mod p,
pв€’1 pв€’1
1 Г— 2 Г— В·В·В· Г— Г— в€’1 Г— в€’2 Г— В· В· В· Г— в€’
2 2
and therefore pв€’1 ! 2 в‰Ў в€’1 mod p.]
2
If в€’1 в‰Ў x2 mod p, then p divides x2 + 1. Now we enter the ring of Gaussian integers
and factor x2 + 1 as (x + i)(x в€’ i). Since p can divide neither factor, it follows that p is
not prime in Z[i], so we can write p = О±ОІ where neither О± nor ОІ is a unit.
Deп¬Ѓne the norm of Оі = a + bi as N (Оі) = a2 + b2 . Then N (Оі) = 1 iп¬Ђ Оі = В±1 or В±i
iп¬Ђ Оі is a unit. (See Section 2.1, Problem 5.) Thus

p2 = N (p) = N (О±)N (ОІ) with N (О±) > 1 and N (ОІ) > 1,

so N (О±) = N (ОІ) = p. If О± = x + iy, then p = x2 + y 2 .

1
2 CHAPTER 7. INTRODUCING ALGEBRAIC NUMBER THEORY

Conversely, if p is an odd prime and p = x2 + y 2 , then p is congruent to 1 mod 4. (If x
is even, then x2 в‰Ў 0 mod 4, and if x is odd, then x2 в‰Ў 1 mod 4. We cannot have x and y
both even or both odd, since p is odd.)
It is natural to conjecture that we can identify those primes that can be represented as
в€љ
x + |d|y 2 , where d is a negative integer, by working in the ring Z[ d]. But the Gaussian
2

integers (d = в€’1) form a Euclidean domain, in particular a unique factorization domain.
On the other hand, unique factorization fails for d в‰¤ в€’3 (Section 2.7, Problem 7), so the
above argument collapses. [Recall from (2.6.4) that in a UFD, an element p that is not
prime must be reducible.] Diп¬ѓculties of this sort led Kummer to invent вЂњideal numbersвЂќ,
which later became ideals at the hands of Dedekind. We will see that although a ring of
algebraic integers need not be a UFD, unique factorization of ideals will always hold.

7.1 Integral Extensions
If E/F is a п¬Ѓeld extension and О± в€€ E, then О± is algebraic over F iп¬Ђ О± is a root of a
polynomial with coeп¬ѓcients in F . We can assume if we like that the polynomial is monic,
and this turns out to be crucial in generalizing the idea to ring extensions.

7.1.1 Deп¬Ѓnitions and Comments
In this chapter, unless otherwise speciп¬Ѓed, all rings are assumed commutative. Let A be
a subring of the ring R, and let x в€€ R. We say that x is integral over A if x is a root of a
monic polynomial f with coeп¬ѓcients in A. The equation f (X) = 0 is called an equation
of integral dependence for x over A. If x is a real or complex number that is integral
в€љ
over Z, then x is called an algebraic integer. Thus for every integer d, d is an algebraic
integer, as is any nth root of unity. (The monic polynomials are, respectively, X 2 в€’ d
and X n в€’ 1.) In preparation for the next result on conditions equivalent to integrality,
note that A[x], the set of polynomials in x with coeп¬ѓcients in A, is an A-module. (The
sum of two polynomials is a polynomial, and multiplying a polynomial by a member of A
produces another polynomial over A.)

7.1.2 Proposition
Let A be a subring of R, with x в€€ R. The following conditions are equivalent:

(i) x is integral over A;
(ii) The A-module A[x] is п¬Ѓnitely generated;
(iii) x belongs to a subring B of R such that A вЉ† B and B is a п¬Ѓnitely generated A-
module.

Proof. (i) implies (ii). If x is a root of a monic polynomial over A of degree n, then xn
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