Structura: /Seriously Applicable/ Math [Did You Know math is *impersonal*?]

[Jeff Rubard]

http://www.math.uiuc.edu/~r-ash/Algebra/Chapter6.pdf

Chapter 6
Galois Theory

6.1 Fixed Fields and Galois Groups
Galois theory is based on a remarkable correspondence between
subgroups of the Galois
group of an extension E/F and intermediate fields between E and F. In
this section
we will set up the machinery for the fundamental theorem. [A remark on
notation:
Throughout the chapter,the composition τ ◦ σ of two automorphisms will
be written as
a product τσ.]
6.1.1 Definitions and Comments
Let G = Gal(E/F) be the Galois group of the extension E/F. If H is a
subgroup of G,
the fixed field of H is the set of elements fixed by every
automorphism in H,that is,
F(H) = {x ∈ E: σ(x) = x for every σ ∈ H}.
If K is an intermediate field,that is, F ≤ K ≤ E,define
G(K) = Gal(E/K) = {σ ∈ G: σ(x) = x for every x ∈ K}.
I like the term “fixing group of K” for G(K),since G(K) is the group
of automorphisms
of E that leave K fixed. Galois theory is about the relation between
fixed fields and fixing
groups. In particular,the next result suggests that the smallest
subfield F corresponds
to the largest subgroup G.
6.1.2 Proposition
Let E/F be a finite Galois extension with Galois group G = Gal(E/F).
Then
(i) The fixed field of G is F;
(ii) If H is a proper subgroup of G,then the fixed field of H properly
contains F.
1
2 CHAPTER 6. GALOIS THEORY
Proof. (i) Let F0 be the fixed field of G. If σ is an F-automorphism
of E,then by
definition of F0, σ fixes everything in F0. Thus the F-automorphisms
of G coincide with
the F0-automorphisms of G. Now by (3.4.7) and (3.5.8), E/F0 is Galois.
By (3.5.9),the
size of the Galois group of a finite Galois extension is the degree of
the extension. Thus
[E : F] = [E : F0],so by (3.1.9), F = F0.
(ii) Suppose that F = F(H). By the theorem of the primitive element
(3.5.12),w e
have E = F(α) for some α ∈ E. Define a polynomial f(X) ∈ E[X] by
f(X) =

σ∈H
(X − σ(α)).
If τ is any automorphism in H,then we may apply τ to f (that is,to the
coefficients of f;
we discussed this idea in the proof of (3.5.2)). The result is
(τf)(X) =

σ∈H
(X − (τσ)(α)).
But as σ ranges over all of H,so does τσ,and consequently τf = f. Thus
each coefficient
of f is fixed by H,so f ∈ F[X]. Now α is a root of f,since X − σ(α) is
0 when X = α
and σ is the identity. We can say two things about the degree of f:
(1) By definition of f,deg f = |H| < |G| = [E : F],and,since f is a
multiple of the
minimal polynomial of α over F,
(2) deg f ≥ [F(α) : F] = [E : F],and we have a contradiction. ♣
There is a converse to the first part of (6.1.2).
6.1.3 Proposition
Let E/F be a finite extension with Galois group G. If the fixed field
of G is F,then E/F
is Galois.
Proof. Let G = {σ1, . . . , σn},where σ1 is the identity. To show that
E/F is normal,
we consider an irreducible polynomial f ∈ F[X] with a root α ∈ E.
Apply each automorphism
in G to α,and suppose that there are r distinct images α = α1 = σ1(α),
α2 = σ2(α), . . . , αr = σr(α). If σ is any member of G,then σ will
map each αi to some
αj,and since σ is an injective map of the finite set {α1, . . . , αr}
to itself,it is surjective as
well. To put it simply, σ permutes the αi. Now we examine what σ does
to the elementary
symmetric functions of the αi,whic h are given by
e1 =
r
i=1
αi, e2 =

i<j
αiαj, e3 =

i<j<k
αiαjαk, . . . ,
er =
r
i=1
αi.
Since σ permutes the αi,it follows that σ(ei) = ei for all i. Thus the
ei belong to the
fixed field of G,whic h is F by hypothesis. Now we form a monic
polynomial whose roots
are the αi:
g(X) = (X − α1) · · · (X − αr) = Xr − e1Xr−1 + e2Xr−2 −· · · +
(−1)rer.
6.1. FIXED FIELDS AND GALOIS GROUPS 3
Since the ei belong to F, g ∈ F[X],and since the αi are in E, g splits
over E. We claim
that g is the minimal polynomial of α over F. To see this,let h(X) =
b0+b1X+· · ·+bmXm
be any polynomial in F[X] having α as a root. Applying σi to the
equation
b0 + b1α + · · · bmαm = 0
we have
b0 + b1αi + · · · bmαm
i = 0,
so that each αi is a root of h,hence g divides h and therefore g
=min(α, F). But our
original polynomial f ∈ F[X] is irreducible and has α as a root,so it
must be a constant
multiple of g. Consequently, f splits over E,pro ving that E/F is
normal. Since the αi,
i = 1, . . . r,are distinct, g has no repeated roots. Thus α is
separable over F,whic h shows
that the extension E/F is separable. ♣
It is profitable to examine elementary symmetric functions in more
detail.
6.1.4 Theorem
Let f be a symmetric polynomial in the n variables X1, . . . , Xn.
[This means that if σ is
any permutation in Sn and we replace Xi by Xσ(i) for i = 1, . . . ,
n,then f is unchanged.]
If e1, . . . , en are the elementary symmetric functions of the
Xi,then f can be expressed
as a polynomial in the ei.
Proof. We give an algorithm. The polynomial f is a linear combination
of monomials
of the form Xr1
1
· · ·Xrn
n ,and we order the monomials lexicographically: Xr1
1
· · ·Xrn
n >
Xs1
1
· · ·Xsn
n iff the first disagreement between ri and si results in ri > si.
Since f is
symmetric,all terms generated by applying a permutation σ ∈ Sn to the
subscripts of
Xr1
1
· · ·Xrn
n will also contribute to f. The idea is to cancel the leading terms
(those
associated with the monomial that is first in the ordering) by
subtracting an expression
of the form
et1
1 et2
2
· · · etn
n = (X1 + · · · + Xn)t1 · · · (X1 · · ·Xn)tn
which has leading term
Xt1
1 (X1X2)t2 (X1X2X3)t3 · · · (X1 · · ·Xn)tn = Xt1+···+tn
1 Xt2+···+tn
2
· · ·Xtn
n .
This will be possible if we choose
t1 = r1 − r2, t2 = r2 − r3, . . . , tn−1 = rn−1 − rn, tn = rn.
After subtraction,the resulting polynomial has a leading term that is
below Xr1
1
· · ·Xrn
n
in the lexicographical ordering. We can then repeat the procedure,whic
h must terminate
in a finite number of steps. ♣
4 CHAPTER 6. GALOIS THEORY
6.1.5 Corollary
If g is a polynomial in F[X] and f(α1, . . . , αn) is any symmetric
polynomial in the roots
α1, . . . , αn of g,then f ∈ F[X].
Proof. We may assume without loss of generality that g is monic. Then
in a splitting
field of g we have
g(X) = (X − α1) · · · (X − αn) = Xn − e1Xn−1 + · · · + (−1)nen.
By (6.1.4), f is a polynomial in the ei,and since the ei are simply ±
the coefficients of g,
the coefficients of f are in F. ♣
6.1.6 Dedekind’s Lemma
The result that the size of the Galois group of a finite Galois
extension is the degree of
the extension can be proved via Dedekind’s lemma,whic h is of interest
in its own right.
Let G be a group and E a field. A character from G to E is a
homomorphism from G
to the multiplicative group E∗ of nonzero elements of E. In
particular,an automorphism
of E defines a character with G = E∗,as does a monomorphism of E into
a field L.
Dedekind’s lemma states that if σ1, . . . , σn are distinct characters
from G to E,then the
σi are linearly independent over E. The proof is given in Problems 3
and 4.
Problems For Section 6.1
1. Express X2
1X2X3 + X1X2
2X3 + X1X2X2
3 in terms of elementary symmetric functions.
2. Repeat Problem 1 forX2
1X2 + X2
1X3 + X1X2
2 + X1X2
3 + X2
2X3 + X2X2
3 + 4X1X2X3.
3. To begin the proof of Dedekind’s lemma,supp ose that the σi are
linearly dependent.
By renumbering the σi if necessary,w e have
a1σ1 + · · · arσr = 0
where all ai are nonzero and r is as small as possible. Show that for
every h and g ∈ G,
we have
r
i=1
aiσ1(h)σi(g) = 0 (1)
and
r
i=1
aiσi(h)σi(g) = 0. (2)
[Equations (1) and (2) are not the same; in (1) we have σ1(h),not
σi(h).]
4. Continuing Problem 3,subtract (2) from (1) to get
r
i=1
ai(σ1(h) − σi(h))σi(g) = 0. (3)
With g arbitrary,reac h a contradiction by an appropriate choice of h.
6.2. THE FUNDAMENTAL THEOREM 5
5. If G is the Galois group of Q( 3
√
2) over Q,what is the fixed field of G?
6. Find the Galois group of C/R.
7. Find the fixed field of the Galois group of Problem 6.
6.2 The Fundamental Theorem
With the preliminaries now taken care of,w e can proceed directly to
the main result.
6.2.1 Fundamental Theorem of Galois Theory
Let E/F be a finite Galois extension with Galois group G. If H is a
subgroup of G,
let F(H) be the fixed field of H,and if K is an intermediate field,let
G(K) be Gal(E/K),
the fixing group of K (see (6.1.1)).
(1) F is a bijective map from subgroups to intermediate fields,with
inverse G. Both maps
are inclusion-reversing,that is,if H1 ≤ H2 then F(H1) ≥ F(H2),and if
K1 ≤ K2,
then G(K1) ≥ G(K2).
(2) Suppose that the intermediate field K corresponds to the subgroup
H under the
Galois correspondence. Then
(a) E/K is always normal (hence Galois);
(b) K/F is normal if and only if H is a normal subgroup of G,and in
this case,
(c) the Galois group of K/F is isomorphic to the quotient group G/H.
Moreover,
whether or not K/F is normal,
(d) [K : F] = [G : H] and [E : K] = |H|.
(3) If the intermediate field K corresponds to the subgroup H and σ is
any automorphism
in G,then the field σK = {σ(x): x ∈ K} corresponds to the conjugate
subgroup
σHσ−1. For this reason, σK is called a conjugate subfield of K.
The following diagram may aid the understanding.
E G
| |
K H
| |
F 1
As we travel up the left side from smaller to larger fields,w e move
down the right side
from larger to smaller groups. A statement about K/F,an extension at
the bottom of
the left side,corresp onds to a statement about G/H,lo cated at the
top of the right side.
Similarly,a statement about E/K corresponds to a statement about H/1 =
H.
Proof. (1) First,consider the composite mapping H →F(H) → GF(H). If σ
∈ H then σ
fixes F(H) by definition of fixed field,and therefore σ ∈ GF(H) =
Gal(E/F(H)). Thus
H ⊆ GF(H). If the inclusion is proper,then by (6.1.2) part (ii) with F
replaced by F(H),
6 CHAPTER 6. GALOIS THEORY
we have F(H) > F(H),a contradiction. [Note that E/K is a Galois
extension for any
intermediate field K,b y (3.4.7) and (3.5.8).] Thus GF(H) = H.
Now consider the mapping K → G(K) → FG(K) = F Gal(E/K). By (6.1.2)
part (i)
with F replaced by K,we have FG(K) = K. Since both F and G are
inclusion-reversing
by definition,the proof of (1) is complete.
(3) The fixed field of σHσ−1 is the set of all x ∈ E such that στσ
−1(x) = x for every
τ ∈ H. Thus
F(σHσ
−1) = {x ∈ E: σ
−1(x) ∈ F(H)} = σ(F(H)).
(2a) This was observed in the proof of (1).
(2b) If σ is an F-monomorphism of K into E,then by (3.5.2) and
(3.5.6), σ extends
to an F-monomorphism of E into itself,in other words (see (3.5.6)),an
F-automorphism
of E. Thus each such σ is the restriction to K of a member of G.
Conversely,the
restriction of an automorphism in G to K is an F-monomorphism of K
into E. By (3.5.5)
and (3.5.6), K/F is normal iff for every σ ∈ G we have σ(K) = K. But
by (3), σ(K)
corresponds to σHσ−1 and K to H. Thus K/F is normal iff σHσ−1 = H for
every σ ∈ G,
i.e., H
G.
(2c) Consider the homomorphism of G = Gal(E/F) to Gal(K/F) given by σ
→ σ|K.
The map is surjective by the argument just given in the proof of (2b).
The kernel is the
set of all automorphisms in G that restrict to the identity on K,that
is,Gal( E/K) = H.
The result follows from the first isomorphism theorem.
(2d) By (3.1.9),[ E : F] = [E : K][K : F]. The term on the left is |G|
by (3.5.9),and
the first term on the right is | Gal(E/K)| by (2a),and this in turn is
|H| since H = G(K).
Thus |G| = |H|[K : F],and the result follows from Lagrange’s theorem.
[If K/F is
normal,the proof is slightly faster. The first statement follows from
(2c). To prove the
second,note that by (3.1.9) and (3.5.9),
[E : K] =
[E : F]
[K : F]
=
|G|
|G/H| = |H|.] ♣
The next result is reminiscent of the second isomorphism theorem,and
is best visualized
via the diamond diagram of Figure 6.2.1. In the diagram, EK is the
composite of
the two fields E and K,that is,the smallest field containing both E
and K.
6.2.2 Theorem
Let E/F be a finite Galois extension and K/F an arbitrary extension.
Assume that E
and K are both contained in a common field,so that it is sensible to
consider the composite
EK. Then
(1) EK/K is a finite Galois extension;
(2) Gal(EK/K) is embedded in Gal(E/F),where the embedding is
accomplished by
restricting automorphisms in Gal(EK/K) to E;
(3) The embedding is an isomorphism if and only if E ∩ K = F.
6.2. THE FUNDAMENTAL THEOREM 7
EK

    
E K
E ∩ K

    
F
Figure 6.2.1
Proof. (1) By the theorem of the primitive element (3.5.12),w e have E
= F[α] for some
α ∈ E,so EK = KF[α] = K[α]. The extension K[α]/K is finite because α
is algebraic
over F,hence over K. Since α,regarded as an element of EK,is separable
over F and
hence over K,it follows that EK/K is separable. [To avoid breaking the
main line of
thought,this result will be developed in the exercises (see Problems 1
and 2).]
Now let f be the minimal polynomial of α over F,and g the minimal
polynomial of α
over K. Since f ∈ K[X] and f(α) = 0, we have g | f,and the roots of g
must belong to
E ⊆ EK = K[α] because E/F is normal. Therefore K[α] is a splitting
field for g over K,
so by (3.5.7), K[α]/K is normal.
(2) If σ is an automorphism in Gal(EK/K),restrict σ to E,th us
defining a homomorphism
from Gal(EK/K) to Gal(E/F). (Note that σ|E is an automorphism of E
because
E/F is normal.) Now σ fixes K,and if σ belongs to the kernel of the
homomorphism,
then σ also fixes E,so σ fixes EK = K[α]. Thus σ is the identity,and
the kernel is trivial,
proving that the homomorphism is actually an embedding.
(3) The embedding of (2) maps Gal(EK/K) to a subgroup H of Gal(E/
F),and we
will find the fixed field of H. By (6.1.2),the fixed field of Gal(EK/
K) is K,and since
the embedding just restricts automorphisms to E,the fixed field of H
must be E ∩ K.
By the fundamental theorem, H = Gal(E/(E ∩ K)). Thus
H = Gal(E/F) iff Gal(E/(E ∩ K)) = Gal(E/F),
and by applying the fixed field operator F,w e see that this happens
if and only if E ∩
K = F. ♣
Problems For Section 6.2
1. Let E = F(α1, . . . , αn),where each αi is algebraic and separable
over F. We are going
to show that E is separable over F. Without loss of generality,w e can
assume that the
characteristic of F is a prime p,and since F/F is separable,the result
holds for n = 0.
To carry out the inductive step,let Ei = F(α1, . . . , αi),so that Ei
+1 = Ei(αi+1).
Show that Ei+1 = Ei(Ep
i+1). (See Section 3.4,Problems 4–8,for the notation.)
2. Continuing Problem 1,sho w that E is separable over F.
8 CHAPTER 6. GALOIS THEORY
3. Let E = F(α1, . . . , αn),where each αi is algebraic over F. If for
each i = 1, . . . , n,all
the conjugates of αi (the roots of the minimal polynomial of αi over
F) belong to E,
show that E/F is normal.
4. Suppose that F = K0 ≤ K1 ≤ ·· · ≤ Kn = E,where E/F is a finite
Galois extension,
and that the intermediate field Ki corresponds to the subgroup Hi
under the Galois
correspondence. Show that Ki/Ki−1 is normal (hence Galois) if and only
if Hi
Hi−1,
and in this case,Gal( Ki/Ki−1) is isomorphic to Hi−1/Hi.
5. Let E and K be extensions of F,and assume that the composite EK is
defined. If A
is any set of generators for K over F (for example, A = K),sho w that
EK = E(A),
the field formed from E by adjoining the elements of A.
6. Let E/F be a finite Galois extension with Galois group G,and let E/
F be a finite
Galois extension with Galois group G. If τ is an isomorphism of E and
E with
τ (F) = F,w e expect intuitively that G
∼=
G. Prove this formally.
7. Let K/F be a finite separable extension. Although K need not be a
normal extension
of F,w e can form the normal closure N of K over F,as in (3.5.11).
Then N/F
is a Galois extension (see Problem 8 of Section 6.3); let G be its
Galois group. Let
H = Gal(N/K),so that the fixed field of H is K. If H is a normal
subgroup of G
that is contained in H,sho w that the fixed field of H is N.
8. Continuing Problem 7,sho w that H is trivial,and conclude that

g∈G
gHg
−1 = {1}
where 1 is the identity automorphism.
6.3 Computing a Galois Group Directly
6.3.1 Definitions and Comments
Suppose that E is a splitting field of the separable polynomial f over
F. The Galois
group of f is the Galois group of the extension E/F. (The extension is
indeed Galois;
see Problem 8.) Given f,ho w can we determine its Galois group? It is
not so easy,but
later we will develop a systematic approach for polynomials of degree
4 or less. Some
cases can be handled directly,and in this section we look at a typical
situation. A useful
observation is that the Galois group G of a finite Galois extension E/
F acts transitively
on the roots of any irreducible polynomial h ∈ F[X] (assuming that
one,hence every,
root of h belongs to E). [Each σ ∈ G permutes the roots by (3.5.1). If
α and β are roots
of h,then by (3.2.3) there is an F-isomorphism of F(α) and F(β)
carrying α to β. This
isomorphism can be extended to an F-automorphism of E by (3.5.2),
(3.5.5) and (3.5.6).]
6.3.2 Example
Let d be a positive integer that is not a perfect cube,and let θ be
the positive cube root
of d. Let ω = ei2π/3 = −1
2 + i 1
2
√
3,so that ω2 = e−i2π/3 = −1
2
− i 1
2
√
3 = −(1 + ω).
The minimal polynomial of θ over the rationals Q is f(X) = X3 − d,b
ecause if f were
6.3. COMPUTING A GALOIS GROUP DIRECTLY 9
reducible then it would have a linear factor and d would be a perfect
cube. The minimal
polynomial of ω over Q is g(X) = X2 + X + 1. (If g were reducible,it
would have a
rational (hence real) root,so the discriminant would be nonnegative,a
contradiction.)
We will compute the Galois group G of the polynomial f(X)g(X),whic h
is the Galois
group of E = Q(θ, ω) over Q.
If the degree of E/Q is the product of the degrees of f and g,w e will
be able to
make progress. We have [Q(θ) : Q] = 3 and,since ω,a complex number,do
es not belong
to Q(θ),we have [Q(θ, ω) : Q(θ)] = 2. Thus [Q(θ, ω) : Q] = 6. But the
degree of
a finite Galois extension is the size of the Galois group by
(3.5.9),so G has exactly 6
automorphisms. Now any σ ∈ G must take θ to one of its
conjugates,namely θ, ωθ or
ω2θ. Moreover, σ must take ω to a conjugate,namely ω or ω2. Since σ is
determined by
its action on θ and ω,w e have found all 6 members of G. The results
can be displayed as
follows.
1: θ → θ, ω → ω,order = 1
τ : θ → θ, ω → ω2,order = 2
σ : θ → ωθ, ω → ω,order = 3
στ : θ → ωθ, ω → ω2,order = 2
σ2 : θ → ω2θ, ω → ω,order = 3
τσ: θ → ω2θ, ω → ω2,order = 2
Note that τσ2 gives nothing new since τσ2 = στ. Similarly, σ2τ = τσ.
Thus
σ3 = τ 2 = 1, τστ
−1 = σ
−1 (= σ2). (1)
At this point we have determined the multiplication table of G,but
much more insight
is gained by observing that (1) gives a presentation of S3 (Section
5.8,Problem 3). We
conclude that G
∼=
S3. The subgroups of G are
{1}, G, σ, τ , τσ, τσ2
and the corresponding fixed fields are
E, Q, Q(ω), Q(θ), Q(ωθ), Q(ω2θ).
To show that the fixed field of τσ = {1, τσ} is Q(ωθ),note that τσ
has index 3 in G,so
by the fundamental theorem,the corresponding fixed field has degree 3
over Q. Now τσ
takes ωθ to ω2ω2θ = ωθ and [Q(ωθ) : Q] = 3 (because the minimal
polynomial of ωθ over
Q is f). Thus Q(ωθ) is the entire fixed field. The other calculations
are similar.
Problems For Section 6.3
1. Suppose that E = F(α) is a finite Galois extension of F,where α is
a root of the
irreducible polynomial f ∈ F[X]. Assume that the roots of f are α1 =
α, α2, . . . , αn.
Describe,as best you can from the given information,the Galois group
of E/F.
2. Let E/Q be a finite Galois extension,and let x1, . . . , xn be a
basis for E over Q.
Describe how you would find a primitive element,that is,an α ∈ E such
that E =
Q(α). (Your procedure need not be efficient.)
10 CHAPTER 6. GALOIS THEORY
3. Let G be the Galois group of a separable irreducible polynomial f
of degree n. Show
that G is isomorphic to a transitive subgroup H of Sn. [Transitivity
means that if i
and j belong to {1, 2, . . . , n},then for some σ ∈ H we have σ(i) =
j. Equivalently,
the natural action of H on {1, . . . , n},giv en by h • x = h(x),is
transitive.]
4. Use Problem 3 to determine the Galois group of an irreducible
quadratic polynomial
aX2 + bX + c ∈ F[X], a = 0. Assume that the characteristic of F is
not 2,so that
the derivative of f is nonzero and f is separable.
5. Determine the Galois group of (X2 − 2)(X2 − 3) over Q.
6. In the Galois correspondence,supp ose that Ki is the fixed field of
the subgroup Hi,
i = 1, 2. Identify the group corresponding to K = K1 ∩ K2.
7. Continuing Problem 6,iden tify the fixed field of H1 ∩ H2.
8. Suppose that E is a splitting field of a separable polynomial f
over F. Show that
E/F is separable. [Since the extension is finite by (3.2.2) and normal
by (3.5.7), E/F
is Galois.]
9. Let G be the Galois group of f(X) = X4 − 2 over Q. Thus if θ is the
positive fourth
root of 2,then G is the Galois group of Q(θ, i)/Q. Describe all 8
automorphisms in G.
10. Show that G is isomorphic to the dihedral group D8.
11. Define σ(θ) = iθ, σ(i) = i, τ (θ) = θ, τ (i) = −i,as in the
solution to Problem 10.
Find the fixed field of the normal subgroup N = {1, στ,σ2, σ3τ} of
G,and verify that
the fixed field is a normal extension of Q.
6.4 Finite Fields
Finite fields can be classified precisely. We will show that a finite
field must have pn
elements,where p is a prime and n is a positive integer. In
addition,there is (up to
isomorphism) only one finite field with pn elements. We sometimes use
the notation
GF(pn) for this field; GF stands for “Galois field”. Also,the field
with p elements will
be denoted by Fp rather than Zp,to emphasize that we are working with
fields.
6.4.1 Proposition
Let E be a finite field of characteristic p. Then |E| = pn for some
positive integer n.
Moreover, E is a splitting field for the separable polynomial f(X) =
Xpn −X over Fp,so
that any finite field with pn elements is isomorphic to E. Not only is
E generated by the
roots of f,but in fact E coincides with the set of roots of f.
Proof. Since E contains a copy of Fp (see (2.1.3),Example 2),w e may
view E as a vector
space over Fp. If the dimension of this vector space is n,then since
each coefficient in a
linear combination of basis vectors can be chosen in p ways,w e have |
E| = pn.
Now let E∗ be the multiplicative group of nonzero elements of E. If α
∈ E∗,then
αpn−1 = 1 by Lagrange’s theorem,so αpn = α for every α ∈ E,including α
= 0. Thus
each element of E is a root of f,and f is separable by (3.4.5). Now f
has at most pn
distinct roots,and as we have already identified the pn elements of E
as roots of f,in
fact f has pn distinct roots and every root of f must belong to E. ♣
6.4. FINITE FIELDS 11
6.4.2 Corollary
If E is a finite field of characteristic p,then E/Fp is a Galois
extension. The Galois group
is cyclic and is generated by the Frobenius automorphism σ(x) = xp, x
∈ E.
Proof. E is a splitting field for a separable polynomial over Fp,so E/
Fp is Galois; see
(6.3.1). Since xp = x for each x ∈ Fp, Fp is contained in the fixed
field F(σ). But
each element of the fixed field is a root of Xp − X,so F(σ) has at
most p elements.
Consequently, F(σ) = Fp. Now Fp = F(Gal(E/Fp)) by (6.1.2),so by the
fundamental
theorem,Gal( E/Fp) = σ. ♣
6.4.3 Corollary
Let E/F be a finite extension of a finite field,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
automorphism τ (x) = xpm, x ∈ E. Furthermore, F is the only subfield
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.
We may then reproduce the proof of (6.4.2) with Fp replaced by F, σ by
τ , xp by xpm,
and Xp by Xpm. Uniqueness of F as a subfield of E with pm elements
follows because
there is only one splitting field over Fp for Xpm − X inside E; see
(3.2.1). ♣
How do we know that finite fields (other than the Fp) exist? There is
no problem.
Given any prime p and positive integer n,w e can construct E = GF(pn)
as a splitting
field for Xpn − X over Fp. We have just seen that if E contains a
subfield 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 finite field is cyclic. More
generally,if G is a finite subgroup
of the multiplicative group of an arbitrary field,then G is cyclic.
Proof. G is a finite 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 ∈ G then the order of x divides r,so
xr = 1. Therefore
each element of G is a root of Xr − 1,so |G| ≤ 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| ≥ 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. ♣
6.4.5 Proposition
GF(pm) is a subfield of E = GF(pn) if and only if m is a divisor of n.
Proof. The “only if” part follows from (6.4.3),so assume that m
divides n. If t is any
positive integer greater than 1,then m | n iff (tm − 1) | (tn − 1). (A
formal proof is not
difficult,but I prefer to do an ordinary long division of tn − 1 by tm
− 1. The successive
12 CHAPTER 6. GALOIS THEORY
quotients are tn−m, tn−2m, tn−3m, . . . ,so the division will be
successful iff n−rm = 0 for
some positive integer r.) Taking t = p,w e see that pm − 1 divides |
E∗|,so by (6.4.4)
and (1.1.4), E∗ has a subgroup H of order pm−1. By Lagrange’s
theorem,eac h x ∈ H∪{0}
satisfies xpm = x. As in the proof of (6.4.1), H ∪ {0} coincides with
the set of roots of
Xpm − X. Thus we may construct entirely inside GF(pn) a splitting
field for Xpm − X
over Fp. But this splitting field is a copy of GF(pm). ♣
In practice,finite fields are constructed by adjoining roots of
carefully selected irreducible
polynomials over Fp. The following result is very helpful.
6.4.6 Theorem
Let p be a prime and n a positive integer. Then Xpn − X is the product
of all monic
irreducible polynomials over Fp whose degree divides n.
Proof. Let us do all calculations inside E = GF(pn) = the set of roots
of f(X) = Xpn−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 α is any root of g,then Fp(α) is a finite field 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 subfield
with pm elements,
and this subfield must be isomorphic to Fp(α). If β ∈ E corresponds to
α under this
isomorphism,then g(β) = 0 (because g(α) = 0) and f(β) = 0 (because β ∈
E). Since g is
the minimal polynomial of β 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. ♣
6.4.7 The Explicit Construction of a Finite Field
By (6.4.4),the multiplicative group E∗ of a finite field E = GF(pn) is
cyclic,so E∗ can
be generated by a single element α. Thus E = Fp(α) = Fp[α],so that α
is a primitive
element of E. The minimal polynomial of α 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 − 1 in α,they are the powers of α. This is
significant in applications to
coding theory. Let’s do an example over F2.
The polynomial g(X) = X4 + X + 1 is irreducible over F2. One way to
verify this is
to factor X16 −X = X16 +X over F2; the factors are the (necessarily
monic) irreducible
polynomials of degrees 1,2 and 4. To show that g is primitive,w e
compute powers of α:
α0 = 1, α1 = α, α2 = α2, α3 = α3, α4 = 1+α (since g(α) = 0),
α5 = α + α2, α6 = α2 + α3, α7 = α3 + α4 = 1+α + α3, α8 = α + α2 + α4 =
1+α2
(since 1+1=0 in F2),
α9 = α+α3, α10 = 1+α+α2, α11 = α+α2+α3, α12 = 1+α+α2+α3, α13 =
1+α2+α3,
α14 = 1+α3,
and at this point we have all 24 − 1 = 15 nonzero elements of GF(16).
The pattern now
repeats,b eginning with α15 = α + α4 = 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 X4+X3+X2+X+1 ∈ F2[X] is not
primitive.
2. Let F be a finite field 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 iff (tm − 1) | (tn − 1) (t = 2,
3, . . . ). Show that an
equivalent condition is (Xm − 1) divides (Xn − 1).
If E is a finite extension of a finite field,or more generally a
finite separable extension
of a field F,then by the theorem of the primitive element, E = F(α)
for some α ∈ E.
We now develop a condition equivalent to the existence of a primitive
element.
4. Let E/F be a finite extension,with E = F(α) and F ≤ L ≤ E. Suppose
that the minimal
polynomial of α over L is g(X) =

r−1
i=0 biXi+Xr,and let K = F(b0, . . . , br−1).
If h is the minimal polynomial of α over K,sho w that g = h,and
conclude that
L = K.
5. Continuing Problem 4,sho w that there are only finitely many
intermediate fields L
between E and F.
6. Conversely,let E = F(α1, . . . , αn) be a finite extension with
only finitely many intermediate
fields 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(α1, . . . , αn−1),sho w that E =
F(β,αn) for some
β ∈ L.
7. Now assume (without loss of generality) that F is infinite. Show
that there are distinct
elements c, d ∈ F such that F(cβ + αn) = F(dβ + αn).
8. Continuing Problem 7,sho w that E = F(cβ + αn). Thus a finite
extension has a
primitive element iff there are only finitely many intermediate
fields.
9. Let α be an element of the finite field GF(pn). Show that α and αp
have the same
minimal polynomial over Fp.
10. Suppose that α 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 αi and αj have the same
minimal
polynomial. Repeat for α an element of order 15 in GF(2n). [Note that
elements of
the specified orders exist,b ecause 13 divides 26 = 33 − 1 and 15 = 24
− 1.]
6.5 Cyclotomic Fields
6.5.1 Definitions and Comments
Cyclotomic extensions of a field F are formed by adjoining nth roots
of unity. Formally,a
cyclotomic extension of F is a splitting field E for f(X) = Xn −1 over
F. The roots of f
are called nth roots of unity,and they form a multiplicative subgroup
of the group E∗ 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∗ is n.
14 CHAPTER 6. GALOIS THEORY
It is tempting to say “obviously,primitiv e nth roots of unity must
exist,just take a
generator of the cyclic subgroup”. But suppose that F has
characteristic p and p divides n,
say n = mp. If ω is an nth root of unity,then
0 = ωn − 1 = (ωm − 1)p
so the order of ω must be less than n. To avoid this difficulty,w e
assume that the
characteristic of F does not divide n. Then f(X) = nXn−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 ω,
and for any such ω,we have E = F(ω).
If σ is any automorphism in the Galois group Gal(E/F),then σ must take
a primitive
root of unity ω to another primitive root of unity ωr,where r and n
are relatively prime.
(See (1.1.5).) We can identify σ 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,b y the fundamental theorem (or (3.5.9)),[ E : F] = | Gal(E/
F)|,whic h is a
divisor of |Un| = ϕ(n).
Cyclotomic fields are of greatest interest when the underlying field F
is Q,the rational
numbers,and from now on we specialize to that case. The primitive nth
roots of unity
are ei2πr/n where r and n are relatively prime. Thus there are ϕ(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 Definition
The nth cyclotomic polynomial is defined by
Ψn(X) =

i
(X − ωi)
where the ωi are the primitive nth roots of unity in the field C of
complex numbers. Thus
the degree of Ψn(X) is ϕ(n).
From the definition,w e have Ψ1(X) = X − 1 and Ψ2(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 −1 =

d|n
Ψd(X).
In particular,if p is prime,then
Ψp(X) = Xp − 1
X − 1
= Xp−1 + Xp−2 + · · · + X + 1.
6.5. CYCLOTOMIC FIELDS 15
Proof. If ω is an nth root of unity,then its order in C
∗ is a divisor d of n,and in this
case, ω is a primitive dth root of unity,hence a root of Ψd(X).
Conversely,if d | n,then
any root of Ψd(X) is a dth,hence an nth,ro ot of unity. ♣
From (6.5.3) we have
Ψ3(X) = X2 + X + 1,
Ψ4(X) = X2 + 1, Ψ5(X) = X4 + X3 + X2 + X + 1,
Ψ6(X) = X6−1
(X−1)(X+1)(X2+X+1) = X6−1
(X3−1)(X+1) = X3+1
X+1 = X2 − X + 1.
It is a natural conjecture that all coefficients of the cyclotomic
polynomials are integers,
and this turns out to be correct.
6.5.4 Proposition
Ψn(X) ∈ Z[X].
Proof. By (6.5.3),w e have
Xn − 1 = [

d|n,d<n
Ψd(X)]Ψn(X).
By definition,the cyclotomic polynomials are monic,and by induction
hypothesis,the
expression in brackets is a monic polynomial in Z[X]. Thus Ψn(X) is
the quotient of two
monic polynomials with integer coefficients. At this point,all we know
for sure is that
the coefficients of Ψn(X) are complex numbers. But if we apply
ordinary long division,
even in C,w e know that the process will terminate,and this forces the
quotient Ψn(X)
to be in Z[X]. ♣
We now show that the nth cyclotomic polynomial is the minimal
polynomial of each
primitive nth root of unity.
6.5.5 Theorem
Ψn(X) is irreducible over Q.
Proof. Let ω be a primitive nth root of unity,with minimal polynomial
f over Q. Since
ω is a root of Xn − 1,w e have Xn − 1 = f(X)g(X) for some g ∈ 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 coefficients of f and g are integers.
If p is a prime that does not divide n,w e will show that ωp is a root
of f. If not,
then it is a root of g. But g(ωp) = 0 implies that ω is a root of
g(Xp),so f(X) divides
g(Xp),sa y g(Xp) = f(X)h(X). As above, h ∈ Z[X]. But by the binomial
expansion
modulo p, g(X)p ≡ g(Xp) = f(X)h(X) mod p. Reducing the coefficients of
a polynomial
k(X) mod p is equivalent to viewing it as an element k ∈ 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 Xn − 1 has a multiple root,con tradicting (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 ≥
ϕ(n) =deg Ψn,and therefore f = Ψn 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 ω is a primitive nth
root of unity where n = 175,then ω72 is a primitive nth root of unity
because 72 and 175
are relatively prime. Moreover,since 72 = 23 × 32,we have
ω72 = (((((ω)2)2)2)3)3
and the result follows. ♣
6.5.6 Corollary
The Galois group G of the nth cyclotomic extension Q(ω)/Q is
isomorphic to the group Un
of units mod n.
Proof. By the fundamental theorem, |G| = [Q(ω) : Q] = degΨn = ϕ(n) = |
Un|. Thus the
monomorphism of G and a subgroup of Un (see (6.5.1)) is surjective. ♣
Problems For Section 6.5
1. If p is prime and p divides n,sho w that Ψpn(X) = Ψn(Xp). (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 fields
associated with the
cyclotomic extension Q7 = Q(ω)/Q where ω = ei2π/7 is a primitive 7th
root of unity.
The Galois group G consists of automorphisms σi, i = 1, 2, 3, 4, 5,
6,where σi(ω) = ωi.
3. Show that σ3 generates the cyclic group G.
4. Show that the subgroups of G are 1 (order 1), σ6 (order 2),
σ2 (order 3),and
G = σ3 (order 6).
5. The fixed field of 1 is Q7 and the fixed field of G is Q. Let K
be the fixed field
of σ6. Show that ω + ω−1 ∈ K,and deduce that K = Q(ω + ω−1) = Q(cos
2π/7).
6. Let L be the fixed field of σ2. Show that ω + ω2 + ω4 belongs to
L but not to Q.
7. Show that L = Q(ω + ω2 + ω4).
8. If q = pr, p prime, r > 0,sho w that
Ψq(X) = tp−1 + tp−2 + · · · + 1
where t = Xpr−1 .
9. Assuming that the first 6 cyclotomic polynomials are available [see
after (6.5.3)],calculate
Ψ18(X) in an effortless 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 field 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 Definitions and Comments
Let f be a polynomial with roots x1, . . . , xn in a splitting field.
Define
Δ(f) =

i<j
(xi − xj).
The discriminant of f is defined by
D(f) = Δ2 =

i<j
(xi − xj)2.
Let’s look at a quadratic polynomial f(X) = X2 + bX + c,with roots 1
2 (−b ±
√
b2 − 4c).
In order to divide by 2,w e had better assume that the characteristic
of F is not 2,and
this assumption is usually made before defining the discriminant. In
this case we have
(x1−x2)2 = b2−4c,a familiar formula. Here are some basic properties of
the discriminant.
6.6.2 Proposition
Let E be a splitting field of the separable polynomial f over F,so
that E/F is Galois.
(a) D(f) belongs to the base field F.
(b) Let σ be an automorphism in the Galois group G of f. Then σ is an
even permutation
(of the roots of f) iff σ(Δ) = Δ,and σ is odd iff σ(Δ) = −Δ.
(c) G ⊆ An,that is, G consists entirely of even permutations,iff D(f)
is the square of
an element of F (for short, D ∈ F2).
Proof. Let us examine the effect of a transposition σ = (i, j) on Δ.
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 − x7 → x3 − x10, x3 − x10 → x3 − x7
x10 − x12 → x7 − x12, x7 − x12 → x10 − x12
x7 − x8 → x10 − x8, x8 − x10 → x8 − x7
x7 − x10 → x10 − x7.
The point of the computation is that the net effect of (i, j) on Δ is
to take xi − xj to
its negative. Thus σ(Δ) = −Δ when σ is a transposition. Thus if σ is
any permutation,
we have σ(Δ) = Δ if Δ is even,and σ(Δ) = −Δ if σ is odd. Consequently,
σ(Δ2) =
18 CHAPTER 6. GALOIS THEORY
(σ(Δ))2 = Δ2,so D belongs to the fixed field of G,whic h is F. This
proves (a),and (b)
follows because Δ = −Δ (remember that the characteristic of F is not
2). Finally G ⊆ An
iff σ(Δ) = Δ for every σ ∈ G iff Δ ∈ F(G) = F. ♣
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
X3 +pX +q is −4p3 −27q2,and the discriminant of the general cubic X3
+aX2 +bX +c
is
a2(b2 − 4ac) − 4b3 − 27c2 + 18abc.
Alternatively,the change of variable Y = X + a
3 eliminates the quadratic term without
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,Problem
3). By direct enumeration, G must be A3 or S3,and by (6.6.2(c)), G =
A3 iff the
discriminant D is a square in F.
If G = A3,whic h is cyclic of order 3,there are no proper subgroups
except {1},so
there are no intermediate fields 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 α1, α2 and α3,then the corresponding fixed
fields are
F(α1), F(α2), F(α3), F(Δ)
where A3 corresponds to F(Δ) because only even permutations fix Δ.
6.6.4 Example
Let f(X) = X3 − 31X + 62 over Q. An application of the rational root
test (Section 2.9,
Problem 1) shows that f is irreducible. The discriminant is
−4(−31)3−27(62)2 = 119164−
103788 = 15376 = (124)2,whic h 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), . . . , (n−1, n).
[Since (1, j − 1)(j − 1, j)(1, j − 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 σ1 = (1, 2) and τ = (1,
2, . . . , n).
[If σ2 = τσ1τ−1,then σ2 is obtained by applying τ to the symbols of σ1
(see Section 5.2,
Problem 1). Thus σ2 = (2, 3). Similarly,
σ3 = τσ2τ
−1 = (3, 4), . . . , σn−1 = τσn−2τ
−1 = (n − 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 α is any root of f,then [F(α) : F] = n,so by the fundamental
theorem, G
contains a subgroup whose index is n. By Lagrange’s theorem, n divides
|G|. If n = p,
then by Cauchy’s theorem, G contains an element σ of order p. We can
express σ as a
product of disjoint cycles,and the length of each cycle must divide
the order of σ. Since
p is prime, σ must consist of disjoint p-cycles. But a single p-cycle
already uses up all the
symbols to be permuted,so σ is a p-cycle. ♣
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 field C,then the Galois group G of f is Sp.
Proof. By (6.6.6), G contains a p-cycle σ. Now one of the elements of
G must be complex
conjugation τ,whic h is an automorphism of C that fixes R (hence Q).
Thus τ permutes
the two nonreal roots and leaves the p − 2 real roots fixed,so τ is a
transposition. Since
p is prime, σk is a p-cycle for k = 1, . . . , p − 1. It follows that
by renumbering symbols if
necessary,w e can assume that (1, 2) and (1, 2, . . . , p) belong to
G. By (6.6.5) part (iii),
G = Sp. ♣
Problems For Section 6.6
In Problems 1–4,all polynomials are over the rational field Q,and in
each case,y ou are
asked to find the Galois group G.
1. f(X) = X3 − 2 (do it two ways)
2. f(X) = X3 − 3X + 1
3. f(X) = X5 − 10X4 + 2
4. f(X) = X3 + 3X2 − 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,
so that the Galois group is A3 or S3 according as
√
D ∈ Q or
√
D /∈ 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 flavor. We are looking for roots of f ∈
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
a ∈ F and our algorithm computes θ = n
√
a in some extension field of F,then θ is a root
of Xn − a,so it is natural to study splitting fields of Xn − a.
6.7.1 Assumptions, Comments and a Definition
Assume
(i) E is a splitting field for f(X) = Xn − a over F,where a = 0.
(ii) F contains a primitive nth root of unity ω.
These are natural assumption if we want to allow the computation of
nth roots. If θ is
any root of f in E,then the roots of f are θ,ωθ, . . . , ωn−1θ. (The
roots must be distinct
because a,hence θ,is nonzero.) Therefore E = F(θ). 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 σ ∈ G; since σ permutes the roots of f by (3.5.1),w e have
σ(θ) = ωu(σ)θ.
[Note that σ fixes ω by (ii).] We identify integers u(σ) with the same
residue mod n. If
σi(θ) = ωu(σi)θ, i = 1, 2,then
σ1(σ2(θ)) = ωu(σ1)+u(σ2)θ,
so
u(σ1σ2) = u(σ1) + u(σ2)
and u is a group homomorphism from G to Zn. If u(σ) is 0 mod n,then
σ(θ) = θ,so σ 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(θ) : F] = deg f =
n. If f is not
irreducible over F,let g be a proper irreducible factor. If β is a
root of g in E,then β is
also a root of f,so E = F(β) and |G| = [E : F] = [F(β) : F] = degg <
n. ♣
6.7. CYCLIC AND KUMMER EXTENSIONS 21
Thus splitting fields of Xn −a give rise to cyclic extensions.
Conversely,w e can prove
that a cyclic extension comes from such a splitting field.
6.7.3 Theorem
Let E/F be a cyclic extension of degree n,where F contains a primitive
nth root of
unity ω. Then for some nonzero a ∈ F, f(X) = Xn − a is irreducible
over F and E is a
splitting field for f over F.
Proof. Let σ be a generator of the Galois group of the extension. By
Dedekind’s lemma
(6.1.6),the distinct automorphisms 1, σ, σ2, . . . , σn−1 are linearly
independent over E.
Thus 1 + ωσ + ω2σ2 + · · · + ωn−1σn−1 is not identically 0,so for some
β ∈ E we have
θ = β + ωσ(β) + · · · + ωn−1σn−1(β) = 0.
Now
σ(θ) = σ(β) + ωσ2(β) + · · · + ωn−2σn−1(β) + ωn−1σn(β) = ω
−1θ
since σn(β) = β. We take a = θn. To prove that a ∈ F,note that
σ(θn) = (σ(θ))n = (ω
−1θ)n = θn
and therefore σ fixes θn. Since σ generates G,all other members of G
fix θn,hence a
belongs to the fixed field of Gal(E/F),whic h is F.
Now by definition of a, θ is a root of f(X) = Xn − a,so the roots of
Xn − a
are θ,ωθ, . . . , ωn−1θ. Therefore F(θ) is a splitting field for f
over F. Since σ(θ) = ω−1θ,
the distinct automorphisms 1, σ, . . . , σn−1 can be restricted to
distinct automorphisms
of F(θ). Consequently,
n ≤ |Gal(F(θ)/F)| = [F(θ) : F] ≤ deg f = n
so [F(θ) : F] = n. It follows that E = F(θ) and (since f must be the
minimal polynomial
of θ over F) f is irreducible over F. ♣
A finite 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 Definition
A Kummer extension is a finite Galois extension with an abelian Galois
group.
6.7.5 Theorem
Let E/F be a finite extension,and assume that F contains a primitive
nth root of unity ω.
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 ∈ F such that E
is a splitting field of
(Xn − a1) · · · (Xn − ar) over F. [For short, E = F( n
√
a1, . . . , n
√
ar).]
22 CHAPTER 6. GALOIS THEORY
Proof. We do the “if” part first. As in (6.7.1),w e have E =
F(θ1, . . . , θr) where θi is a
root of Xn − ai. If σ ∈ Gal(E/F),then σ maps θi to another root of Xn
− ai,so
σ(θi) = ωui(σ)θi.
Thus if σ and τ are any two automorphisms in the Galois group G,then
στ = τσ and G
is abelian. [The ui are integers,so ui(σ) + ui(τ) = ui(τ) + ui(σ).]
Now restrict attention
to the extension F(θi). By (6.7.2),the Galois group of F(θi)/F has
order dividing n,so
σn(θi) = θi for all i = 1, . . . , r. Thus σn is the identity,and the
exponent of G is a divisor
of n. For the “only if” part,observ e that since G is a finite 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
∼=
Ci by the first isomorphism theorem.
(Consider the projection mapping x1 · · · xr → xi ∈ Ci.) Let Ki be the
fixed field 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
i root
of unity,namely ωn/di . We conclude that Ki = F(θi),where θdi
i is a nonzero element
bi ∈ F. But θn
i = θdi(n/di)
i = bn/di
i = ai ∈ F.
Finally,in the Galois correspondence,the intersection of the Hi is
paired with the
composite of the Ki,whic h is F(θ1, . . . , θr); see Section
6.3,Problem 7. But

r
i=1 Hi = 1,
so E = F(θ1, . . . , θr),and the result follows. ♣
Problems For Section 6.7
1. Find the Galois group of the extension Q(
√
2,
√
3,
√
5,
√
7) [the splitting field of (X2−
2)(X2 − 3)(X2 − 5)(X2 − 7)] over Q.
2. Suppose that E is a splitting field for f(X) = Xn − 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 field for Xn − 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 field for f(X) = Xp − c over
F,where c = 0, p is
prime and the characteristic of F is not p. Let ω 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 ≤ d < p. Let θ be a root of g in E.
4. Let g0 be the product of the roots of g. (Since g0 is ± the
constant term of g, g0 ∈ F.)
Show that gp
0 = θdp = 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 Xp − c is not irreducible over F,then it must
have a root
in F.
6.8. SOLVABILITY BY RADICALS 23
6. Continuing Problem 5,sho w that if Xp −c is not irreducible over
F,then E = F(ω).
7. Continuing Problem 6,sho w that if Xp − c is not irreducible over
F,then Xp − 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 σ be a generator of G = Gal(E/F). It is a consequence of Hilbert’s
Theorem 90 (see
the Problems for Section 7.3) that there is an element θ ∈ E such that
σ(θ) = θ + 1.
Prove the Artin-Schreier theorem:
8. E = F(θ).
9. θ is a root of f(X) = Xp − X − a for some a ∈ F.
10. f is irreducible over F (hence a = 0).
Conversely,Let F be a field of prime characteristic p,and let E be a
splitting field for
f(X) = Xp − X − a,where a is a nonzero element of F.
11. If θ is any root of f in E,sho w that E = F(θ) 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,sho w that the Galois group of f is
cyclic of order p.
6.8 Solvability By Radicals
6.8.1 Definitions and Comments
We wish to solve the polynomial equation f(X) = 0, f ∈ F[X],under the
restriction that
we are only allowed to perform ordinary arithmetic operations
(addition,subtraction,
multiplication and division) on the coefficients,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 ≤ F(α1) ≤ F(α1, α2) ≤ ·· · ≤ F(α1, . . . , αr) = E
where αn1
1
∈ F and αni
i
∈ F(α1, . . . , αi−1), i = 2, . . . , r. Equivalently,w e have
F = F0 ≤ F1 ≤ ·· · ≤ Fr = E
where Fi = Fi−1(αi) and αni
i
∈ Fi−1, i = 1, . . . , r. We say that E is a radical extension
of F. It is convenient (and legal) to assume that n1 = · · · = nr = n.
(Replace each ni
by the product of all the ni. To justify this,observ e that if αj
belongs to a field L,then
αmj ∈ L,m = 2, 3, . . . .) Unless otherwise specified,w e 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,w e assume that the base field
contains a primitive
nth root of unity.)
24 CHAPTER 6. GALOIS THEORY
We say that the polynomial f ∈ 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 finite,but it need not be normal or
separable. We
will soon specialize to characteristic 0,whic h 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 α1, . . . ,
αr,where αi is the nth
root of an element in Fi−1. On the other hand, N is obtained from F by
adjoining
not only the αi,but their conjugates αi1, . . . , αim(i). For any
fixed i and j,there is an
automorphism σ ∈ Gal(N/F ) such that σ(αi) = αij (see (3.2.3),(3.5.5)
and (3.5.6)).
Thus
αn
ij = σ(αi)n = σ(αn
i )
and since αn
i belongs to F(α1, . . . , αi−1),it follows from (3.5.1) that σ(αn
i ) belongs to
the splitting field Ki of

i−1
j=1min(αj, F) over F. [Take K1 = F,and note that since
αn
1 = b1 ∈ F,we have σ(αn
1) = σ(b1) = b1 ∈ F. Alternatively,observ e that by (3.5.1), σ
must take a root of Xn − b1 to another root of this polynomial.] Thus
we can display N
as a radical extension of F by successively adjoining
α11, . . . , α1m(1), . . . , αr1, . . . , αrm(r). ♣
6.8.3 Preparation for the Main Theorem
If F has characteristic 0,then a primitive nth root of unity ω can be
adjoined to F to
reach an extension F(ω); see (6.5.1). If E is a radical extension of F
and F = F0 ≤
F1 ≤ ·· · ≤ Fr = E,w e can replace Fi by Fi(ω), i = 1, . . . , r,and
E(ω) will be a radical
extension of F. By (6.8.2),w e can pass from E(ω) to its normal
closure over F. Here is
the statement we are driving at:
Let f ∈ F[X],where F has characteristic 0. If f is solvable by
radicals,then there
is a Galois radical extension N = Fr ≥ ·· · ≥ F1 ≥ F0 = F containing a
splitting field
K for f over F,suc h that each intermediate field Fi, i = 1, . . . ,
r,con tains a primitive
nth root of unity ω. We can assume that F1 = F(ω) and for i > 1, Fi is
a splitting
field for Xn − bi over Fi−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),eac h Fi/Fi−1, i =
2, . . . , r is a cyclic
(Galois) extension.
6.8. SOLVABILITY BY RADICALS 25
We now do some further preparation. Suppose that K is a splitting
field for f over F,
and that the Galois group of K/F is solvable,with
Gal(K/F) = H0  H1  · · ·  Hr = 1
with each Hi−1/Hi abelian. By the fundamental theorem (and Section
6.2,Problem 4),
we have the corresponding sequence of fixed fields
F = K0 ≤ K1 ≤ · · · ≤ Kr = K
with Ki/Ki−1 Galois and Gal(Ki/Ki−1) isomorphic to Hi−1/Hi. Let us
adjoin a primitive
nth root of unity ω to each Ki,so that we have fields Fi = Ki(ω) with
F ≤ F0 ≤ F1 ≤ ·· · ≤ Fr.
We take n = | Gal(K/F)|. Since Fi can be obtained from Fi−1 by
adjoining everything
in Ki \ Ki−1,we have
Fi = Fi−1Ki = KiFi−1
the composite of Fi−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 Fi−1,and on the bottom Ki ∩ Fi−1 ⊇ Ki−1 (see Figure
6.8.1). We conclude
that Fi/Fi−1 is Galois,with a Galois group isomorphic to a subgroup of
Gal(Ki/Ki−1).
Since Gal(Ki/Ki−1) ∼=
Hi−1/Hi,it follows that Gal(Fi/Fi−1) is abelian. Moreover,the
exponent of this Galois group divides the order of H0,whic h coincides
with the size of
Gal(K/F). (This explains our choice of n.)
Fi

    
Ki Fi−1
Ki ∩ Fi−1

    
Ki−1
Figure 6.8.1
6.8.4 Galois’ Solvability Theorem
Let K be a splitting field 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.
26 CHAPTER 6. GALOIS THEORY
Proof. If f is solvable by radicals,then as in (6.8.3),w e have
F = F0 ≤ F1 ≤ ·· · ≤ Fr = N
where N/F is Galois, N contains a splitting field K for f over F,and
each Fi/Fi−1
is Galois with an abelian Galois group. By the fundamental theorem
(and Section 6.2,
Problem 4),the corresponding sequence of subgroups is
1 = Hr
Hr−1
· · ·
H0 = G = Gal(N/F )
with each Hi−1/Hi abelian. Thus G is solvable,and since
Gal(K/F) ∼=
Gal(N/F )/Gal(N/K)
[map Gal(N/F ) → Gal(K/F) by restriction; the kernel is Gal(N/
K)],Gal( K/F) is solvable
by (5.7.4).
Conversely,assume that Gal(K/F) is solvable. Again as in (6.8.3),w e
have
F ≤ F0 ≤ F1 ≤ ·· · ≤ Fr
where K ≤ Fr,eac h Fi contains a primitive nth root of unity,with n =
| Gal(K/F)|,
and Gal(Fi/Fi−1) is abelian with exponent dividing n for all i =
1, . . . , r. Thus each
Fi/Fi−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),eac h Fi/Fi−1 is a radical
extension. By transitivity
(see (6.8.1)), Fr is a radical extension of F. Since K ⊆ Fr, f is
solvable by radicals. ♣
6.8.5 Example
Let f(X) = X5 − 10X4 + 2 over the rationals. The Galois group of f is
S5,whic h is not
solvable. (See Section 6.6,Problem 3 and Section 5.7,Problem 5.) Thus
f is not solvable
by radicals.
There is a fundamental idea that needs to be emphasized. The
significance of Galois’
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,w e will sketch another classical problem,that of
constructions with ruler
and compass. In Euclidean geometry,w e start with two points (0, 0)
and (1, 0),and we
are allowed the following constructions.
(i) Given two points P and Q,w e can draw a line joining them;
(ii) Given a point P and a line L,w e can draw a line through P
parallel to L;
(iii) Given a point P and a line L,w e can draw a line through P
perpendicular to L;
(iv) Given two points P and Q,w e can draw a circle with center at P
passing through Q;
6.8. SOLVABILITY BY RADICALS 27
(v) Let A,and similarly B,b e a line or a circle. We can generate new
points,called
constructible points,b y forming the intersection of A and B. If (c,
0) (equivalently
(0, c)) is a constructible point,w e call c a constructible number. It
follows from (ii)
and (iii) that (a, b) is a constructible point iff a and b are
constructible numbers. It
can be shown that every rational number is constructible,and that the
constructible
numbers form a field. 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 iff there are real fields Q = F0 ≤ F1 · · · ≤ Fr
such that c ∈ Fr and
each [Fi : Fi−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 α = cos 20◦
is constructible.
Using the identity
ei3θ = cos 3θ + i sin 3θ = (cos θ + i sin θ)3,
reach a contradiction.
2. (Duplicating the cube) Show that it is impossible to construct,with
ruler and compass,
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 π,
then π would be algebraic over Q. (It is known that π is
transcendental over Q.)
To construct a regular n-gon,that is,a regular polygon with n sides, n
≥ 3,we must
be able to construct an angle of 2π/n; equivalently,cos 2π/n must be a
constructible
number. Let ω = ei2π/n,a primitive nth root of unity.
4. Show that [Q(ω) : Q(cos 2π/n)] = 2.
5. Show that if a regular n-gon is constructible,then the Euler phi
function ϕ(n) is a
power of 2.
Conversely,assume that ϕ(n) is a power of 2.
6. Show that Gal(Q(cos 2π/n)/Q) is a 2-group,that is,a p-group with p
= 2.
7. By Section 5.7,Problem 7,ev ery nontrivial finite 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. ¿From the preceding,a regular n-gon is constructible if and only if
ϕ(n) is a power
of 2. Show that an equivalent condition is that n = 2sq1 · · · 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.
9. Show that if 2m + 1 is prime,then m must be a power of 2. The only
known Fermat
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 Xa + 1 in Z[X]; the
quotient is Xa−1 −
Xa−2 + · · ·−X + 1 (since a − 1 is even).]
Let F be the field of rational functions in n variables e1, . . . , en
over a field K with
characteristic 0,and let f(X) = Xn − e1Xn−1 + e2Xn−2 − ·· · + (−1)nen
∈ F[X]. If
28 CHAPTER 6. GALOIS THEORY
α1, . . . , αn are the roots of f in a splitting field over F,then the
ei are the elementary
symmetric functions of the αi. Let E = F(α1, . . . , αn),so that E/F
is a Galois
extension and G = Gal(E/F) is the Galois group of f.
10. Show that G
∼=
Sn.
11. What can you conclude from Problem 10 about solvability of
equations?
6.9 Transcendental Extensions
6.9.1 Definitions and Comments
An extension E/F such that at least one α ∈ 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 profitable 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| ≤ m. We are going to replace linear independence by algebraic
independence
and spanning by algebraic spanning. We will find 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 defined 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 development
is given in a sequence of exercises. See also Morandi,“Fields and
Galois Theory”,
pp. 173–182.
Let E/F be an extension. The elements t1, . . . , tn ∈ E are
algebraically dependent
over F (or the set {t1, . . . , tn} is algebraically dependent over F)
if there is a nonzero
polynomial f ∈ F[X1, . . . , Xn] such that f(t1, . . . , tn) = 0;
otherwise the ti are algebraically
independent over F. Algebraic independence of an infinite set means
algebraic
independence of every finite 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
“linear” by
“algebraic”,w e say that the element t ∈ E depends algebraically on T
over F if t is
algebraic over F(T),the field 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,w e will frequently regard F as fixed and drop the phrase “over
F”.)
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.
6.9. TRANSCENDENTAL EXTENSIONS 29
Thus by (ii), S is a transcendence basis for E/F iff S is
algebraically independent and E
is algebraic over F(S).
Proof. (i) implies (ii): If S ⊂ T where T is algebraically
independent,let u ∈ T \ S.
Then u cannot depend on S algebraically (by algebraic independence of
T),so S cannot
span E algebraically.
(ii) implies (i): If S does not span E algebraically,then there exists
u ∈ E such
that u does not depend algebraically on S. But then S ∪{u} is
algebraically independent,
contradicting maximality of S.
(i) implies (iii): If T ⊂ S and T spans E algebraically,let u ∈ S \ T.
Then u depends
algebraically on T,so T ∪ {u},hence S,is algebraically dependent,a
contradiction.
(iii) implies (i): If S is algebraically dependent,then some u ∈ S
depends algebraically
on T = S \ {u}. But then T spans E algebraically,a contradiction. ♣
6.9.3 Proposition
Every transcendental extension has a transcendence basis.
Proof. The standard argument via Zorn’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,whic h is a
transcendence basis
by (6.9.2). ♣
For completeness,if E/F is an algebraic extension,w e can regard ∅ as
a transcendence
basis.
6.9.4 The Steinitz Exchange
If {x1, . . . , xm} spans E algebraically and S ⊆ E is algebraically
independent,then
|S| ≤ 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,sa y 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,sa y
xi+1 for convenience. (Otherwise, S would be algebraically dependent.)
Then xi+1 depends
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 y’s than x’s,ev entually the x’s disappear,and
y1, . . . , ym span E
algebraically. But then ym+1 depends algebraically on y1, . . . ,
ym,con tradicting the algebraic
independence of S. ♣
30 CHAPTER 6. GALOIS THEORY
6.9.5 Corollary
Let S and T be transcendence bases of E. Then either S and T are both
finite or they
are both infinite; in the former case, |S| = |T|.
Proof. Assume that one of the transcendence bases,sa y T,is finite. By
(6.9.4), |S| ≤ |T|,
so S is finite also. By a symmetrical argument, |T| ≤ |S|,so |S| = |
T|. ♣
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),w e may assume that S and T are both infinite. Let T
= {yi : i ∈ I}.
If x ∈ S,then x depends algebraically on finitely many elements
yi1, . . . , yir in T. Define
I(x) to be the set of indices {i1, . . . , ir}. It follows that I =
∪{I(x): x ∈ 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,con tradicting (6.9.2) part (iii). Now an element
of ∪{I(x): x ∈ S}
is determined by selecting an element x ∈ S and then choosing an index
in I(x). Since
I(x) is finite,w e have |I(x)| ≤ ℵ0. Thus
|I| = |

{I(x): x ∈ S}| ≤ |S|ℵ0 = |S|
since S is infinite. Thus |T| ≤ |S|. By symmetry, |S| = |T|. ♣
6.9.7 Example
Let E = F(X1, . . . , Xn) be the field of rational functions in the
variables X1, . . . , Xn
with coefficients 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 = {Xu1
1 , . . . , Xun
n
},where u1, . . . , un are arbitrary positive integers. We
claim that T is also a transcendence basis. As above, T is
algebraically independent.
Moreover,eac h Xi is algebraic over F(T). To see what is going on,lo
ok at a concrete
example,sa y T = {X5
1,X3
2,X4
3
}. If f(Z) = Z3−X3
2
∈ F(T)[Z],then X2 is a root of f,so
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 ⊆ T,sho w that there is a transcendence basis B such that S ⊆ B
⊆ 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} ⊆ E,that the
following conditions are equivalent.
6.9. TRANSCENDENTAL EXTENSIONS 31
(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, . . . , ti−1) (where the statement
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 ∈ E\S,
then t is transcendental over F(S) if and only if S ∪ {t} is
algebraically independent
over F.
[Problems 3 and 4 suggest the reasoning that is involved in
formalizing the results of this
section.]
5. Let F ≤ K ≤ 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∪T
is algebraically
independent over F,and S ∩ T = ∅.
6. Let F ≤ K ≤ E,with S a transcendence basis for K/F and T a
transcendence basis
for E/K. Show that S∪T 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 finite subset
of E. Show that
F(T) is F-isomorphic to the rational function field F(X1, . . . , Xn)
if and only if T is
algebraically independent over F.
8. An algebraic function field F in one variable over K is a field F/K
such that there
exists x ∈ F transcendental over K with [F : K(x)] < ∞. If z ∈ F,sho w
that z is
transcendental over K iff [F : K(z)] < ∞.
9. Find the transcendence degree of the complex field 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 Definition
If x1, . . . , xn (n ≥ 2) are arbitrary elements of a field,the
Vandermonde determinant of
the xi is
det V =

1 1 · · · 1
x1 x2 · · · xn
...
xn−1
1 xn−1
2
· · · xn−1
n

32 CHAPTER 6. GALOIS THEORY
A6.2 Proposition
det V =

i<j
(xj − xi).
Proof. det V is a polynomial h of degree 1 + 2 + · · · + (n − 1) = (n2
) in the variables
x1, . . . , xn,as is g =

i<j(xj − xi). If xi = xj for i < j,then the determinant is 0,so
by the remainder theorem (2.5.2),eac h 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
and g,i.e.,those terms in which xn appears to as high a power as
possible,and subject
to this constraint, xn−1 appears to as high a power as possible,etc.
In both cases,the
leading term is x2x23
· · · xn−1
n ,and therefore c must be 1. (For this step it is profitable to
regard the xi as abstract variables in a polynomial ring. Then
monomials xr1
1
· · · xrn
n with
different sequences (r1, . . . , rn) of exponents are linearly
independent.) ♣
A6.3 Corollary
If f is a polynomial in F[X] with roots x1, . . . , xn in some
splitting field over F,then the
discriminant of f is (det V )2.
Proof. By definition of the discriminant D of f (see 6.6.1),w e have D
= Δ2 where
Δ = ±det V . ♣
A6.4 Computation of the Discriminant
The square of the determinant of V is det(V V t),whic h is the
determinant of


1 1 · · · 1
x1 x2 · · · xn
...
xn−1
1 xn−1
2
· · · xn−1
n



 1
x1 ·
·
·
x
n−1
1
1 x2 · · · xn−1
2
...
1 xn . . . xn−1
n


and this in turn is

t0 t1 · · · tn−1
t1 t2 · · · tn
...
tn−1 tn · · · t2n−2

where the power sums tr are given by
t0 = n, tr =
n
i=1
xri
, r ≥ 1.
We must express the power sums in terms of the coefficients of the
polynomial f. This
will involve,improbably ,an exercise in differential calculus. We have
F(z) =
n
i=1
(1 − xiz) =
n
i=0
cizi with c0 = 1;
6.9. TRANSCENDENTAL EXTENSIONS 33
the variable z ranges over real numbers. Take the logarithmic
derivative of F to obtain
F(z)
F(z)
= d
dz
log F(z) =
n
i=1
−xi
1 − xiz
= −
n
i=1
∞
j=0
xj+1
i zj = −
∞
j=0
tj+1zj .
Thus
F
(z) + F(z)
∞
j=0
tj+1zj = 0,
that is,
n
i=1
icizi−1 +
n
i=0
cizi
∞
j=1
tjzj−1 = 0.
Equating powers of zr−1,w e have,assuming that n ≥ r,
rcr + c0tr + c1tr−1 + · · · + cr−1t1 = 0; (1)
if r > n,the first summation does not contribute,and we get
tr + c1tr−1 + · · · + cntr−n = 0. (2)
Our situation is a bit awkward here because the roots of F(z) are the
reciprocals of the xi.
The xi are the roots of

n
i=0 aizi where ai = cn−i (so that an = c0 = 1). The results can
be expressed as follows.
A6.5 Newton’s Identities
If f(X) =

n
i=0 aiXi (with an = 1) is a polynomial with roots x1, . . . , xn,then
the power
sums ti satisfy
tr + an−1tr−1 + · · · + an−r+1t1 + ran−r = 0, r ≤ n (3)
and
tr + an−1tr−1 + · · · + a0tr−n = 0, r>n. (4)
A6.6 The Discriminant of a Cubic
First consider the case where the X2 term is missing,so that f(X) = X3
+pX +q. Then
n = t0 = 3, a0 = q, a1 = p, a2 = 0 (a3 = 1). Newton’s identities yield
t1 + a2 = 0, t1 = 0; t2 + a2t1 + 2a1 = 0, t2 = −2p;
t3 + a2t2 + a1t1 + 3a0 = 0, t3 = −3a0 = −3q;
t4 + a2t3 + a1t2 + a0t1 = 0, t4 = −p(−2p) = 2p2
D =

3 0 −2p
0 −2p −3q
−2p −3q 2p2

= −4p3 − 27q2.
34 CHAPTER 6. GALOIS THEORY
We now go to the general case f(X) = X3 + aX2 + bX + c. The quadratic
term can be
eliminated by the substitution Y = X + a
3 . Then
f(X) = g(Y ) = (Y − a
3
)3 + a(Y − a
3
)2 + b(Y − a
3
) + c
= Y 3 + pY + q where p = b − a2
3 , q =
2a3
27
− ba
3
+ c.
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 = −4p3 − 27q2,whic h
simplifies to
D = a2(b2 − 4ac) − 4b3 − 27c2 + 18abc.
We now consider the Galois group of a quartic X4 + aX3 + bX2 + cX +
d,assumed
irreducible and separable over a field F. As above,the translation Y =
X + a
4 eliminates
the cubic term without changing the Galois group,so we may assume that
f(X) =
X4 + qX2 + 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 verification (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 ∩ G
G by the second isomorphism theorem.
A6.7 Lemma
F(V ∩ 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 fixes u, v and w,so GF(u, v,w) ⊇ V ∩ G. If
σ ∈ G
but σ /∈ V ∩ G then (again by direct verification) σ 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 σ /∈ GF(u,
v,w). Therefore
GF(u, v,w) = V ∩ G,and an application of the fixed field operator F
completes the
proof. ♣
A6.8 Definition
The resolvent cubic of f(X) = X4 + qX2 + rX + s is g(X) = (X − u)(X −
v)(X − w).
To compute g,w e must express its coefficients in terms of q, r and s.
First note that
u−v = −(x1 −x4)(x2 −x3), u−w = −(x1 −x3)(x2 −x4), v−w = −(x1 −x2)(x3
−x4).
Thus f and g have the same discriminant. Now
X4 + qX2 + rX + s = (X2 + kX + l)(X2 − kX + m)
where the appearance of k and −k is explained by the missing cubic
term. Equating
coefficients gives l +m− k2 = q, k(m− l) = r, lm = s. Solving the
first two equations for
m and adding,w e have 2m = k2 + q + r/k,and solving the first two
equations for l and
6.9. TRANSCENDENTAL EXTENSIONS 35
adding,w e get 2l = k2 + q − r/k. Multiply the last two equations and
use lm = s to get
a cubic in k2,namely
k6 + 2qk4 + (q2 − 4s)k2 − r2 = 0.
(This gives a method for actually finding the roots of a quartic.) To
summarize,
f(X) = (X2 + kX + l)(X2 − kX + m)
where k2 is a root of
h(X) = X3 + 2qX2 + (q2 − 4s)X − r2.
We claim that the roots of h are simply −u,−v,−w. For if we arrange
the roots of f so
that x1 and x2 are the roots of X2+kX +l,and x3 and x4 are the roots
of X2−kX +m,
then k = −(x1 + x2),−k = −(x3 + x4),so −u = k2. The argument for −v
and −w
is similar. Therefore to get g from h,w e 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) = X3 − 2qX2 + (q2 − 4s)X + r2.
We need some results concerning subgroups of Sn, n ≥ 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 first assertion of (i),see Section 5.6,Problem 4. For
the second assertion
of (i),note that
(a, b)(a, c)(a, b)−1(a, c)−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,
S
n
≤ H,and since An also has index 2,the same argument gives S
n
≤ An. By (i),
An ≤ S
n,so An = S
n
≤ H. Since An and H have the same finite number of elements
n!/2,it follows that H = An. ♣
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∩V ).
Then
(a) If m = 6,then G = S4;
36 CHAPTER 6. GALOIS THEORY
(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,equalit y is up to isomorphism.]
Proof. If m = 6 or 3,then since |G| = m|G∩ 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 ∩ V )| = |S4/V | = 24/4 = 6
and
|A4/(A4 ∩ V )| = |A4/V | = 12/4 = 3
proving both (a) and (b). If m = 1,then G = G∩V ,so G ≤ V ,and since |
G| is a multiple
of 4 and |V | = 4,we have G = V ,pro ving (c).
If m = 2,then |G| = 2|G ∩ V |,and since |V | = 4, |G ∩ V | is 1, 2 or
4. If it is 1,
then |G| = 2× 1 = 2,con tradicting the hypothesis. If it is 2,then |G|
= 2× 2 = 4, and
G = Z4 or V (the only groups of order 4). Finally,assume |G∩ 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,eac h 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∩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,pro ving (e). ♣
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 ∩ 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 field of a separable polynomial,so L/F is Galois.
Consequently,[ L : F] = m
by (3.5.9). To apply (A6.11),w e 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
6.9. TRANSCENDENTAL EXTENSIONS 37
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 = G∩V = Gal(E/
L) by (A6.7).
[E is a splitting field for f over F.] Since V is transitive,for each
i, j = 1, 2, 3, 4, i = j,
there is an L-automorphism τ of E such that τ (xi) = xj . Applying τ
to the equation
h(xi) = 0,where h is the minimal polynomial of xi over L,w e see that
each xj is a root
of h,and therefore f | h. But h | f by minimality of h,so h = f,pro
ving that f is
irreducible over L.
Finally,assume m = 2 and G = Z4,whic h we take as {1, (1, 2, 3, 4),
(1, 3)(2, 4),
(1, 4, 3, 2)}. Then G ∩ V = {1, (1, 3)(2, 4)},whic h 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. ♣
A6.13 Example
Let f(X) = X4 + 3X2 + 2X + 1 over Q,with q = 3, r = 2, s = 1. The
resolvent cubic is,
by (A6.9), g(X) = X3 −6X2+5X +4. To calculate the discriminant of g,w
e can use the
general formula in (A6.6),or compute g(X+2) = (X +2)3 −6(X +2)2 +5(X
+2)+4 =
X3 −7X −2. [The rational root test gives irreducibility of g and
restricts a factorization
of f to (X2 + aX ± 1)(X2 − aX ± 1), a ∈ Z,whic h is impossible. Thus f
is irreducible
as well.] We have D(g) = −4(−7)3 − 27(−2)2 = 1264,whic h is not a
square in Q. Thus
m = 6,so the Galois group of f is S4.

[Jeff Rubard]

Post by Jeff Rubard
X3 −7X −2. [The rational root test gives irreducibility of g and
restricts a factorization
of f to (X2 + aX ± 1)(X2 − aX ± 1), a ∈ Z,whic h is impossible. Thus f
is irreducible
as well.] We have D(g) = −4(−7)3 − 27(−2)2 = 1264,whic h is not a
square in Q. Thus
m = 6,so the Galois group of f is S4.

Add: Film at 11.

[Jeff Rubard]

Post by Jeff Rubard

Post by Jeff Rubard
X3 −7X −2. [The rational root test gives irreducibility of g and
restricts a factorization
of f to (X2 + aX ± 1)(X2 − aX ± 1), a ∈ Z,whic h is impossible. Thus f
is irreducible
as well.] We have D(g) = −4(−7)3 − 27(−2)2 = 1264,whic h is not a
square in Q. Thus
m = 6,so the Galois group of f is S4.

Add: Film at 11.

/Superadd/: the mathematician I admire most -- 'father, son, and holy
ghost' /in this respect/ --
http://aleph0.clarku.edu/~djoyce/hilbert/problems.html
[Like that - Ed]

[Jeff Rubard]

Post by Jeff Rubard

Post by Jeff Rubard
Add: Film at 11.

/Superadd/: the mathematician I admire most -- 'father, son, and holy
ghost' /in this respect/ --http://aleph0.clarku.edu/~djoyce/hilbert/problems.html
[Like that - Ed]

An *allusion couverte* [!!] to Gentzen -- probably /both/ of them --