1. No category

Galois Theory III.
3.1. Splitting fields.
We know how to construct a field extension L of a given field K where a given
irreducible polynomial P (X) ∈ K[X] has a root. We need a field extension of K
where P (X) has all its roots.
Definition. Let E be a field and f (X) ∈ E[X]. A field extension F ⊃ E
is called a splitting field for f (X) over E if F = E(α1 , . . . , αn ) with f (X) =
c(X − α1 ) . . . (X − αn ), where c ∈ K ∗ is the (non-zero) leading coefficient of F (X).
Notice that F must contain “all” roots of f (X) and it is a minimal field with
this property.
Examples:
(1) C is a spliting field for X 2 + 1 over R;
(2) C is a splitting field for X 4 + 1 over R;
(3) C is not a splitting field for X 2 + 1 over Q (C is too large);
(4) Q(i) is a splitting field for X 2 + 1 over Q;
(5) Q is a splitting field for X 2 − 4 over Q;
√ √
(6) Q( 3 2, −3) is a splitting field for X 3 − 2 over Q.
Theorem (Existence of splitting fields). For any field E and any polynomial
f (X) ∈ E(X) there is a splitting field F ⊃ E for f (X) over E and [F : E] 6 n!,
where deg f (X) = n.
Proof. Use induction on n = deg f (X).
If n = 1 then f (X) is linear and F = E.
Let n > 1 and assume our theorem is proved for polynomials of degree < n.
Choose an irreducible factor m(X) for f (X) in E[X]. Consider the field extension
E1 = E(α1 ) ⊃ E, where α1 is a root of m(X). Notice that [E(α1 ) : E] = deg m(X)
and f (X) = (X − α1 )f1 (X), where f1 (X) ∈ E1 [X] and deg f1 (X) = n − 1 < n.
By inductive assumption, there is a splitting field for f1 (X) over E1 . In other
words, there is a field extension F ⊃ E1 such that f1 (X) = c(X − α2 ) . . . (X − αn )
and F = E1 (α2 , . . . , αn ). But E1 = E(α1 ) implies that F = E(α1 , α2 , . . . , αn ) and
f (X) = (X − α1 )f1 (X) = c(X − α1 )(X − α2 ) . . . (X − αn ). So, F is a splitting field
for f (X) over E. It remains to notice that
[F : E] = [F : E1 ][E1 : E] 6 (n − 1)! deg m(X) 6 (n − 1)!n = n!.
1
e be fields and let σ :
Theorem (Uniqueness of splitting fields). Let E and E
e be a field isomorphism. Let f (X) ∈ E[X] and σ(f (X)) = f˜(X) ∈ E[X].
e
E −→ E
Let K be a splitting field for f (X) over E.
e be a splitting field for f˜(X) over E.
e
Let K
e such that τ |E = σ : E −→ E.
e
Then there is a field isomorphism τ : K −→ K
Proof. Use induction on n = deg f (X) = deg f˜(X).
e =E
e and we can take τ = σ.
If n = 1 then K = E and K
Suppose n > 1 and choose an irreducible factor m(X) of f (X) in E[X]. Then
m(X)
˜
= σ(m(X)) is an irreducible factor of f˜(X). Choose a root α ∈ K of m(X)
e of m(X).
and a root α
˜∈K
˜
Because the polynomial m(X)
˜
is obtained from m(X)
e α) such
via the field isomorphism σ there is a field isomorphism τ1 : E(α) −→ E(˜
that τ1 |E coincides with σ and τ1 (α) = α
˜ . (Use the uniqueness of a minimal field
extension where a given polynomial has a root.) But then notice that K became
e is
a splitting field of f1 (X) over E(α), where f (X) = (X − α)f1 (X). Similarly, K
a splitting field of f˜1 (X) over E(˜
α), where f˜(X) = (X − α
˜ )f˜1 (X). It remains to
˜
notice that the degree of f1 (X) is n − 1 < n and f1 (X) is obtained from f1 (X) via
e α). Therefore, by inductive assumption τ1
the field isomorphism τ1 : E(α) −→ E(˜
e
can be extended to required field isomorphism τ from K to K.
3.2. Normal extensions.
Definition. Let F/E be a field extension. It is normal if the following condition
is satisfied: whenever P (X) ∈ E[X] is irreducible and has a root in F then P (X)
splits completely in F [X].
In other words, a field F is normal over its subfield E if any irreducible polynomial P (X) from E[X] with a root in F has all its roots in F .
Examples;
a) If [F : E] = 1 then F is normal over E;
b) Prove that any quadratic extension F/E is normal. Indeed, suppose P (X) ∈
E[X] is irreducible and let θ ∈ F is such that P (θ) = 0. Then E ⊂ E(θ) ⊂ F and
because [F : e] = 2 we have [E(θ) : E] = deg P (X) is either 1, or 2.
If deg P (X) = 1 there is nothing to prove.
If deg P (X) = 2, then P (X) = c(X − θ)(X − θ0 ) in F [X]. This implies that
c(θ + θ0 ) ∈ F and, therefore, θ0 ∈ F . So, P (X) splits completely in F [X].
c) Not every cubic extension is normal. Indeed, take E = Q and F = Q(θ),
where θ is a root of X 3 − 2 ∈√Q[X]. Then two remaining roots of X 3 − 2 are
θζ3 and θζ32 , where ζ3 = (−1 + −3)/2. If θζ3 ∈ Q(θ) then ζ3 ∈ Q(θ). Therefore,
Q ⊂ Q(ζ3 ) ⊂ Q(θ). This is impossible because this would imply that 2 = [Q(ζ3 ) : Q]
divides 3 = [Q(θ) : Q].
d) Give an example of cubic normal extension of Q. Let f (X) = X 3 − 3X − 1 ∈
Q[X]. It is irreducible and if θ1 ∈ C is its root then F = Q(θ1 ) is cubic extension
of Q.
Apply Cardano formulas to find explicitly all roots of f (X) in C. Any root of
f (X) appears in the form u + v, where uv = 1 and u3 + v 3 = 1. Therefore, u3
and v 3 are roots exp(±iπ/3) = cos(π/3) ± i sin(π/3) of the quadratic polynomial
2
T 2 − T + 1 = 0. This gives the following three pairs of solutions (u, v): (exp(iπ/9),
exp(−iπ/9), (exp(7iπ/9), exp(−7iπ/9)) and (exp(13iπ/9), exp(−13iπ/9)). Therefore, the roots of our cubic polynomial f (X) are θ1 = 2 cos(π/9), θ2 = cos(7π/9)
and θ3 = cos(13π/9). This implies that θ2 = −2 cos(2π/9) = 2(1 − 2θ12 ) ∈ Q(θ1 ).
And θ3 ∈ Q(θ1 ) because θ1 + θ2 + θ3 = 0.
Theorem (Criterion of normality). A finite field extension F/E is normal if
and only if there is a polynomial G(X) ∈ E[X] such that F is a splitting field for
G(X) over E.
Proof. Suppose, first, that F is normal over E. Choose a finite number of elements
α1 , . . . , αr ∈ F such that F = E(α1 , . . . , αr ). (Explain, why such finite set of elements exists.) For 1 6 i 6 r, let Pi (X) ∈ E[X] be the minimal polynomial for αi
over E. Each polynomial Pi (X) is irreducible and has a root αi in F , therefore,
Pi (X) splits completely in F [X]. (Use that F/E is normal.) Let G(X) be the
product P1 (X) . . . Pr (X). Then G(X) splits completely in F [X], therefore, F contains the splitting field EG for G(X) over E. On the other side, F = E(α1 , . . . , αr )
contains EG because all αi are roots of G(X) and EG is generated over E by all
roots of G(X). So, F = EG .
Prove our theorem in the opposite direction. Suppose F = EG is the splitting
field for some polynomial G(X) ∈ E[X]. Take any irreducible P (X) ∈ E[X] such
that P (a) = 0 for some a ∈ F . We must prove that P (X) splits completely over F .
Let q(X) be an irreducible factor of P (X) in F [X]. Consider the minimal field
extension F (b) of F , where b is a root of q(X). WE must prove that b ∈ F , i.e.
F = F (b).
Step 1. a and b are roots of the same irreducible polynomial P (X) ∈ E[X].
Therefore, there is a field isomorphism σ : E(a) −→ E(b) such that σ(a) = b and
σ|E = id. Notice that [E(a) : E] = [E(b) : E] = deg P (X).
Step 2. We know that F is generated by all roots of G(X) ∈ E[X] over E.
This iomplies that the roots of G(X) generate the field extensions F (a)/E(a) and
F (b)/E(b). In particular, F (a) is the splitting field for G(X) over E(a) and F (b)
is the splitting field for F (b) over E(b).
Step 3. By the uniqueness property of splitting fields, the field isomorphism σ
from step 1 can be extended to a field isomorphism τ : F (a) −→ F (b). In particular,
[F (a) : E(a)] = [F (b) : E(b). Remind that a ∈ F and, therefore, F = F (a). It
remains to count the degrees: [F : E] = [F (a) : E] = [F (a) : E(a)][E(a) : E] =
[F (b) : E(b)][E(b) : E] = [F (b) : E] = [F (b) : F ][F : E]. Therefore, [F (b) : F ] = 1,
i.e. F = F (b) and b ∈ F .
Our Theorem is completely proved.
Examples.
√ √ √ √
1) Q( 2, 3, 5, 7) is normal over Q. (It is a splitting field of (X 2 − 2)(X 2 −
3)(X 2 − 5)(X 2 − 7) ∈ Q[X].);
√ √
2) Q( 3 2, −3) is the splitting field of X 3 − 2 ∈ Q[X] and is, therefore, normal
over Q;
√
3) Q( 3 2) is not normal
over Q. (Consider X 3 − 2 ∈ Q[X] and prove that it has
√
3
only one root in Q( 2).);
3
√
√
4) Q( 4 2) is not normal but Q( 4 2, i) is normal over Q.
5) If α is a root of X 3 + X + 1 ∈ F2 [X] then F2 (α) is a normal extension of F2
of degree 3; (Later we shall prove that any finite field extension of Fp , where p is a
prime number, is automatically normal.)
6) Let K = Fp (T ) = Frac Fp [T ] be a field of rational functions in one variable T
over prime field Fp . Then the polynomial F (X) = X p − T ∈ K[X] is irreducible
and L = K(θ), where θ is a root of P (X), is normal over K.
3.3. Why normal extensiona are good but not good enough?
Consider a finite field extension. The Galois Theory deals with “symmetries” of L
over K. These symmetries are field automorphisms σ : L −→ L such that σ|K = id.
For a given field extension L/K we shall denote the set of such “symmetries” by
AutK L.
Examples:
a) AutR C contains (at least) two elements: the identity map id : C −→ C and
the complex conjugation σ : C −→ C given by the correspondence a + bi 7→ a − bi;
√
√
√
b) AutQ Q( 2) again contains (at least) two elements a + b 2 7→ a ± b 2.
The above examples give as a matter of fact precise information about the corresponding sets of symmetries due to the following result.
Proposition. Suppose L = K(θ), where θ is a root of an irreducible polynomial
P (X) ∈ K[X]. Then | AutK L| 6 deg P (X).
Proof. We know that any element of α ∈ L can be written (uniquely) as a linear
combination α = b0 + b1 θ + · · · + bn−1 θn−1 , where n = deg P (X) and all coefficients
bi ∈ K. If σ ∈ AutK L then
σ(α) = b0 + b1 σ(θ) + · · · + bn−1 σ(θ)n−1
(Use that σ is compatible with addition and multiplication and σ|K = id.) In other
words, the knowledge of the whole map σ is equivalent to the knowledge of just the
image σ(θ) of θ.
But σ(θ) must be again a root of our polynomial P (X). Indeed, if P (X) =
n
X + b1 X n−1 + · · · + bn−1 X + bn , then
P (σ(θ))σ(θ)n + b1 σ(θ)n−1 + · · · + bn−1 σ(θ) + bn = σ(P (θ)) = 0
(use again that σ is compatible with operations and acts as identity on K.) This
implies our proposition because the number of distinct roots of P (X) is 6 n.
Taking into account results of section 2 we obtain
Corollary. | AutK L| = n is maximal if and only if P (X) splits completely in L[X]
(i.e. L is normal over K) and all roots of P (X) are different (i.e. each root of
P (X) appears with multiplicity 1).
The above properties can be generalized as follows:
4
Theorem. For any finite field extension L/K, | AutK L| 6 [L : K] and this inequality becomes equality if and only if L/K is normal and the minimal polynomials
of all elements of L over K have no multiple roots.
3.4. Separable extensions.
Definition. Let F ⊃ E be an algebraic field extension. An element θ ∈ F is
separable over E if its minimal polynomial over E has no multiple roots (in its
splitting field).
Definition. F/E is separable if any element of F is separable over E.
So, the last theorem of n.3.3 states that if L/K is a finite field extension then it
has a maximal possible number of field automorphisms if and only if it is normal
and separable.
Inseparable extensions do not appear very often. More precisely, we have the
following theorem.
Theorem. A finite field extension F/E is separable if it satisfies to one of the
following conditions (where p is a prime number):
a) char E = 0;
b) char E = p and [F : E] is not divisible by p;
c) char E = p and E = E p (i.e. any element of E is p-th power of element of
E)
Proof. We must prove that under one of above conditions a)-c) the minimal polynomial P (X) of any element θ ∈ F has no multiple roots.
Let Fe be a splitting field of P (X) over E. Then
Y
P (X) = c
(X − θi ),
16i6n
with c ∈ E ∗ and θ1 = θ.
Notice, for any i, P (X) is the minimal polynomial for θi over E. Indeed, P (θi ) =
0 and deg P (X) = [E(θ1 ) : E] = [E(θi ) : E].
For any polynomial
f (X) = an X n + an−1 X n−1 + · · · + a1 X + a0 ∈ Fe[X]
we can introduce its formal derivative by explicit formula
f 0 (X) = an nX n−1 + an−1 (n − 1)X n−2 + · · · + a1 .
For any polynomials f (X), g(X) ∈ Fe[X] and a ∈ Fe, one can verify the usual rules:
(f (X) ± g(X))0 = f 0 (X) ± g 0 (X)
(af (X))0 = af 0 (X)
(f (X)g(X))0 = f 0 (X)g(X) + f (X)g 0 (X).
5
Suppose P (X) has multiple roots, i.e. for some different indices i 6= j, θi = θj ,
i.e. P (X) = (X − θi )2 P1 (X) in Fe[X]. Then
P 0 (X) = 2(X − θi )P1 (X) + (X − θi )2 P10 (X)
and, therefore, P 0 (θi ) = 0. But P (X) is the minimal polynomial for θi and P 0 (X) ∈
E[X] has degree < deg P (X). Therefore, P 0 (X) = 0 is the zero polynomial, i.e.
the polynomial with zero coefficients.
So, if P (X) = X n + an−1 X n−1 + · · · + a1 X + a0 then n = 0, an−1 (n − 1) =
0, . . . , a1 = 0.
If char E = 0 then it is impossible. This proves the case a) of our theorem.
If char E = p then n ≡ 0 mod p. This proves the case b) of our theorem. Even
more, P (X) can have non-zero coefficients ai only if i is divisible by p. In other
words, P (X) = X mp + bm−1 X (m−1)p + · · · + b1 X p + b0 . (Verify that P 0 (X) = 0!)
Under assumption c) there are c0 , c1 , . . . , cm−1 ∈ E such that b0 = cp0 , b1 = cp1 ,...,
bm−1 = cpm−1 . Therefore,
P (X) = X mp + cpm−1 X (m−1)p + · · · + cp0 = (X m + cm−1 X m−1 + · · · + c0 )p
(Use that char E = p.) This means that P (X) is not irreducible in E[X]. Contradiction.
The theorem is completely proved.
The above theorem explains why it is not easy to construct an example of inseparable field extension. In the case of fields of characteristic 0 any algebraic field
extension is separable. The simplest field extensions in characteristic p are extensions of the finite field Fp , but any element of a ∈ Fp satisfies tha condition ap = a
and, therefore any finite field extension of Fp is separable. (This also implies that
any algebraic extension of any finite field is automatically separable.) So, the first
chance to get an example of inseparable extension is to take an infinite field of
characteristic p.
Example of inseparable extension.
Suppose E = F2 (t) = Frac F2 [t] is the field of rational functions in one variable
t with coefficients in F2 . Let P (X) = X 2 − t ∈ E[X]. Then P (X) is irreducible
(use that E[X] is a unique factorisation domain). Let F = E(θ), where θ is a root
of P (X). Then in F [X] we have that P (X) = (X − θ)2 (use that F is a field of
characteristic 2). So, P (X) has θ as a root with multiplicity 2. Therefore, F is an
inseparable extension of E of degree 2. Notice that replacing everywhere 2 by a
prime number p we shall obtain an example of an inseparable extension of degree
p.
There is also the following general criterion.
Theorem. A finite field extension F = E(α), where α is a root of an irreducible
polynomial f (X) ∈ E[X], is separable if and only if f (X) has no multiple roots (in
its splitting field over E).
Proof. The “only if” part is easy: f (X) is the minimal polynomial for α over E
and, therefore, if our extension is separable then f (X) can’t have multiple roots.
Prove the “if” part of the theorem. (It is very far from to be straightforward!)
6
Suppose θ ∈ F and P (X) has multiple roots. As earlier, this implies that
for some prime number p, char E = p and P (X) = Q(X p ) for some polynomial
Q(X) ∈ E(X). Let K = E(θ) and K1 = E(θp ). Then K and K1 are subfields in F ,
K ⊃ K1 , [K : E] = deg P (X), [K1 : E] = deg Q(X) and, therefore, [K : K1 ] = p.
Notice that K p ⊂ K1 , i.e. p-th power of any element of K is an element of the
smaller field K1 .
Now take our α ∈ F and consider its minimal polynomials fK (X) over K and
fK1 (X) over K1 . Then both fK (X) and fK1 (X) are factors of f (X) and, therefore,
have no multiple roots. In addition, fK1 (X) is divisible by fK (X), deg fK (X) =
[F : K] and deg fK (X) = [F : K1 ] = p[F : K].
Finally, consider the p-th power g(X) = fK (X)p of fK (X). Then g(X) has
coefficients in K p ⊂ K1 , it has α as a root and its degree is p[F : K] = [F : K1 ].
Therefore, g(X) satisfies all conditions for the minimal polynomial for α over K1 .
Therefore, g(X) = fK1 (X). But this is impossible because then fK1 (X) is a p-th
power of fK (X) and, therefore, has multiple roots.
The theorem is completely proved.
Finally notice that above methods allow to prove the following general property:
Suppose K ⊂ L ⊂ F are finite field extensions. Then F/K is separable if and
only if both F/L and L/K are separable.
3.5. Galois extensions and their Galois groups.
Definition. A finite field extension E/K is Galois if | AutK E| = [E : K].
Theorem. A finite field extension is Galois if and only if it is normal and separable.
We do not prove this theorem, but notice that it was completely proved in the
previous section in the case of simple extensions E/K, i.e. such that E = K(θ) for
some θ ∈ E.
Let E be any field. Consider the set Aut E of all field automorphismsm of the
field E. By definition, Aut E consists of all injective and surjective maps ψ : E −→
E such that for any α, β ∈ E, it holds ψ(α + β) = ψ(α) + ψ(β) and ψ(αβ) =
ψ(α)ψ(β). Notice that automatically, ψ(0) = 0, ψ(1) = 1, ψ(−α) = −ψ(α) and for
α 6= 0, ψ(α−1 ) = ψ(α)−1 .
One can prove that:
a) Aut Q = {id};
√
√
√
b) Aut Q( 2) = {id, σ}, where for any a, b ∈ Q, σ(a + b 2) = a − b 2;
c) Aut R = {id}.
Remark. The properties a) and b) can be easily proved. In order to prove c),
prove that for any ψ ∈ Aut R, if x, y ∈ R and x < y then ψ(x) < ψ(y).
Proposition. Aut E is a group, where the operation is the composition of automorphisms.
Proof. Suppose ψ, ϕ ∈ Aut E. Then for any a ∈ E, (ψϕ)(a) = ϕ(ψ(a)). Then
standard set-theoretic arguments prove that ψϕ is injective and surjective. In
addition, for any α, β ∈ E,
(ψϕ)(α + β) = ψ(ϕ(α + β)) = ψ(ϕ(α)) + ψ(ϕ(β)) = (ψϕ)(α) + (ψϕ)(β)
7
and, similarly, (ψϕ)(αβ) = (ψϕ)(α)(ψϕ)(β). This proves that the set Aut E is
closed under operation given by the composition of morphisms. Then we must
verify group axioms:
associativity
Indeed, if ψ, ϕ, χ ∈ Aut E then for any a ∈ E,
((ψϕ)χ)(a) = χ(ϕ(ψ(a)) = (ψ(ϕχ))(a)
;
existence of identity element
Indeed, the identity map idE ∈ Aut E and for any a ∈ E,
(ψ idE )(a) = ψ(a) = (idE ψ)(a).
existence of inverse map
One can easily see (do this!) that for any ψ ∈ Aut E, one can define the map
ψ 0 : E −→ E by the following rule: if α, β ∈ E and ψ(α) = β then ψ 0 (β) = α. Also
then (check it!) ψ 0 ∈ Aut E and ψψ 0 = ψ 0 ψ = idE .
The proposition is proved.
Examples: a) Aut Q and Aut R are trivial groups, i.e. groups which consist of
only one (identity!) element;
√
b) Aut Q( 2) is the cyclic group of order 2, i.e. it equals {id, σ} where σ 2 = id.
Suppose now that K is a subfield of E. Then (as earlier) AutK E is the subset
of all ψ ∈ Aut E such that ψ|K = id, i.e. for any a ∈ K, ψ(a) = a. Clearly (prove
this!), AutK E is a subgroup in Aut E.
Definition. If E is a finite Galois extension of K then AutK E is the Galois
group of the extension E/K which will be denoted by Gal(E/K).
√
Example. Gal(K/K) is the trivial group, Gal(Q( 2)/Q) is the cyclic group of
order 2.
3.6. Basic results of Galois theory.
Suppose E/K is a finite Galois extension and G = Gal(E/K) is its Galois group.
Let
E G = {a ∈ E | ∀σ ∈ G, σ(a) = a}.
In other words, E G is the subset of all invariant elements of E with respect to
the action of G on E. Clearly, E G ⊃ K.
Example. If E = C and K = R then C/R is Galois, Gal(C/R) = G = {id, σ},
where σ is the complex conjugation and CG = R.
8
Lemma. E G is a subfield in E.
Proof. Suppose α, β ∈ E G . Then for any σ ∈ G, σ(α) = α and σ(β) = β. Because
σ is a field automorphism of E, σ(α ± β) = σ(α) ± σ(β) = α ± β and σ(αβ) =
σ(α)σ(β) = αβ. Therefore, α ± β and αβ belong to E G . In other words, E G is
closed with respect to operations on E and, therefore, is a subfield in E.
Notice that the subfield E G of E contains K, i.e. K ⊂ E G ⊂ E.
The following statement is he first basic result of Galois theory. (We are not
going to prove it.)
Theorem A. With above notation, E G = K.
Suppose H is any subgroup in G. Consider
E H = {a ∈ E | ∀σ ∈ H, σ(a) = a}.
As earlier, E H is a subfield in E and E H ⊃ K. Therefore, the correspondence
H 7→ E H gives a map
{subgroups in G} −→ {fields L such that K ⊂ L ⊂ E}.
This map is a very important component of Galois Theory, it is called “the Galois
correspondence”.
Theorem B. The Galois correspondence is a one-one correspondence between the
sets of all subgroups of G = Gal(E/K) and all subfields of E which contain K.
Notice that if a subgroup is “bigger” then the corresponding subfield is “smaller”:
for subgroups {e} ⊂ H1 ⊂ H2 ⊂ G we have E = E {e} ⊃ E H2 ⊃ E H1 ⊃ E G = K.
Proposition. If E/K is Galois and H is a subgroup in Gal(E/K) then E is Galois
over E H and Gal(E/E H ) = H.
Example. If H = {e} then E H = E and Gal(E/E) = {e}.
Proof of the proposition.
Let E H = L, then K ⊂ L ⊂ E and:
• E/K is normal implies that E/L is normal;
• E/K is separable impolies that E/L is separable.
So, E/L is Galois and we can introduce H1 = Gal(E/L). Then by theorem A,
E = L = E H and by theorem B, H = H1 , i.e. Gal(E/E H ) = H.
The proposition is proved.
H1
Corollary. The correspondence L 7→ Gal(E/L) gives the inverse to the Galois
correspondence and, therefore, defines a one-one map from the set of all fields
between K and E to the set of all subgroups of G = Gal(E/K).
Corollary. If, as earlier, E/K is finite Galois, G = Gal(E/K) and H ⊂ G is a
|G|
[E : K]
=
= (G : H) — the
subgroup then [E : E H ] = |H| and [E H : K] =
[E : E H ]
|H|
index of H in G.
Problem. If H is a subgroup of Gal(E/K) then E/E H is Galois but, generally,
E H /K is not Galois. (Give an example!) How we can characterize the subgroups
H such that E H /K is Galois? What will be then Gal(E H /K)?
The answer to this question is given by third main result of Galois theory.
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Theorem C. In the above notation, E H /K is Galois if and only if H is a normal
subgroup of G. In this case, Gal(E H /K) is the quotient group G/H.
Reminder. A subgroup H of G is normal if for any g ∈ G, g −1 Hg = H. Equivalently, for any g ∈ G, the left coset gH must coincide with the right coset Hg. (In
particular, if G is abelian group then any its subgroup is automatically normal.)
Then the quotient group G/H appears as a set of all cosets gH with group operation induced by the group operation on G. This means that the natural projection
from G to G/H given by the correspondence g 7→ gH is a group epimorphism.
Remark. There is a natural map pH from G = Gal(E/K) to Gal(E H /K). It
is given by the restriction g 7→ g|E H of any g ∈ G to E H . (One can verify that
if H is a normal subgroup in G then g(E H ) = E H .) Finally, with respect to the
identification Gal(E H /K) = G/H from above Theorem C, the map pH is just the
natural projection from G to G/H.
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