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8.1.1. Proposition. Let F be an extension field of K. The set of all automorphisms : F -> F such that (a) = a for all a in K is a group under composition of functions.
8.1.2. Definition. Let F be an extension field of K. The set
{ Aut(F) | (a) = a for all a K }
is called the Galois group of F over K, denoted by Gal(F/K).8.1.3. Definition. Let K be a field, let f(x) K[x], and let F be a splitting field for f(x) over K. Then Gal(F/K) is called the Galois group of f(x) over K, or the Galois group of the equation f(x) = 0 over K.
8.1.4. Proposition. Let F be an extension field of K, and let f(x) K[x]. Then any element of Gal(F/K) defines a permutation of the roots of f(x) that lie in F.
8.1.5. Lemma. Let f(x) K[x] be a polynomial with no repeated roots and let F be a splitting field for f(x) over K. If : K -> L is a field isomorphism that maps f(x) to g(x) L[x] and E is a splitting field for g(x) over L, then there exist exactly [F:K] isomorphisms : F -> E such that (a) = (a) for all a in K.
8.1.6. Theorem. Let K be a field, let f(x) K[x], and let F be a splitting field for f(x) over K. If f(x) has no repeated roots, then |Gal(F/K)| = [F:K].
8.1.7. Corollary. Let K be a finite field and let F be an extension of K with [F:K] = m. Then Gal(F/K) is a cyclic group of order m.
f(x) = (x - r_{1})^{m1} (x - r_{2})^{m2} · · · (x - r_{t})^{mt}
over F, then we say that the root r_{i} has multiplicity m_{i}.8.2.2. Definition. Let f(x) K[x], with f(x) = a_{k} x^{k}. The formal derivative f'(x) of f(x) is defined by the formula
f'(x) = k a_{k} x^{k}^{-1},
where k a_{k} denotes the sum of a_{k} added to itself k times.8.2.3. Proposition. The polynomial f(x) in K[x] has no multiple roots if and only if gcd(f(x),f'(x)) = 1.
8.2.4. Proposition. Let f(x) be an irreducible polynomial over the field K. Then f(x) has no multiple roots unless chr(K) = p 0 and f(x) has the form
f(x) = a_{0} + a_{1} x^{p} + a_{2} x^{2}^{p} + · · · + a_{n} x^{n}^{p}.
8.2.5. Definition. A polynomial f(x) over the field K is called separable if its irreducible factors have only simple roots.8.2.6. Theorem. Any field of characteristic zero is perfect. A field of characteristic p>0 is perfect if and only if each of its elements has a pth root.
8.2.7. Corollary. Any finite field is perfect.
8.2.8. Theorem. Let F be a finite extension of the field K. If F is separable over K, then it is a simple extension of K.
{ a F | (a) = a for all G }
is a subfield of F.8.3.2. Definition. Let F be a field, and let G be a subgroup of Aut (F). Then
{ a F | (a) = a for all G }
is called the G-fixed subfield of F, or the G-invariant subfield of F, and is denoted by F^{G}.8.3.3. Proposition. If F is the splitting field over K of a separable polynomial and G = Gal(F/K), then F^{G} = K.
8.3.4. Lemma. [Artin] Let G be a finite group of automorphisms of the field F, and let K = F^{G}. Then
[F:K] | G |.
8.3.5. Definition. Let F be an algebraic extension of the field K. Then F is said to be a normal extension of K if every irreducible polynomial in K[x] that contains a root in F is a product of linear factors in F[x].8.3.6. Theorem. The following conditions are equivalent for an extension field F of K:
Example. 8.3.1. The Galois group of GF(p^{n}) over GF(p) is cyclic of order n, generated by the automorphism defined by (x) = x^{p}, for all x in GF(p^{n}). This automorphism is usually known as the Frobenius automorphism of GF(p^{n}).
8.3.8. Theorem. [Fundamental Theorem of Galois Theory] Let F be the splitting field of a separable polynomial over the field K, and let G = Gal(F/K).
H = Gal(F/F^{H}).
(ii) If E is a subfield of F that contains K, then the corresponding subgroup of G is H = Gal(F/E), and
E = F^{H}.
[F:F^{H}] = | H | and [F^{H}:K] = [G:H].
Gal(E/K) Gal(F/K) / Gal(F/E).
8.3.9. Proposition. Let F be the splitting field of a separable polynomial over the field K, and let E be a subfield such that K E F, with H = Gal(F/E). If Gal(F/K), then
Gal(F/(E)) = H ^{-1}.
8.3.10. Theorem. [Fundamental Theorem of Algebra] Any polynomial in C[x] has a root in C.8.4.1. Definition. An extension field F of K is called a radical extension of K if there exist elements u_{1}, u_{2}, ... , u_{m} in F such that
(ii) u_{1}^{n1} K and u_{i}^{ni} K ( u_{1}, ... , u_{i-1} ) for i = 2, ... , m and n_{1}, n_{2}, ... , n_{m} Z.
8.4.2. Proposition. Let F be the splitting field of x^{n} - 1 over a field K of characteristic zero. Then Gal(F/K) is an abelian group.
8.4.3. Theorem. Let K be a field of characteristic zero that contains all nth roots of unity, let a K, and let F be the splitting field of x^{n}-a over K. Then Gal(F/K) is a cyclic group whose order is a divisor of n.
8.4.4. Theorem. Let p be a prime number, let K be a field that contains all pth roots of unity, and let F be an extension of K. If [F:K] = |Gal(F/K)| = p, then F = K(u) for some u F such that u^{p} K.
8.4.5. Lemma. Let K be a field of characteristic zero, and let E be a radical extension of K. Then there exists an extension F of E that is a normal radical extension of K.
8.4.6. Theorem. Let f(x) be a polynomial over a field K of characteristic zero. The equation f(x) = 0 is solvable by radicals if and only if the Galois group of f(x) over K is solvable.
Theorem 7.7.2 shows that S_{n} is not solvable for n 5, and so to give an example of a polynomial equation of degree n that is not solvable by radicals, we only need to find a polynomial of degree n whose Galois group over Q is S_{n}.
8.4.7. Lemma. Any subgroup of S_{5} that contains both a transposition and a cycle of length 5 must be equal to S_{5} itself.
8.4.8. Theorem. There exists a polynomial of degree 5 with rational coefficients that is not solvable by radicals.
_{n} (x) = _{k} (x - ^{k}),
where k belongs to the set of positive integers less than n and relatively prime to n, is called the nth cyclotomic polynomial.8.5.2. Proposition. Let n be a positive integer, and let _{n}(x) be the nth cyclotomic polynomial. The following conditions hold:
(b) x^{n} - 1 = _{d | n} _{d} (x);
(c) _{n} (x) is monic, with integer coefficients.
8.5.4. Theorem. For every positive integer n, the Galois group of the nth cyclotomic polynomial _{n}(x) over Q is isomorphic to Z_{n}^{×}.
Example. 8.5.2. A regular n-gon is constructible if and only if (n) is a power of 2. If p is an odd prime, and (p) is a power of 2, then p must have the form p = 2^{k} + 1, where k is a power of 2. Such primes are called Fermat primes. The only known examples are 3, 5, 17, 257, and 65537. This implies, for example, that a regular 17-gon is constructible.
A set that satisfies all the axioms of a field except for commutativity of multiplication is called a division ring or skew field.
8.5.6. Theorem. [Wedderburn] Any finite division ring is a field.
8.6.2. Proposition. Let f(x) be a separable polynomial over the field K, with roots r_{1} , ... , r_{n} in its splitting field F. Then f(x) is irreducible over K if and only if Gal(F/K) acts transitively on the roots of f(x).
8.6.3. Lemma. Let p be a prime number, and let G be a transitive subgroup of S_{p}. Then any nontrivial normal subgroup of G is also transitive.
8.6.4. Lemma. Let p be a prime number, and let G be a solvable, transitive subgroup of S_{p}. Then G contains a cycle of length p.
8.6.5. Proposition. Let p be a prime number, and let G be a solvable, transitive subgroup of S_{p}. Then G is a subgroup of the normalizer in S_{p} of a cyclic subgroup of order p.
Let f(x) be a polynomial of degree n over the field K, and assume that f(x) has roots r_{1}, r_{2}, ... , r_{n} in its splitting field F. The element of F defined by
= (r_{i} - r_{j})^{2},
where the product is taken over all i,j with 1 i < j n, is called the discriminant of f(x).
It can be shown that the discriminant of any polynomial f(x)
can be expressed as a polynomial in the coefficients of f(x),
with integer coefficients.
This requires use of elementary symmetric functions,
and lies beyond the scope of what we have chosen to cover in the book.
We have the following properties of the discriminant:
(ii) K;
(iii) If 0, then a permutation S_{n} is even if and only if it leaves unchanged the sign of
_{} (r_{i}-r_{j}) .
We now restrict our attention to polynomials with rational coefficients. The next lemma shows that in computing Galois groups it is enough to consider polynomials with integer coefficients. Then a powerful technique is to reduce the integer coefficients modulo a prime and consider the Galois group of the reduced equation over the field GF(p).
8.6.7. Lemma.
Let f(x) = x^{n} + a_{n-1} x^{n-1}
+ · · · + a_{1} x + a_{0}
Q[x],
and assume that
a_{i}
= b_{i} / d
for d, b_{0},
b_{1}, ... ,
b_{n-1}
Z.
Then
d^{n}
f(x/d) is monic with integer coefficients,
and has the same splitting field over Q as f(x).
If p is a prime number, we have the natural mapping : Z[x] -> Z_{p}[x] which reduces each coefficient modulo p. We will use the notation ( f(x) ) = f_{p}(x).
Theorem [Dedekind]. Let f(x) be a monic polynomial of degree n, with integer coefficients and Galois group G over Q, and let p be a prime such that f_{p}(x) has distinct roots. If f_{p}(x) factors in Z_{p}[x] as a product of irreducible factors of degrees n_{1}, n_{2}, ... , n_{k}, then G contains a permutation with the cycle decomposition
(1,2, ... ,n_{1}) (n_{1}+1, n_{1}+2, ... , n_{1}+n_{2}) · · · (n-n_{k}+1, ... ,n),
relative to a suitable ordering of the roots.