Galois Theory

Binary operation, associative, identity element, inverses

(R, +) is a commutative group, x is associative and distributive

Ring with no zero divisors and a non-zero identity element for x

A commutative domain

An integral domain with inverses for x/(R, +) and (R\0, x) commutative groups and distributive

H is non-empty, x-y and xy in H

H is non-empty, H is closed under +, x and their respective inverses

Smallest n such that n.1 = 0 (if no such n exists char F = 0, otherwise prime)

The smallest subfield of a field (isomorphic to Q if char F = 0 and Z_p is Char F = p)

(E:F) is finite (E, F fields)

F<B<E with E:B and B:F finite extensions. Then E:F is a finite extension and (E:F) = (E:B)(B:F)

The size of the basis for E as seen as a vector space over F

d/f and d and if e/f and e then e/d

Consider f and g and let g not equal zero. Then there exists unique q and r such that f=qg+r where deg r < deg g or r=0

Let f belong to F[x]. Then a belonging to F is a root of f iff x-a/f

A non-constant polynomial f with f=gh and g,h are non-constant (irreducible otherwise)

Polynomial cannot be factored into non-trivial polynomials over the same field

All non-constant polynomials are a product of irreducible polynomials and these are unique up to order and constant polynomials (all polynomial rings over fields are UFDs)

Every non-constant polynomial in C[x] has a zero in C

A polynomial with integer coefficients is irreducible over Q iff it cannot be expressed as a product of two non-constant polynomials in Z[x]

Let f(x) = a^nx_n + ... + a_0 in Z[x], n>=1, a_n =/ 0. Let p be a prime such that p does not divide a_n, p/a_i, p^2 does not dicide a_0. Then f is irreducible over Q

There exists f in F[x] such that f=/0 and f(a)=0

f in F[x] monic and deg f minimal subject to f(a) = 0

Let f a non-constant polynomial belong to F[x]. Then there exists an extension field E such that f has at least one root in E

An extension E of F such that p splits in E but does not split in any intermediate field

p can be factored into linear factor in E[x]

All functions from G to F (F^G is a vector space over F)

There exists a_i in F not all zero such that a_1f_1(x) + ... + a_nf_n(x) = 0 for all x in G (otherwise independent)

A function f from F to G such that f(xy) = f(x)f(y) for all x,y in G and f(x) =/ for all x in G

injective homomorphism

an isomorphism from a set to itself

f_i(a) = f_j(a)

f_i(a) = a for all i

The set of all fixed points for S

Set of all field automorphisms of E (it is a group)

Set of f in Aut(E) such that f is a F-automorphism

(E:F) is finite and Fix(Aut(E:F)) = F

Each irreducible factor of f in F[x] does not have repeated roots

a is the root of a separable polynomial in F[x]

Each element of E is separable over F

a set of elements of V such that every element of V can be written as a linear combination of the basis elements with coefficients in V and all basis elements are linearly independent

D(a_ix^i) = ia_ix^(i-1) (follows linearity, product rule and chain rule)

Suppose E is a normal extension of F and G = Aut(E:F). Let H' denote all the subgroups of G and B' all the intermediate fields of E:F

There is a bijection f between H' and B' given by f(H) = Fix(H) and f^-1(B) = Aut(E:B)

(E:B) = |Aut(E:B)| and (B:F) = (G:Aut(E:B)) (so |G| = (E:F))

B is a normal extension of F iff Aut(E:B) is a normal subgroup of G. In this case, Aut(B:F) is isomorphic to G/Aut(E:B)

a^n = 1 for some n in N

The generators of the cyclic subgroup of all the nth roots of units

Splitting field of (x^n-a_1)...(x^n-a_r), where a_i are in F*

There exist intermediate fields such that C_i=C_i-1(a_i) (i=1,...,r), where a_i is a root of x^n_i - b_i in C_i-1[x], C_0 = F and C_r=E

There exists subgroups H_i such that H_i is a normal subgroup of H_i-1, H_i-1/Hi is abelian, H_r is the trivial group and H_0 is G

Let H be a subgroup of G and N a normal subgroup. If G is soluble then H is soluble and G/N is soluble; if N and G/N are soluble then G is soluble; if G is abelian then it is soluble; if G is a non-abelian simple group then it is not soluble

x^-1y^-1xy

<[x,y]|x,y in G>

<r,s | r^n=s^2=(sr)^2=1>

The splitting field of f is contained in some radical extension of F

let charF = 0 and f be in F[x]. then the polynomial equation f(x) = 0 is solvable by radicals iff Gal(f) is a soluble group