# Fundamentals. Algebra.

A group is a nonempty set G together with a binary operation on G, denoted as ◦, ·, * or simply omitted, satisfying the following axioms:

1. Associativity, (ab)c = a(bc) ∀a, b, c ∈ G.
2. Identity, ∃ e ∈ G: ae = ea = a ∀a ∈ G.
3. Inverses, ∀ a ∈ G, ∃ a-1, aa-1 = a-1a = e.

The group is called Abelian if ab = ba ∀a, b ∈ G.

A group is finite if it has a finite number of elements. Otherwise, it is said to be infinite. The number of elements in a group G is called the order of the group, and is denoted by |G|, e.g., ℤ10 and U10 = {1, 3, 7, 9} -positive integers less than 10 that are coprime to 10- are finite group of orders 10 and 4 respectively. The order of an element a ∈ G is the smallest positive integer n such that xn = 1, e.g., |0| = 1, |1| = 6, |2| = 3, |3| = 2, |4| = 3, |5| = 6 in Z6.

Proposition. Let (G, ∘) be a group.

1. The identity of G is unique.
2. ∀a ∈ G, a-1 is uniquely determined.
3. ∀a, b ∈ G, (a-1)-1 = a, (ab)-1 = b-1a-1
4. ∀a1, a2,…, an ∈ G, the value of ∀a1 ∘ a2∘…∘an is independent of how the expression is bracketed.
5. The equations ax = b and ya = b have unique solutions for x, y ∈ G. The left and right cancellation laws hold in G, i.e., if au = av, then u = v; and if ub = vb, then u = v.

The group ℤn consists of the elements {0, 1,2, . . . , n−1} with addition mod n as the operation. You can also multiply elements of Zn, but you do not obtain a group. For each n > 1, we define Un to be the set of all positive integers less than n that have a multiplicative inverse modulo n, that is, there is an element b∈ Zn such that ab = 1. Un = {a ∈ Zn: a is relatively prime to n, a < n} = {a ∈ Zn: gcd(k, n)=1}.

The dihedral group Dn is the group of symmetries of a regular polygon with n vertices, which includes rotations and reflections or, alternatively, the right motions taking a regular n-gon back to itself. Dn = {e, r, r2,…, rn-1, s, rs, r2s,…, rn-1s} = ⟨r, s | rn = s2 = e, rs = srn-1⟩.

1. 1, r, r2,…, rn-1 are all distinct, rn = 1, |r| = n.
2. |s| = 2.
3. s ≠ ri ∀i, sri ≠ srj, ∀i, j: 0 ≤ i, j ≤ n-1, i ≠ j.
4. rs = sr-1
5. ris = sr-i ∀i, j: 0 ≤ i ≤ n

Definition. Let G be a group. A subgroup is a subset of a group that itself is a group. Formally, H ⊂ G (H is a subset of G), ∀a, b ∈ H, a·b ∈ H, and (H, ·) is a group (closure, identity, and inverses).

H ≤ G if H is closed (∀a, b ∈ H, a·b ∈ H) and every element in H has an inverse in H (∀a ∈ H, a-1 ∈ H). Besides, H ≤ G iff ∀a, b ∈ H, a·b-1 ∈ H.A non empty finite subset H ⊂ G closed under the operation of G (∀a, b ∈ H, a·b ∈ H) is a subgroup of G.

The center of a group, G, is the subset of elements that commute with every element of G. Z(G) = {a ∈ G | ab = ba ∀ b in G} ≤ G. Let G be a group, H ≤ G be a subgroup of G. Then the conjugate of H by g is a subgroup of G, that is, g-1Hg = {g-1hg | h ∈ H} ≤ G.

# Cyclic Groups

Let G be a group and a be any element in G (a ∈ G). Then, the set ⟨a⟩={an | n ∈ Z} is a subgroup of G. We call it the cyclic subgroup generated by a.

Definition. Let G be a group. A cyclic group G is a group that can be generated by a single element a. More formally, G is cyclic if there exists at least an element a ∈ G: G = ⟨a⟩ = {an | n ∈ ℤ}

Definition. Let G be a group, H ≤ G. A left coset of H in G is a subset of the form aH = [ah | h ∈ H] for some a ∈ G. This is a equivalent relation, so aH = [a], and a is the representative of the coset aH. The collection of left cosets is denoted by G/H.

• The left (or right) coset of H containing a does contain a, a ∈ aH.
• A coset H absorbs an element if and only if the element belongs to it, aH = H iff a ∈ H.
• A left (or right) coset is uniquely determined or represented by any one of its elements, i.e., aH = bH iff a ∈ bH.
• Two left (or right) cosets are either identical or disjoint, i.e., aH = bH or aH ∩ bH = ∅.
• aH = bH iff a-1b ∈ H.
• Any two left (or right) cosets have the same size or number of elements, i.e., |aH| = |bH|
• aH = Ha iff H = aHa-1.
• H itself is the only coset of H that is a subgroup of G, i.e. aH ≤ G iff a ∈ H.

Lemma. Let G be a group, H ≤ G, g1, g2 ∈ G. The following statements are equivalent:

1. g1H = g2H
2. Hg1-1 = Hg2-1
3. g1H ⊆ g2H
4. g1 ∈ g2H
5. g1-1g2 ∈ H.

Definition. Let G be a group, H < G a subgroup. The set of cosets of H in G is denoted as G/H and called quotient set of G by H. G/H = {aH | a ∈ G}. The index of H in G is denoted as [G:H] = |G/H| where we say [G:H] = ∞ if G/H is infinite.

Let (A, *) and (B, ⋄) be two binary algebraic structures. A homomorphism is a structure-preserving map between two algebraic structures of the same type or, in other words, a map ϕ: A → B, such that ∀x, y ∈ S : ϕ(x ∗ y) = ϕ(x) ⋄ ϕ(y).

Let (A, *) and (B’, ⋄) be two binary algebraic structures. An isomorphism is a bijective homomorphism, i.e., one-to-one and onto.

• For each prime p there is one group of order p up to isomorphism, namely the cyclic group Z/(p).
• There are two groups of order 4 up to isomorphism: the cyclic group ℤ/4ℤ, and the Klein 4-group, {1, a, b, ab}. Order(a) = Order(b) = 2, ab = ba, (ab)2 = 1.
• There are two groups of order 6 up to isomorphism: the cyclic group of order 6: the cyclic group ℤ/6ℤ, and S3, the non-Abelian symmetric group = {(), (12), (13), (23), (123), (132)}

Cayley’s Theorem. Every group G is isomorphic to a a subgroup of a symmetric group. In particular, every finite group is isomorphic to a subgroup of Sn. Lagrange’s Theorem. Let G be a finite group and let H ≤ G. Then, |H| | |G|. In particular, the order of an element (the order of the subgroup generated by a) of a finite group divides the order of the group.

Cauchy’s Theorem. Let G be a finite group and let p be a prime that divides the order of G. Then, G has an element of order p.

A group is called Abelian if ab = ba ∀a, b ∈ G. Every group of prime order (in fact, it is cyclic), order five or smaller, or cyclic is Abelian. $\mathbb{S}_3$ is the smallest non-commutative group.

A subgroup H of a group G is called a normal subgroup of G if aH = Ha, ∀a ∈ G. The alternating group of even permutations A3 = {(), (123) -(13)(12)-, (132) -(12)(13)-} is a normal subgroup of S3. In general, An ◁ Sn.

A group G is simple if it has no trivial, proper normal subgroups, e.g., p, cyclic group of prime order, and An the alternating group, that is the group of even permutation of a finite set, for n ≥ 5.

A ring R is a non-empty set with two binary operations. It is an Abelian group under addition. A ring is associative, but not necessarily commutative under multiplication. When it is, we say that the ring is commutative. A ring does not necessarily have an identity under multiplication. When it does have an identity, it is called as unity or identity. Examples: ℤ, ℤ/ℤn.

A field F is a commutative ring with identity in which every nonzero element has an inverse or Fx = F -{0} is an Abelian group under multiplication. Examples: ℚ, ℝ, and ℂ are fields. $\mathbb{F_p}$ = ℤ/pℤ with p prime is a field. ℤ/ℤn is a field ↭ n is prime.

A subring A of a ring R is called a (two-sided) ideal of R if it absorbs multiplication from the left and right from R, that is, ∀r ∈ R, a ∈ A, ra ∈ A and ar ∈ A. ↭ ∀r ∈ R, rA = {ra | a ∈ A} ⊆ A and Ar = {ar | a ∈ A} ⊆ A ↭ ∀a, b ∈ A, ∀r ∈ R, a ∈ A, a - b ∈ A, ra ∈ A, ar ∈ A. Examples: nℤ is an ideal of ℤ.

A prime ideal I of a commutative ring R is a proper ideal of R such that a, b ∈ R, ab ∈ I ⇒ a ∈ I or b ∈ I ↭ R/I is an integral domain. Example: ⟨x⟩ ⊆ ℤ[x] is a prime ideal, but ⟨x2⟩ is not, because x2 ∈ ⟨x2⟩ but x ∉ ⟨x2⟩.

A maximal ideal A of a commutative ring R is a proper ideal of R that is maximal with respect to set inclusion amongst all proper ideals, i.e., whenever B is an ideal of R, A ⊆ B ⊆ R ⇒ B = A or B = R. Let R be a commutative ring with unity and let I be an ideal of R. Then, R/I is a field iff I is maximal, e.g., ℤ[x]/⟨x⟩ ≋ ℤ is not a field, ⟨x⟩ is not a maximal ideal.

Let R be a commutative ring with unity. A principal ideal is an ideal generated by a single element a of R through multiplication by every element of R, ⟨a⟩ = Ra = {ra | r ∈ R} = {ar | r ∈ R}, e.g., ⟨x⟩ is a principal ideal of ℤ[x], but ⟨x, 3⟩ is not.

F field, p(x) ∈ F[x], ⟨p(x)⟩ is the set of all multiples of p(x), the principal ideal generated by p(x). F[x]/⟨p(x)⟩ is the quotient ring. If a(x) ∈ F[x], then a(x) + ⟨p(x)⟩ is the coset of ⟨p(x)⟩ represented by a(x). a(x) = b(x) (mod (p(x))) to mean that p(x) | a(x) - b(x).

Let R be a commutative ring with unity, let a1, a2, ···, an be a family of elements of R. Then, the ideal generated by the family a1, a2, ···, an is the set of element of R of the form r1a1 + r2a2 + ··· + rnan, i.e., I = ⟨a1, a2, ···, an⟩ = ⟨r1a1 + r2a2 + ··· + rnan | ri ∈ R⟩

Let R be a commutative ring. A polynomial ring is a ring formed from the set of polynomials in one or more indeterminates or variables with coefficients in R. R[x] = {anxn + an-1xn-1+ ··· + a1x + a0 | ai ∈ R, n ∈ ℤ+}, e.g. 23x2 + 4x -7 ∈ ℚ[x].

Let D be an integral domain. An irreducible polynomial is a polynomial that cannot be factored into the product of two non-constant polynomials. A non-zero and non-unit polynomial f(x) ∈ D[x] is said to be irreducible if it cannot be written as the product of two non-units. If f(x) = g(x)h(x), g(x) and h(x) ∈ D[x], then either g(x) or h(x) is a unit in D[x].

A non-constant polynomial f(x) ∈ F[x], F field, is said to be irreducible if f(x) cannot be expressed a as a product of two polynomials g(x), h(x) ∈ F[x] of lower degree, that is, f(x) = g(x)h(x), deg(g(x)) < deg(f(x)) and deg(h(x)) < deg(f(x)).

Definition. We say an integral domain D is a unique factorization domain, UFD for short, if:

1. Every non-zero, non-unit element of D can be written as a product of irreducibles of D.
2. This factorization into irreducibles is unique in the following sense. If a = up1p2···pr = w11q2···qs, u and w units, then m = n and there exists a bijective map Φ:{1, 2, …, n} → {1, 2, …, n} such that pi is associated to qΦ(i) ∀i ∈ {1, 2, …, n}. In other words, ∃Φ ∈ Sr: pi = qΦ(i)ui and ui is a unit.

If D is a unique factorization domain, then D[x] is a unique factorization domain. Examples: ℤ is a UFD (fundamental theorem of arithmetic) ⇒ ℤ[x] is a UFD. A field K is a UFD ⇒ K[x] is a UFD

A ring R is a principal ideal domain (PID) if it is an integral domain such that every ideal of R is a principal ideal, that is, has the form ⟨a⟩ = Ra = {ra | r ∈ R} = {ar | r ∈ R}, e.g., , and if F is a field, both F and F[x]. PID ⇒ UFD, but the reverse is not true.

Reducibility test for polynomials with degrees two or three. Let F be a field. If f(x) ∈ F[x] and deg(f(x)) = 2 or 3, then f(x) is reducible over F iff f has a zero in F.

Let D be an integral domain. The content of a non-zero polynomial f(x) = anxn + an-1xn-1 + ··· + a0 ∈ D[x] is the greatest common divisor of its coefficients an, an-1, ··· a0. A primitive polynomial is an element of D[x] with content 1, that is, gcd(a0, a1, ···, an) = 1, e.g., 3x2 -6x +9 is not primitive, 3x2 -6x +10 is primitive.

Gauss’s Lemma. Let D be a unique factorization domain. Let f(x), g(x) ∈ D[x]. Then, the content of f(x)g(x) equals the product of the content of f(x) and the content of g(x). In particular, if f(x), g(x) ∈ D[x] are primitive, then f(x)g(x) is also primitive.

Theorem. A non-constant polynomial in Z[x] is irreducible in Z[x] if and only if it is both irreducible in Q[x] and primitive in Z[x].

Eisenstein Criterion. p prime, f(x) = anxn + an-1xn-1 + ··· + a0 ∈ ℤ[x], n ≥ 1, p | a0, p | a1, ···, p | an-1, but p ɫ anand p2 ɫ a0. Then, f(x) is irreducible in ℚ[x].

For any prime p, the pth cyclotomic polynomial Φp(x) = $\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+···+x+1$ is irreducible over ℚ.

Mod p Irreducibility Test. Let p be a prime, f(x) ∈ ℤ[x], and deg(f(x)) ≥ 1. Let f'(x) be the polynomial in ℤp[x] obtained from f(x) by reducing all the coefficients of f(x) modulo p. If f'(x) is irreducible over ℤp and deg(f'(x)) = deg(f(x)), then f(x) is irreducible over ℚ.

A field F is a commutative ring with identity in which every nonzero element has an inverse or Fx = F -{0} is an Abelian group under multiplication, e.g., ℚ, ℝ, ℂ. The characteristic of a field F, char(F), as the smallest positive integer p such that p·1F = 1 + 1 + ··· + 1 (p times) = 0 if such a p exists, or char(F) = 0 if 1 + 1 + ··· + 1 ≠ 0 ∀n ∈ N, e.g., char(ℝ) = char(ℚ) = 0; and for any prime p, $\mathfrak{F}_p=ℤ/pℤ$ has characteristic p.

If F is a subfield of a field E, then E is called an extension field of F, denoted E/F or E:F. Let p(x) be an irreducible polynomial in F[x] and let u be a root of p(x) in an extension E of F. Then,

1. F(u), the subfield of E generated by F and u is the smallest field containing F and u. It is indeed F[u], the smallest subring containing both F and u = {b0 + b1u + ··· + bmum ∈ E | b0 + b1x + ··· + bmxm ∈ F[x]}

2. The degree of p(x) is n, the set (1, u, ···, un-1) forms a basis of F(u) over F, i.e., each element of F(u) can be written uniquely as c0 + c1u + ··· + cnun-1 where ci ∈ F, and [F(u):F] = n.

An element α ∈ E is algebraic over F if there exists a non-constant polynomial p(x) ∈ F[x] such that p(α)=0, i.e., α is the root of p(x). Example: E = ℂ, F = ℚ, α = $\sqrt{2}$ is algebraic over ℚ, p(x) = x2 -2, and p($\sqrt{2}$)=0. Similarly, i is algebraic over ℚ, p(x) = x2 + 1, g(i) = 0. An element that is not algebraic, i.e., there no exists a non-constant polynomial p(x) such that p(α)=0 is called transcendental, e.g., π.

An extension E of F is called algebraic if each element of E is algebraic over F. An extension that is not algebraic, it is called transcendental extension, e.g, the extension ℝ over ℚ is a transcendental extension, and ℂ = ℝ(i) and ℚ($\sqrt{5}$) are algebraic extensions over ℝ and ℚ respectively.

Theorem. Let E be an extension field of F, and let u ∈ E be algebraic over F. Let p(x) ∈ F[x] be a polynomial of least degree such that p(u) = 0, we call p the minimal polynomial of u over E, Then,

(i) p(x) is irreducible over F
(ii) if g(x) ∈ F[x] is such that g(u) = 0, then p(x) | g(x)
(iii) There is exactly one monic polynomial p(x) ∈ F[x] of least degree such that p(u) = 0.

Let E be an extension of F,

1. Then, E is a vector space over F.
2. E will have a basis set.
3. The dimension of E considered as a vector space over F is called the degree of E over F and denoted as [E:F].

If E is a finite extension of F, then E is an algebraic extension of F.

If p(x) is the minimal polynomial of α over F and degree(p(x))= n. Then, [F(α):F] = n, and in particular it is finite, so F(α) = F[α] is an algebraic extension. More formally, if E is an extension of F and α ∈ E is algebraic over F, then F(α) is an algebraic extension of F and deg(F(α):F)=deg(p(x)) where p is the minimal polynomial of α.

Multiplicative property of degree extensions. Let F ⊆ E ⊆ K be a tower of fields. If [K:E] < ∞ and [E:F] < ∞ then, (i) [K:F] < ∞, and (ii) [K:F] =[K:E][E:F]

A field homomorphism between two fields E and F is a function f: E → F such that ∀x, y ∈ E: f(x + y) = f(x) + f(y), f(xy) = f(x)f(y), f(1) = 1. It implies: f(0) = 0, f(x-1) = f(x)-1, and f is injective.

Let K/F and L/F be two extensions of F. An F-homomorphism of extensions of F is a field homomorphism σ: K → L such that σ(a) = a ∀a ∈F.

Let f(x) be a polynomial in F[x] of degree ≥ 1. Then, an extension K of F is called a splitting field of f(x) over F if:

1. f(x) can be factored as a product of linear factors or polynomials in K[x], i.e., f(x) = c(x -α1)(x -α2) ··· (x -αn), where αi ∈ K.
2. K = F(α1, ···, αn), i.e., K is generated over F by the roots α1, ···, αn of f(x) in K.

Theorem. Existence of Splitting Fields. Let F be a field and let f(x) be a non-constant polynomial of F[x]. Then, there exists a splitting field E for f(x) over F,e.g., x2 + 1 ∈ ℚ[x] splits over ℂ and ℚ(i).

A finite field is a field that contains a finite number of elements,e.g., $\mathbb{F_p}=ℤ/pℤ$ The characteristic of a finite field have to be a prime number, say p. |F| = pn for some n ∈ ℕ.

Structure theorem for finite fields. Let p be a prime, n be a positive integer, q = pn. Then,

1. There exists a unique finite field of order q, up to isomorphism (Classification of finite fields)
2. Let F be a field of order q. Then, Fx = F \ {0} is a cyclic group under the multiplication operation (The Multiplicative Group of Finite Subgroups of a Field)
3. Let F be a finite field of order q = pn. F is the splitting field of the polynomial xq-x ∈ $\mathbb{F}_p[x]$.
4. A field of order pn contains a field of order pm ($\mathbb{F}$pm ⊆ $\mathbb{F}$pn) iff m|n.
5. Irreducible factors of xq -x ∈ $\mathbb{F_p}$[x] are the irreducible polynomials in $\mathbb{F_p}$[x] whose order divides r.

A field K is called algebraically closed if every non-constant polynomial f(x) ∈ K[x] has a root in K ↭ any irreducible polynomial of positive degree in K[x] has degree 1, e.g, ℂ is algebraically closed.

Theorem. Every field has an algebraic closure

Let F be a field and let p(x) ∈ F[x] be an irreducible polynomial over F, deg(p(x)) = n. If a is a root of p(x) in some extension E of F, then F(a) is isomorphic to F[x]/⟨p(x)⟩, F(a)≈F[x]/⟨p(x)⟩. Moreover, F(a) is a vector space over F with a basis {1, a, ···, an-1}, i.e., every member of F(a) can be written in the form cn-1an-1 + cn-2an-2 + ··· + c1a + c0, where c0, c1, ···, cn-1 ∈ F (Figure A)

Let ϕ: E → F be an isomorphism of fields. Let K be an extension field of E and α ∈ K be algebraic over E with minimal polynomial p(x). Suppose that L is an extension field of F such that β(∈L) is root of Φ(p(x)) in F[x]. Then, ϕ extends to a unique isomorphism $\barϕ :E(α)→F(β)$ such that $\barϕ(α)=β~ and~ \barϕ$ agrees with Φ on E (Figure B.)

Theorem. Let ϕ: E→ F be an isomorphism of fields and let p(x) be a non-constant polynomial in E[x] and q(x) the corresponding polynomial in F[x] under ϕ. If K and L are splitting field of p(x) and q(x) respectively, then ϕ extends to an isomorphism ψ: K → L (Figure C)

Theorem. Let F be a field, let E = F(α) be an extension of F, where α is algebraic over F and let f ∈ F[x] be the irreducible polynomial of α over F. Let Φ: F → K be a homomorphism from F to a field K, and let L be an extension of K. If β ∈ L is a root of Φ(f), then there is a unique extension of Φ to a homomorphism Φ: E → L such that ϕ(α) = β and fixes F.

Theorem. Let L/K be an algebraic extension. Every field homomorphism Φ: K → C where C is an algebraically closed field, can be extended to a homomorphism L → C

# Galois Theory

A multiplicative character, linear character, or simply character of a group G in a field F is a group homomorphism from G to the multiplicative group of a field, σ: G → Fx By homomorphism is meant a mapping σ such that ∀a, b ∈ G: σ(a)·σ(β) = σ(α·β).

The characters σ12, ···, σn are independent if a1σ1 + a2σ2 + ··· + anσn ≡ 0 (≡ denotes equality as a function from G → Fx) for some a1, a2,..., an ∈ F ⇒ a1 = a2 = ... = an = 0.

Theorem. If G is a group and σ1, σ2, ···, σn are n mutually distinct characters of G in a field F, then σ1, σ2, ···, σn are independent.

Independence of field homomorphisms. Let K, L be two fields, let σ1, σ2, ···, σn: K → L be distinct field homomorphisms (σi≠σj, ∀i, j: i≠j). If a1σ1(α) + a1σ1(α)+ ··· + anσn(α) = 0 ∀α ∈ K, then σ1, σ2, ···, σn are independent, i.e., a1 = a2 = ··· = an = 0.

Let K, L be two fields. If σ1,···, σn: K → L are mutually distinct field homomorphisms of K into L and let F be the fixed field of K, F = {a ∈ K| σ1(a) = ··· = σn(a)}.

Let K be any field, and let σ1, ···, σn: K → K be distinct field automorphisms. Suppose that σ1, ···, σn forms a group under composition. If F is a fixed field of 1, ···, σn. Then, [K: F] = n ⇒ [K:F] = |G| = n.

The Galois group of the extension K/F, denoted by Gal(K/F), is the group of all F-automorphisms of K (the group of all automorphisms that fix the base field, Gal(K/F) = {σ ∈ Aut(K)| σ(a) = a ∀a ∈ F} e.g., $Gal(\mathbb{Q}(i)/\mathbb{Q})≋\mathbb{Z}/2\mathbb{Z},~ Gal(\mathbb{Q}(i, \sqrt{2})/\mathbb{Q})≋\mathbb{Z}/2\mathbb{Z} x \mathbb{Z}/2\mathbb{Z},~ Gal(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})≋$ {e}, $Gal(\mathbb{F_{p^r}}/\mathbb{F_p})≋\mathbb{Z}/r\mathbb{Z}$

Degree of the fixed field extension is at least the number of homomorphisms. Let K, L be two fields. If σ1,···, σn: K → L are mutually distinct field homomorphisms of K into L and let F be the fixed field of K, F = {a ∈ K| σ1(a) = ··· = σn(a)}. Then, [K : F] ≥ n.

Degree of Fixed Fields. Let K be any field, and let σ1, ···, σn: K → K be distinct field automorphisms. Suppose that σ1, ···, σn forms a group under composition. If F is a fixed field of σ1, ···, σn. Then, [K: F] = n

Proposition. Let K be any field and let G be a finite group of automorphisms of K. Suppose F is the fixed field of G. Then G = Gal(K/F) = Gal(K/KG).

To sum up. Let K be any field. S is any set of automorphisms of F. G is a group of automorphisms of K. [K : KS] ≥ |S|, [K : KG] = |G|, Gal(K, KG) = G. F ⊆ KGal(K/F) ⊆ K. K/F is Galois if F = KGal(K/F)

Theorem. A finite extension K/F is Galois iff the order of the Galois group equals the degree of the extension, i.e., |Gal(K/F)| = [K : F]

Equivalence of Definition of Normal Extensions. Let K/F be a finite extension of fields. Let $\bar \mathbb{F}$ be an algebraic closure of F that contains K. Then, a field extension is called normal if it satisfies one of the following equivalent statements:

1. Every embedding of K in $\bar \mathbb{F}$, σ: K → $\bar \mathbb{F}$, induces an F-automorphism from K to itself.
2. Every irreducible polynomial f ∈ F[x] with at least one root in K, splits (into a product of linear factors) completely in K[x].
3. K is the splitting field of a polynomial f ∈ F[x].

Let K be any extension of ℚ of degree 2. Then, we claim that Gal(K/Q) = ℤ/2ℤ, K/Q is normal and Galois.

Properties. If K is a normal extension of F, and L is an intermediate extension (that is, F ⊂ L ⊂ K) then K is normal extension of L. However, L/F is not necessarily normal.

Let K ⊆ L ⊆ F be finite extensions. Suppose K/F is a normal extension (⇒ K/L is normal). Then, L/F is a normal extension if and only if σ(L) ⊆ L ∀σ ∈ Gal(K/F).

Let F be any field.

1. An irreducible polynomial f ∈ F[x] is separable iff the greatest common factor of f and f' is 1, it is expressed or written as (f, f') = 1.
2. If F has characteristic zero, F is a finite field or perfect, then every polynomial in F[x] is separable.
3. If F has characteristic zero, F is a finite field or perfect, then any finite extension K/F is separable.

Characterization of Galois extensions. Let K/F be a finite extension. Then, K/F is Galois ↭ K is the splitting field of a separable polynomial over F.

Theorem. Let K/F be a finite extension. The following statements are equivalent:

1. K/F is Galois.
2. F is the fixed field of Aut(K/F), i.e., F = KGal(K/F)
3. A finite extension K/F is Galois iff the order of the Galois group equals the degree of the extension, i.e., |Gal(K/F)| = [K : F]
4. K/F is a normal, finite, and separable extension.
5. K is the splitting field of a separable polynomial f ∈ F[x] over F.
6. Let Char(F)=0 or F be finite (or more generally F is perfect). Then, a finite extension K/F is Galois if and only if K/F is normal.

Fundamental Theorem of Galois Theorem. Let K/F be a finite Galois extension. Let G = Gal(K/F) be the Galois group. Then, the following statements holds, (1) There is an inclusion-reversing bijective correspondence between subgroups of G and intermediate fields of K/F, given by H → KH, its inverse is defined by L → Gal(K/L). H1 ⊇ H2 ⇒ KH1 ⊆ KH2 and L1 ⊆ L2 ⇒ Gal(K/L1) ⊇ Gal(K/L2). Futhermore, it satisfies the following equality |H| = [K : KH] and [G : H] = [KH : F]

This map bijection fails when K/F is not Galois, e.g. K = $\mathbb{Q}(\sqrt[3]{2}), F = \mathbb{Q}$ is not Galois, G = Gal(K/F) = {1}. There is exactly one subgroup of G, but there are two intermediate fields, F and K.

(2) An intermediate fieldL (F ⊆ L ⊆ K) is Galois over F ↭ Gal(K/L) is a normal subgroup of G, Gal(K/L) ◁ G In this case, Gal(L/F) ≋ G/Gal(K/L)

Let K/F be a finite extension. An element α ∈ K is separable over F (A finite extension is always algebraic, and thus α is algebraic) if its irreducible polynomial over F is separable. A finite extension K/F is separable if every element of K is separable over F.

Definition. A field F is called perfect if either has characteristic zero or when F has characteristic p > 0, every element of F is a pth power.

Proposition. Let F be any field.

1. An irreducible polynomial f ∈ F[x] is separable iff the greatest common factor of f and f' is 1, it is expressed or written as (f, f') = 1 ↭ f and f' have no common roots
2. If F has characteristic zero, F is a finite field or perfect, then every polynomial in F[x] is separable.
3. If F has characteristic zero, F is a finite field or perfect, then any finite extension K/F is separable.

Theorem. Suppose f ∈ F[x] is irreducible. Then, f is separable if and only if f’ ≠ 0

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