The initial state of density was so insanely massive that you can only grasp it by considering that was the only time that a physics magnitude could get somehow closer to the carbon-based ape-descendants’ stupidity, that’s why is called singularity and for good reasons, Apocalypse, Anawim, #justtothepoint.

A field L is said to be an extension field, denoted L/F, of a field F if F is a subfield of L, e.g., ℝ/ℚ and ℂ/ℝ. The degree of L/F, denoted [L:F] = dim_{F}L is the dimension of L considered as a vector space over F, e.g., [ℂ : ℝ] = 2, [ℝ : ℚ] = ∞.

L/K, α is algebraic over K if α is the root of a polynomial p(x), with coefficients in the field K[x]. Otherwise, α is called transcendental. If α is algebraic, there is a smallest irreducible polynomial that it is a root of, and the degree of α is naturally the degree of such polynomial, e.g., α = $\sqrt[5]{2}$ is algebraic root of x^{5} -2 = 0 and α has degree 5. However, π and e ∈ ℝ are transcendental.

Let f be the irreducible polynomial of α over F, the roots of f are called conjugates of α.

Exercise. is α = cos(^{2π}⁄_{7}) transcendental? No, it is not. α is the root of the polynomial 8x^{3} +4x^{2} -4x -1 ∈ ℚ[x] with rational coefficients, and the other roots or conjugates are cos(^{2π}⁄_{7}·2) and cos(^{2π}⁄_{7}·3).

Consider that ℂ = ℝ(i), ℚ($\sqrt{2}$), ℚ($\sqrt{2}, \sqrt{3}$) are **algebraic extensions**. The elements which has been adjoint to these fields are the root of some irreducible polynomials, namely x^{2} + 1 in ℝ, x^{2} -2 in ℚ, and x^{2} -2, x^{2} - 3 in ℚ respectively. i, $\sqrt{2}$, and $\sqrt{3}$ are called **algebraic elements**, they are the zeros or roots of some nonzero polynomial in F[x] where F is a field (e.g., ℝ or ℂ).

Algebraic number | Minimal polynomial | Field | Degree |
---|---|---|---|

a/b ∈ Q, gcd(a, b) = 1 | bx − a | ℚ | 1 |

i | x^{2} + 1 |
ℚ(i) ≋ℚ [x]/⟨x^{2}+1⟩ |
2 |

$\sqrt{2}$+1 | (x-1)^{2}-2 |
ℚ($\sqrt{2}$) ≋ ℚ[x]/⟨x^{2} -2⟩ |
2 |

$\sqrt[3]{2}$ | x^{3}-2 |
ℚ($\sqrt[3]{2}$) ≋ ℚ[x]/⟨x^{3} -2⟩ |
3 |

ξ_{p} |
$Φ_p(x)=\frac{x^p-1}{x-1}$ | ℚ(ξ_{p}) ≋ ℚ[x]/⟨Φp(x)⟩ |
p-1 |

Notice that ξ_{p} is a primitive p-th root of unity for a prime p, Φ_{p}(x) is the p-th cyclotomic polynomial.

Lemma. If α in E is algebraic over F, then it minimal polynomial is prime.

Proof.

Let α ∈ E, suppose that its minimal polynomial f(x), deg(f(x)) = n is not prime ⇒ it factors f(x) = g(x)h(x) where f(α) = 0 ⇒ g(α) = 0 or h(α) = 0 ⇒ you can use g or h as a generator of the ideal Ker(Φ) = ⟨f(x)⟩ leading to a generator of lower degree ⊥ F(α) ≋ F[x]/⟨f(x)⟩, i.e., [F(α): F] = degree of minimal polynomial over α since {1, α, ···, α^{n-1}} is a basis of F(α) as a vector space over F.

Let E be an extension of the field F and let α ∈ E. If α is transcendental over F, then F(α) ≋ F(x). If α is algebraic over F, then F(α) ≋ F[x]/⟨p(x)⟩, where p(x) is a unique monic irreducible over F (ker(Φ)=⟨p(x⟩), and a polynomial in F[x] of minimum degree such that p(α) = 0.

p(x) is called the minimal polynomial of α, for f(x) ∈F[x], f(α)=0 ↭ p(x)|f(x).

Proof.

Let Φ:F[x] → F(α), defined by Φ(f(x)) = f(α) be a natural ring homomorphism.

If α is **transcendental** over F, i.e., it is not the zero of some nonzero polynomial in F[X] ⇒ the evaluation homomorphism Φ:F[X] → F(a), defined by f(x) → f(a) has trivial kernel, Ker(Φ) = {0}, hence Φ is injective and F(α) = Im(Φ) ≋ F[x]/Ker(Φ) = F[x]

Let F(x) be the field of quotients of F[x] = {$\frac{f(x)}{g(x)}|f(x),g(x)∈F[x],g(x)≠0$}. F(α) = {$\frac{a_0+a_1α+···+a_nα^n}{b_0+b_1α+···+b_mα^m}:m,n∈ℕ, a_i, b_j ∈ F$, b_{j} not all zero}, this is an infinite-dimensional vector space over F and is naturally isomorphic to the field of rational fractions F(x) = Frac(F[x]) in the indeterminate x. All simple transcendental extension of a given field F are therefore isomorphic.

We could extend Φ to an isomorphism $\bar Φ$:F(x) → F(α), defined by $\bar Φ(\frac{f(x)}{g(x)})=\frac{f(α)}{g(α)}$

Another way of putting it (this paragraph maybe too repetitive for the avid reader), F(α) = Im(Φ) ≋ F[x]/Ker(Φ) = F[x]. **This is merely an integral domain, not a field.** A simple transcendental extension F ≤ F(α) ≤ E is the subfield of rational expressions in α: F(α) = {$\frac{a_0+a_1α+···+a_nα^n}{b_0+b_1α+···+b_mα^m}:m,n∈ℕ, a_i, b_j ∈ F$, b_{j} not all zero} ≤ E. This is an infinite-dimensional vector space over F and is naturally isomorphic to the field of rational functions F(x) = Fract(F[x]) in the indeterminate x. And yet again, we get the same result, all simple transcendental extensions of a given field F are isomorphic.

If α is **algebraic** over F, then Ker(Φ) ≠ {0}, the kernel is a non-trivial ideal ⇒ F[x]/Ker(Φ) ≋ Im(Φ) = F(α). Since Im(Φ) is a subring containing 1 ⇒ [Theorem. **Any subring of a field containing 1 is an integral domain**, that is, a ring with no zero divisors] it is an integral domain.

If Φ:R ⇾ S is a ring homomorphism, then Φ(r)=Φ(1·r)=Φ(1)·Φ(r), so Φ(1) is a unity with respect to Φ(R) which is a subring of S ⇒ If Φ is surjective, then Φ(1) is the unity of S.

Every ideal in F[x] is principal, and since Ker(Φ) is a non-trivial ideal ⇒ [A principal ideal is an ideal that is generated by a single element of the ring] Ker(Φ) = ⟨p(x)⟩ for some non-constant p(x) ∈ F[x].

⟨p(x)⟩ = ⟨f(x)⟩ ↭ f(x) | p(x) and p(x) | f(x) ↭ f and p are associates ↭ if one can be obtained from the other by multiplying by some unit. Thus, we may divide p(x) by its leading coefficient to obtain the unique monic generator of ker(Φ). p(x) ≠ 0, p(x) is primitive, and F[x]/⟨p(x)⟩ ≋ Im(Φ) = F(α) is a field containing F and α, p(x) is a polynomial of minimum degree such that p(α)=0.

If degree of p(x) is n, the elements of F(α) look like a_{n-1}α^{n-1} + a_{n-1}α^{n-2} + ··· + a_{0}. [F(α):F] = -finite- n = degree of the minimal polynomial- with basis {1, α, α^{2}, ···, α^{n-1}}.

Conversely, if [E:F] = m, ∀b∈ E, then the set {1, b, b^{2},···, b^{m}} is linearly dependent, that is, m+1 elements of a m-dimensional vector space, so c_{m}b^{m} + c_{m-1}b^{m-1} + ··· + c_{0} = 0 with not all c_{i} zeros, then b is the zero of the polynomial p(x) = c_{m}x^{m} + c_{m-1}x^{m-1} + ··· + c_{0}, i.e., ∀b ∈ E, b is algebraic over F. To sum up, a finite extension, say [E:F] = m, implies E is algebraic extension of F. In other words, we can associate b with its minimal polynomial.

Theorem. Let F be a field, p(x) ∈ F[x] - F (a non constant polynomial). Then, the ring F[x]/⟨p(x)⟩ is a field ↭ p is irreducible

Proof.

Recall:

- A field is an integral domain, too, i.e., it has no zero-divisors (the product of two non-zero numbers is always non-zero) Suppose for the sake of contradiction a, b ∈ R - {0} s.t. a·b = 0. However, 1 = 1·1 = (a
^{-1}a)(bb^{-1}) = a^{-1}(ab)b^{-1}= a^{-1}0b^{-1}= 0⊥ - Let F be a field, p(x) ∈ F[x] -{0}. Then, (F[x]/⟨p(x)⟩, +, ∘) is a commutative ring.

⇒) Suppose for the sake of contradiction, the ring F[x]/⟨p(x)⟩ is a field, but p is reducible ⇒ ∃a, b ∈ F[x] - F, p = ab. Hence, a ∘ b = a·b + ⟨p(x)⟩ =[a·b = p ↭ ab + ⟨p(x)⟩ =_{Absorption} ⟨p(x)⟩] ⟨p(x)⟩ ⇒ F[x]/⟨p(x)⟩ is not a field because it has zero-divisors, namely a and b.

⇐) Suppose p is irreducible. We know (Recall 2) that (F[x]/⟨p(x)⟩, +, ∘) is a commutative ring. The unity of F is also the unity of F[x]/⟨p(x)⟩, namely 1 + ⟨p(x)⟩. The only problem is the existence of inverses.

Let a(x) ∈ F[x]/⟨p(x)⟩ - {0} ⇒[Recall {0} = ⟨p(x)⟩ = 0 + ⟨p(x)⟩] 0 ≤ deg(a) < deg(p). Let g = gcd(a, p).

g = gcd(a, p) ⇒ g | a ⇒ ∃b ∈ F[x], a = gb ⇒[a(x) ∈ F[x]/⟨p(x)⟩ - {0}] g ≠ 0 Λ b ≠ 0 ⇒ 0 ≤ deg(g) Λ 0 ≤ deg(b) ⇒[deg(a) = deg(g) + deb(b)] 0 ≤ deg(g) ≤ deg(a) < deg(p) (i)

g = gcd(a, p) ⇒ g | p ⇒ ∃h ∈ F[x], p = gh ⇒[p(x) ∈ F[x] - F] g ≠ 0 Λ h ≠ 0 ⇒[By assumption, p is irreducible, and p = gh] g ∈ F- {0} or h ∈ F- {0} ⇒ deg(g) = 0 or deg(h) = 0 ⇒[Recall p = gh] deg(g) = 0 or deg(g) = deg(p) ⇒[deg(g) ≤ deg(a) < deg(p) (i)] deg(g) = 0 ⇒ g = 1 (g is monic)

By the gcd theorem, 1 = gcd(a, p) = sa + tp. (s + ⟨p(x)⟩) ∘ (a + ⟨p(x)⟩) = sa + ⟨p(x)⟩ = (1 -tp) + ⟨p(x)⟩ =[By absorption] 1 + ⟨p(x)⟩ ⇒[Abusing notation, we refer to cosets by their representatives] a^{-1} = s ∎

Theorem. Let L/K be a field extension. Let α ∈ L, α is algebraic over K ↭ α is in a finite extension of K.

Proof.

⇐) Suppose α ∈ L and α is in a finite field extension of K, say L’, that is, [L’ : K] = n < ∞ ⇒ 1, α, α^{2}, ···, α^{n} are n+1 elements of a n-dimensional vector space ⇒ ∃linear relation: a_{0} + a_{1}α + ··· + a_{n}α^{n} = 0, a_{i}∈ K, not all zeros ⇒[α is a root of p(x) = a_{0} + a_{1}x + ··· + a_{n}x^{n}, a_{i} ∈ K] α is algebraic over K.

⇒) Conversely, suppose α ∈ L, α is algebraic over K. Then we want to prove that α is in a finite field extension.

Recall. Suppose p(x) is an irreducible polynomial in K[x], then K[x]/⟨p⟩ is a field.

α is the root of p(x) ∈ K[x], p irreducible (Figure 1), Φ: x + ⟨p(x)⟩ → α

And Φ(K[x]/⟨p(x)⟩) in L is a field containing α, and it is also a finite extension of K (its dimension equals deg(p))∎ Recall, K[x]/⟨p(x)⟩ ≋ Φ(K[x]/⟨p(x)⟩) = K(α), and a basis as a vector space over K is {1, α, α^{2}, ···, α^{n-1}} where n = deg(p).

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn (Abstract Algebra), and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- Andrew Misseldine: College Algebra and Abstract Algebra.