Their carbon-based ape-descended life forms are so astonishingly primitive and comprehensively stupid that bleeding was considered a universal remedy for centuries and they have believed from immemorial times that the eternal, almighty, all-knowing, all-powerful, and omnipresent God cannot handle money, so he always needs more money; dreams equals reality; some bubbles somehow will never burst; they can tax like there’s no tomorrow, hyper-regulate, borrow like a drunken sailor, spend their way to prosperity, fiat money will never go to its intrinsic value, and they are all very intelligent or, at least, more intelligent than average, 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 degree of α is the degree of such polynomial, e.g., α = $\sqrt[5]{2}$ is algebraic root of x^{5} -2 = 0 and α has degree 5, π and e ∈ ℝ are transcendental.

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

Theorem. **Primitive Element Theorem**. If F is a field, char(F) = 0, i.e.,a field of characteristic zero, and a and b are algebraic over F then ∃c ∈ F(a, b): F(a, b) = F(c).

An element “c” with the property that E = F(c) is called a primitive element of E.

Proof.

Let p(x) and q(x) be the minimal polynomials over F for and b respectively. In some extension K of F, let a_{1} (= a), a_{2}, ···, a_{n} and b_{1} (= b), b_{2}, ···, b_{n} be the distinct zeros or roots of p(x) and q(x) respectively.

Let’s choose an element d ∈F, d ≠ (a_{i} -a)/(b -b_{j}) ∀i, j, i ≥ 1, j > 1 ⇒ d(b -b_{j}) ≠ (a_{i} -a) ⇒ a_{i} ≠ a + d(b -b_{j})

Claim: Let c = a + db, F(a, b) = F(c).

- c = a + db ⇒ F(c) ⊆ F(a, b).
- Let’s prove that F(a, b) ⊆ F(c).

It suffices to prove that b ∈ F(c) because if b ∈ F(c), a = c -bd ⇒ a ∈ F(c) ⇒ F(a, b) ⊆ F(c).

Let’s consider the polynomials q(x) and r(x)= p(c -dx) over F(c) ⇒ [q(b)=0 and r(b)=p(c-db)=p(a)=0] q and r are divisible by the minimal polynomial s(x) for b over F(c) (s(x)∈F(c)[x]).

Since s(x) is a common divisor of q(x) and r(x), the only possible zeros of s(x) in K are the zeros of q(x) (-b_{j}-) that are also zeros of r(x). However, r(b_{j}) = p(c-db_{j}) = p(a + db -db_{j}) = p(a + d(b -b_{j})), and by election d is such that a_{i} ≠ a + d(b -b_{j}) for j > 1 ⇒ b = b_{1} is the only zero of s(x) in K[x] ⇒ s(x) = (x -b)^{u} ⇒ [s(x) is the minimal/irreducible polynomial common divisor of q and r and F has characteristic 0. If f is an irreducible polynomial over a field of characteristic zero, then f(x) has no multiples zeroes] u = 1, s(x) = (x -b) ∈ F(c)[x] ⇒ b ∈ F(c)∎

Theorem. **Algebraic over Algebraic is Algebraic.** If K is an algebraic extension of E and E is an algebraic extension of F, then K is an algebraic extension of F.

- $\frac{ℚ[x]}{x^2-2} ≋ ℚ(\sqrt{2})$
- $\frac{ℚ(\sqrt{2})[x]}{x^2-3} ≋ ℚ(\sqrt{2}, \sqrt{3})$
- $\frac{ℚ(\sqrt{2}, \sqrt{3})[x]}{x^2-5} ≋ ℚ(\sqrt{2}, \sqrt{3}, \sqrt{5})$
Basically, we are building new fields, $ℚ → ℚ(\sqrt{2}) → ℚ(\sqrt{2}, \sqrt{3}) → ℚ(\sqrt{2}, \sqrt{3}, \sqrt{5})$

Proof.

An extension K of a field F is said to be algebraic if every element of K is algebraic over F. We want to show that every element a ∈ K belongs to a finite extension of F (Recall. A finite extension of a field is an algebraic extension of F) ⇒ a is algebraic over F.

By assumption, K is an algebraic extension of E ⇒ ∀a∈ K, a is algebraic over E ⇒ a is the zero of some minimal polynomial in E[x] (this polynomial is monic, irreducible, and prime and [E(a):E] = deg(p), the basis is {1, a, a^{2}, ···, a^{n-1}}), say p(x) = b_{n}x^{n} + ··· + b_{0}, b_{i} ∈ E. Then [By assumption, E is algebraic over F, each b_{i} is algebraic over F], we construct a series of field extensions of F, as follows, F_{0} = F(b_{0}), F_{1} = F_{0}(b_{1}), ···, F_{n} = F_{n-1}(b_{n}) ⇒ F_{n} = F(b_{0}, b_{1}, ···, b_{n}), and a is the zero of the minimal polynomial p(x) ∈ F_{n}[x] ⇒ [F_{n}(a):F_{n}] = n. Each b_{i} is algebraic over F, we know that each [F_{i+1}:F_{i}] is finite, and therefore [F_{n}(a):F] = [F_{n}(a):F_{n}] [F_{n}:F_{n-1}] ··· [F_{1}:F_{0}][F_{0}:F] is finite. Therefore, a ∈ K belongs to a finite extension of F, namely F_{n}(a) ⇒[A finite extension of a field is an algebraic extension of F] a is algebraic over F.

Theorem. **Subfield of Algebraic Elements**. Given a field extension [E:F], then the elements of E that are algebraic over F form a subfield.

Proof. If a, b ∈ E are algebraic over F, b ≠ 0, we want to show that a + b, a -b, ab, and a/b are algebraic over F.

a + b, a -b, ab, and a/b ∈ F(a, b), and [F(a, b):F] = [Notice that a is algebraic over F ⇒ a is algebraic over F(b) ⇒ [F(a, b):F(b)]< ∞. Besides, b is algebraic over F, then [F(b):F] < ∞] [F(a, b):F(b)][F(b):F] < ∞ ⇒ A finite extension of a field is an algebraic extension of F ⇒ **F(a, b) is a subfield that is an algebraic extension of F** ⇒ a + b, a -b, ab, and a/b are algebraic over F ∎

Definition. For any extension E of a field F, the subfield of E of the elements that are algebraic over F is called the algebraic closure of F in E.

Theorem. Let L/K be an extension field. If α ∈ L is a root of p(x) with algebraic coefficients in K, then α is algebraic.

Proof.

Let α ∈ L, root of p(x), so it satisfies α^{n} + a_{n-1}α^{n-1} + ··· + a_{0} = 0 with a_{i} ∈ L algebraic in K.

Consider K ⊆ K(a_{0}) ⊆ K(a_{0}, a_{1}) ⊆ ··· ⊆ K(a_{0}···a_{n-1}) ⊆ K(a_{0}···a_{n-1}, **α**), all the extensions are finite because a_{i}, 0 ≤ i ≤ n-1, are algebraic, and the last one is also finite because we’ve got a polynomial p(x)∈K(a_{0}···a_{n-1}) with roots, namely α, in the field K(a_{0}···a_{n-1}, **α**) ⇒ α belongs to a finite extension of K, namely K(a_{0}···a_{n-1}, **α**) -sometimes I wonder if being too verbose, makes things seem more complicated that they really are-, hence α is algebraic ∎

Exercise. Is e + π transcendental? Is e·π transcendental? These questions are open problems, however we can tell that e + π or e + π is transcendental.

Consider the polynomial, x^{2} - (e + π)x + eπ, its roots are e and π. If e + π and e + π are both algebraic ⇒[Theorem. Let L/K be an extension field. If α ∈ L is root of p(x) with algebraic coefficients in K, then α is algebraic] the polynomial's roots (e and π) are algebraic ⊥

Definition. 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 (Fundamental Theorem of Algebra), but ℝ and ℚ are not algebraically closed, x^{2} +1 has no root in ℝ or ℚ.

Theorem. Let f(x) ∈ F[x] be a polynomial of degree n. Let K be a splitting field of f(x) over F. Then [K:F] ≤ n!

Proof.

K is a splitting field of f(x) over F. Let α_{1} be a root of f(x) ⇒ f(x) = (x -α_{1})g(x), i.e., f(x) factors in F(α_{1})[x], where g(x) ∈ F(α_{1})[x] (**K is a s.f. of f over F, g(x)∈ F[x]⊆ F(α _{1})[x]**). Thus, the minimal polynomial of α

- F = ℚ, f = x
^{3}-1 = (x-1)(x^{2}+ x +1), K = ℚ(w) where w is the primitive third root of unity, [ℚ(w) : ℚ] = 2 < 3! - F = ℚ, f = x
^{3}-2, K = ℚ($\sqrt[3]{2}$, w) where w is the primitive third root of unity, [ℚ($\sqrt[3]{2}$ : ℚ] = 6 = 3!

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.