Algebraic Extensions II. Primitive Element Theorem

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Recall

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_FL 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 x⁵ -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 a₁ (= a), a₂, ···, a_n and b₁ (= b), b₂, ···, 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 = b₁ 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.

Examples

• $\frac{ℚ[x]}{x^2-2} ≋ ℚ(\sqrt{2})$
• $\frac{ℚ(\sqrt{2})[x]}{x^2-3} ≋ ℚ(\sqrt{2}, \sqrt{3})$
• $\frac{ℚ(\sqrt{2}, \sqrt{3})[x]}{x^2-5} ≋ ℚ(\sqrt{2}, \sqrt{3}, \sqrt{5})$

  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, a², ···, a^n-1}), say p(x) = b_nxⁿ + ··· + b₀, 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, F₀ = F(b₀), F₁ = F₀(b₁), ···, F_n = F_n-1(b_n) ⇒ F_n = F(b₀, b₁, ···, 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] ··· [F₁:F₀][F₀: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 αⁿ + a_n-1α^n-1 + ··· + a₀ = 0 with a_i ∈ L algebraic in K.

Consider K ⊆ K(a₀) ⊆ K(a₀, a₁) ⊆ ··· ⊆ K(a₀···a_n-1) ⊆ K(a₀···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(a₀···a_n-1) with roots, namely α, in the field K(a₀···a_n-1, α) ⇒ α belongs to a finite extension of K, namely K(a₀···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, x² - (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, x² +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 α₁ be a root of f(x) ⇒ f(x) = (x -α₁)g(x), i.e., f(x) factors in F(α₁)[x], where g(x) ∈ F(α₁)[x] (K is a s.f. of f over F, g(x)∈ F[x]⊆ F(α₁)[x]). Thus, the minimal polynomial of α₂ (≠ α₁) working over F(α₁) must divide g(x) which has degree n-1, so [F(α₁, α₂):F(α)] ≤ deg(g(x)) = n-1. By induction, the result follows.

Examples

• F = ℚ, f = x³-1 = (x-1)(x² + x +1), K = ℚ(w) where w is the primitive third root of unity, [ℚ(w) : ℚ] = 2 < 3!
• F = ℚ, f = x³-2, K = ℚ($\sqrt[3]{2}$, w) where w is the primitive third root of unity, [ℚ($\sqrt[3]{2}$ : ℚ] = 6 = 3!

Bibliography

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