# 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] = dimFL 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, ···, an and b1 (= b), b2, ···, bn be the distinct zeros or roots of p(x) and q(x) respectively.

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

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) (-bj-) that are also zeros of r(x). However, r(bj) = p(c-dbj) = p(a + db -dbj) = p(a + d(b -bj)), and by election d is such that ai ≠ a + d(b -bj) 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, ···, an-1}), say p(x) = bnxn + ··· + b0, bi ∈ E. Then [By assumption, E is algebraic over F, each bi is algebraic over F], we construct a series of field extensions of F, as follows, F0 = F(b0), F1 = F0(b1), ···, Fn = Fn-1(bn) ⇒ Fn = F(b0, b1, ···, bn), and a is the zero of the minimal polynomial p(x) ∈ Fn[x] ⇒ [Fn(a):Fn] = n. Each bi is algebraic over F, we know that each [Fi+1:Fi] is finite, and therefore [Fn(a):F] = [Fn(a):Fn] [Fn:Fn-1] ··· [F1:F0][F0:F] is finite. Therefore, a ∈ K belongs to a finite extension of F, namely Fn(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 FF(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 + an-1αn-1 + ··· + a0 = 0 with ai ∈ L algebraic in K.

Consider K ⊆ K(a0) ⊆ K(a0, a1) ⊆ ··· ⊆ K(a0···an-1) ⊆ K(a0···an-1, α), all the extensions are finite because ai, 0 ≤ i ≤ n-1, are algebraic, and the last one is also finite because we’ve got a polynomial p(x)∈K(a0···an-1) with roots, namely α, in the field K(a0···an-1, α) ⇒ α belongs to a finite extension of K, namely K(a0···an-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

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.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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