JustToThePoint English Website Version
JustToThePoint en español

Maximize your online presence with our exclusive offer: Get a stunning hero banner, the hero you need and deserve, at an unbeatable price! Bew, 689282782, bupparchard@gmail.com

Algebraic Extensions. Characterization of field extensions.

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] = 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 the degree of α is naturally the degree of such polynomial, e.g., α = $\sqrt[5]{2}$ is algebraic root of x5 -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(7) transcendental? No, it is not. α is the root of the polynomial 8x3 +4x2 -4x -1 ∈ ℚ[x] with rational coefficients, and the other roots or conjugates are cos(7·2) and cos(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 x2 + 1 in ℝ, x2 -2 in ℚ, and x2 -2, x2 - 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 1
i x2 + 1 ℚ(i) ≋ℚ [x]/⟨x2+1⟩ 2
$\sqrt{2}$+1 (x-1)2-2 ℚ($\sqrt{2}$) ≋ ℚ[x]/⟨x2 -2⟩ 2
$\sqrt[3]{2}$ x3-2 ℚ($\sqrt[3]{2}$) ≋ ℚ[x]/⟨x3 -2⟩ 3
ξp $Φ_p(x)=\frac{x^p-1}{x-1}$ ℚ(ξp) ≋ ℚ[x]/⟨Φp(x)⟩ p-1

Notice that ξp is a primitive p-th root of unity for a prime p, Φp(x) is the p-th cyclotomic polynomial.

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


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, ···, αn-1} is a basis of F(α) as a vector space over F.

Characterization of field extensions

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).


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) = {$\frac{a_0+a_1α+···+a_nα^n}{b_0+b_1α+···+b_mα^m}:m,n∈ℕ, a_i, b_j ∈ F$, bj 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) = {$\frac{a_0+a_1α+···+a_nα^n}{b_0+b_1α+···+b_mα^m}:m,n∈ℕ, a_i, b_j ∈ F$, bj 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 an-1αn-1 + an-1αn-2 + ··· + a0. [F(α):F] = -finite- n = degree of the minimal polynomial- with basis {1, α, α2, ···, αn-1}.

Conversely, if [E:F] = m, ∀b∈ E, then the set {1, b, b2,···, bm} is linearly dependent, that is, m+1 elements of a m-dimensional vector space, so cmbm + cm-1bm-1 + ··· + c0 = 0 with not all ci zeros, then b is the zero of the polynomial p(x) = cmxm + cm-1xm-1 + ··· + c0, 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



  1. 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-1a)(bb-1) = a-1(ab)b-1 = a-10b-1 = 0⊥
  2. 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.


⇐) Suppose α ∈ L and α is in a finite field extension of K, say L’, that is, [L’ : K] = n < ∞ ⇒ 1, α, α2, ···, αn are n+1 elements of a n-dimensional vector space ⇒ ∃linear relation: a0 + a1α + ··· + anαn = 0, ai∈ K, not all zeros ⇒[α is a root of p(x) = a0 + a1x + ··· + anxn, ai ∈ 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, α2, ···, αn-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.
  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.
Bitcoin donation

JustToThePoint Copyright © 2011 - 2024 Anawim. ALL RIGHTS RESERVED. Bilingual e-books, articles, and videos to help your child and your entire family succeed, develop a healthy lifestyle, and have a lot of fun. Social Issues, Join us.

This website uses cookies to improve your navigation experience.
By continuing, you are consenting to our use of cookies, in accordance with our Cookies Policy and Website Terms and Conditions of use.