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Field Theory

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Recall

  1. A ring (R, +, ·). (i) (R, +) is an Abelian group: closure, associative, identity, inverses, and commutative. (ii) (R, .) closure and associative. (iii) Distributive laws.
  2. A non-empty subset S of a ring R is called ideal of R is: ∀a, b ∈ S, r ∈ R ⇒ a - b ∈ S, ar ∈ S, ra ∈ S. Examples: {0} and R are trivial ideals.
  3. An ideal A in a ring R is maximal if (i) A ≠ R, (ii) ∀B ideal B ⊇ A, either B = A or B = R.
  4. Th. R/A is a field iff A is a maximal ideal.
  5. Th. ⟨p(x)⟩ is maximal ↭ p(x) is irreducible. Let F be a field, p(x) be an irreducible polynomial over F, then F[x]/⟨p(x)⟩ is a field.

Characteristic of a field

A field F is a commutative ring with identity in which every nonzero element has an inverse or Fx = F -{0} is an Abelian group under multiplication.

Let 0F and 1F denote the additive and multiplicative identity of F respectively, then F is closed under addition by definition, and therefore, it contains 1F + 1F, 1F + 1F+ 1F, and so on. These elements may not all be distinct. Let’s define n·1F = 1F + ··· + 1F (n times).

We define the characteristic of a field F, char(F), as the smallest positive integer p such that p·1F = 1 + 1 + ··· + 1 (p times) = 0 if such a p exists, or char(F) = 0 if 1 + 1 + ··· + 1 ≠ 0 ∀n ∈ N, e.g., char(ℝ) = char(ℚ) = 0; and for any prime p, $\mathfrak{F}_p=ℤ/pℤ$ has characteristic p.

We know that ∀n, m ∈ ℤ+,

  1. n·1F + m·1F = (m + n)·1F
  2. (n·1F)(m·1F) = mn·1F

Theorem. Let F be a field. Then, the characteristic of F is either zero or a prime number,

Proof.

Suppose for the sake of contradiction, char(F) = n, n ≠ 0 and n is not prime ⇒ ∃a, b ∈ F: n = ab, with n·1F = 1 + 1 + ··n·· + 1 + 1 = 0 ⇒ ab·1F = 0 ⇒ (a·1F)(b·1F) = 0 ⇒ [A field is a ring with no zero divisors.] a·1F = 0 or b·1F = 0 ⊥ (char(F) = n). Therefore, the smallest positive integer p such that p·1F = 1 + 1 + ··· + 1 (p times) = 0 if such a p exists, it is necessarily a prime. Futhermore, if n·1F = 0, then n is divisible by char(F) = p, and p·n = n + n + ··· + n (p times) = n·1F + n·1F + ··· + n·1F (p times) =[By assumption, n·1F = 0] 0.

Let’s define (-n)·1F = -(n·1F) for n > 0, 0·1F = 0, then we have the following ring homomorphism Φ: ℤ → F, defined as Φ(n) = n·1F where ker(Φ) = char(F)ℤ.

If we take the quotient by the kernel gives us an injection of either ℤ (char(F) = 0) or ℤ/pℤ (char(F) = p). Since F is a field, it contains a subfield isomorphic either to ℚ (the field of fractions of ℤ) or to $\mathfrak{F}_p=ℤ/pℤ$ (the field of fractions of ℤ/pℤ). In either case, it is the smallest subfield of F containing 1F, the prime subfield, that is, the field generated by the multiplicative identity 1F of F, e.g., the prime subfield of both ℚ and ℝ is ℚ.

Field Extension

Definition. If F is a subfield of a field E, then E is called an extension field of F, denoted E/F or E:F. (Figure 1.a.) Image 

Examples.

Suppose that a field F does not contain any element x such that x2 = -1 (as in ℝ), then the polynomial x2 + 1 would be irreducible in F[x] ⇒ The ideal ⟨x2 + 1⟩ would be maximal and E = F[x]/⟨x2 + 1⟩ is an extension field of F that contains a root for x2 = -1. Figure 1.b.

Similarly, we can construct splitting fields of any polynomial in F[x]. Let E be an extension of F, the given polynomial splits into a product of linear factors, e.g., a0 + a1x + ··· + anxn = (x -α1)(x -α2) ··· (x -αn)

Let E be an extension of F,

  1. Then, E is a vector space over F.
  2. E will have a basis set.
  3. The dimension of E considered as a vector space over F is called the degree of E over F and denoted as [E:F]. Figure 1.c.

The extension is said to be finite if [E:F] is finite, its base could be counted. Otherwise, if [E:F] is infinite, the extension is said to be ∞.

Examples.

  1. {1, i} is a basis of ℂ, dim(ℂ) = [ℂ : ℝ] = 2.
  2. [ ℝ : ℚ] = [ ℂ : ℚ] = Infinite
  3. {1, $\sqrt{2}$} is a basis of ℚ($\sqrt{2}$) and [ℚ($\sqrt{2}$):ℚ] = 2
  4. {1, $\sqrt{2}, \sqrt{3}, \sqrt{6}$} is a basis of ℚ($\sqrt{2}, \sqrt{3}$) and [ℚ($\sqrt{2}, \sqrt{3}$):ℚ] = 4, and [ℚ($\sqrt{2}, \sqrt{3}$):ℚ($\sqrt{2}$] = 2.

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 (x2 + 1 in ℝ, x2 -2 in ℚ, (x2 -2)(x2 - 3) in ℚ respectively), and they are called algebraic elements.

Multiplicative property of degree extensions. Let F ⊆ E ⊆ K be a tower of fields. If [K:E] < ∞ and [E:F] < ∞ then, (i) [K:F] < ∞, and (ii) [K:F] =[K:E][E:F]

Proof:

Let [K:E] = m and [E:F] = n, and consider the basis of K over E be {v1, v2, ···, vm}, and the basis of E over F be {w1, v2, ···, wn}.

∀u ∈K, then u = $\sum_{i=1}^m a_iv_i, ~a_i~∈~E$ because K is a vector space over E with basis {v1, v2, ···, vm}, and E is a vector space over F with basis {w1, v2, ···, wn}, so ai ∈ E, then $a_i=\sum_{j=1}^n b_{ij}w_j, b_{ij}∈F$

u = $\sum_{i=1}^m a_iv_i=\sum_{i=1}^m\sum_{j=1}^n b_{ij}(v_iw_j),b_{ij}∈F$

And therefore, the set {viwj} where i=1,···, m and j=1, ···, n form a set generator with nm elements of the vector space K over F, so [K:F] ≤ mn, and therefore [K:F] < ∞ (i).

To prove the set {viwj} forms a basis, we need to show these vectors are linearly independent.

Assume $\sum_{i=1}^m\sum_{j=1}^n b_{ij}(v_iw_j)=0, b_{ij}∈F$. We need to prove that bij = 0. This sum can be rewritten as $\sum_{i=1}^m\sum_{j=1}^n (b_{ij}w_j)v_i=0$ [Commutativity and Associativity] and $\sum_{i=1}^m c_iv_i=0, where~ c_i=\sum_{j=1}^n (b_{ij}w_j)∈E$ is a linear combination of vectors of E with scalars of F.

As {v1, v2, ···, vm} is a basis of K over E ⇒ ci = 0, i = 1, 2, ···, m ⇒ $c_i=\sum_{j=1}^n (b_{ij}w_j)=0$ where bij ∈ F and {w1, v2, ···, wn} is a basis of E over F ⇒ bij = 0 ∎ Therefore, (ii) [K:F] = n·m = [K:E]·[E:F]

Exercise. Show that $\sqrt{2}∉ℚ(\sqrt[5]{2})$. The irreducible polynomial of $\sqrt[5]{2}$ is x5 - 2 ∈ ℚ[x] (Eisenstein criteria, p = 2) ⇒[x5 - 2 irr. ⇒ ⟨x5-2⟩ maximal ⇒ ℚ($\sqrt[5]{2}) = \frac{ℚ[x]}{⟨x^5-2⟩}$] [ℚ($\sqrt[5]{2}$) : ℚ] = 5. For the sake of contradiction suppose $\sqrt{2}∈ℚ(\sqrt[5]{2})$, then there is a tower of extensions $ℚ ⊆ ℚ(\sqrt{2}) ⊆ ℚ(\sqrt[5]{2})$ ⇒ [ℚ($\sqrt[5]{2}$) : ℚ] = 5 = [ℚ($\sqrt[5]{2}$) : ℚ($\sqrt{2}$)] · [ℚ($\sqrt{2}$) : ℚ] = α · 2, α ∈ ℕ ⊥. Observe that [ℚ($\sqrt{2}$) : ℚ] = 2, x2 -2 is an irreducible polynomial of ℚ with root $\sqrt{2}$.

Definition. An embedding of F into E is an injecting homomorphism of a field F into a field E, σ: F → E.

Theorem. Let E and F be two fields, and let σ: F → E be an embedding of a field F into a field E. Then there exist a field K such that F is a subfield of K and σ can be extended to an isomorphism of K onto E.

Notice that (i) K is an extension of F. (ii) σ can be extended to σ*: K → E, and σ* is an isomorphism of K onto E (Figure 1.e.).

Image 

Proof.

Notice that σ: F → E is an embedding ⇒ σ is homomorphism, 1-1, σ(F)⊂E (it is not onto). We are going to define a new set S such that cardinality(S) = cardinality(E -σ(F)), and S ∩ F = ∅. Then, let f: S → E -σ(F) be a bijection (1-1 and onto), and let K = F ∪ S.

Define σ*: K → E, σ*(a)$ = \begin{cases} σ(a), &a ∈ F\\\\ f(a), &a ∈ S \end{cases}$

We claim that σ* is an isomorphism of K onto E.

  1. It is well-defined. If a ∈ F, a is “landing” on E because σ is an embedding of a field F into E. If a ∈ S, a goes to E - σ(F) ⊆ E.
  2. σ* is one-to-one, i.e., σ*(a) = σ*(b) ⇒ a = b. If a, b ∈ F, σ*(a) = σ*(b) ⇒ σ(a) = σ(b) ⇒ [σ is one-to-one] a = b. If a, b ∈ S, σ*(a) = σ*(b) ⇒ f(a) = f(b) ⇒ [f is bijection] a = b. Notice that S ∩ F = ∅, so σ*(a) = σ*(b) ⇒ a, b ∈ S or a, b ∈ F.
  3. σ* is onto. Notice that E is composed of σ(F) and E - σ(F). Every element of σ(F) has a pre-image element because σ(F) is obviously onto, and every element of E - σ(F) has a pre-image element because f is bijective.
  4. σ* is homomorphism. It is left to the reader to prove it, but considering that,

We have defined K = F ∪ S. The field structure on K is defined ∀x, y ∈ K, as

Therefore, if σ: F → E is an embedding, we can identify F with the corresponding image σ(F) in E, basically we write 'x' in place of σ(x). Futhermore, E (E ≋ K), hence it can be considered as an extension of F.

Theorem. Let p(x) be an irreducible polynomial in F[x]. Then, there exists an extension E of F in which p(x) has a root.

Proof. Suppose p(x) is an irreducible polynomial in F[x] ⇒ The ideal generated by p(x), i.e., ⟨p(x)⟩ is a maximal ideal ⇒ E = F[x]/⟨p(x)⟩ is a field.

Define σ: F → E by σ(a) = $\bar a=a+⟨p(x)⟩$ It is easy to prove that σ is one-to-one and homomorphism, so σ is an embedding of F into E. Using the previous result, E is regarded as an extension of F.

Let p(x) = a0 + a1x + ··· + anxn, be a polynomial in F[x], an ≠ 0, ai ∈ F

Then, in E, x + ⟨p(x)⟩ is a zero or root of p. $p(x + ⟨p(x)⟩)=a_0 + a_1(x + ⟨p(x)⟩) + ··· + a_n(x + ⟨p(x)⟩)^n=$ a0 + a1x + ⟨p(x)⟩ + a2x2 + ⟨p(x)⟩ ··· + anxn + ⟨p(x)⟩ = p(x) + ⟨p(x)⟩ =[By absorption] 0 + ⟨p(x)⟩.

Krnoecker’s Theorem. Existence of extension. Let F be a field and let f(x) be a non-constant polynomial in F[x]. Then, there exists an extension E of F in which f(x) has a root or zero. (Figure 1.f)

Proof.

There are two cases:

  1. Case I. f(x) ∈ F[x] has a root in F. Thus, we can say, there exists an extension F of F, the trivial extension, in which f(x) has a root.
  2. Case II. Let f(x) ∈ F[x] does not have a root in F.

Every field F is a UFD (every non-zero non-unit element can be written as a product of prime or irreducible elements, uniquely up to order and units) because it is an integral domain and everything non-zero is a unit. The polynomial ring over F is a UFD, that is, F[x] is a UFD, too, and therefore f(x) can be factored into the product of irreducible polynomials. Let p(x) be an irreducible factor of f(x) in F[x].

Let p(x) be an irreducible polynomial in F[x] ⇒ We know that the ideal generated by any irreducible polynomial is a maximal ideal, ⟨p(x)⟩ is a maximal ideal. Moreover, the quotient ring E = F[x]/⟨p(x)⟩ is a field.

Let’s define σ: F → E by σ(a) = $\overline a$ = a + ⟨p(x)⟩. σ is a one-to-one homomorphism, and therefore E has a subfield isomorphic to F. It is reasonable to think of E as containing F and identify the coset a + ⟨p(x)⟩ with its coset representative that belongs to F as just “a” and vice versa or, in other words, σ is an embedding of F into E.

Finally, let’s prove that p(x) = a0 + a1x + ··· + anxn has a root in E, where ai ∈ F, an ≠ 0

Then, in E, x + ⟨p(x)⟩ is a zero or root of p. p(x + ⟨p(x)⟩) = a0 + a1(x + ⟨p(x⟩) + a2(x + ⟨p(x⟩)2 + ··· + an(x + ⟨p(x⟩)n = a0 + a1(x + ⟨p(x⟩) + ··· + an(xn + ⟨p(x⟩) = p(x) + ⟨p(x⟩) = 0 + ⟨p(x⟩)∎

Theorem. Let p(x) be an irreducible polynomial in F[x] and let u be a root of p(x) in an extension E of F. Then,

  1. F(u), the subfield of E generated by F and u is the smallest field containing F and u. It is indeed F[u], the smallest subring containing both F and u = {b0 + b1u + ··· + bmum ∈ E | b0 + b1x + ··· + bmxm ∈ F[x]}

  2. If the degree of p(x) is n, the set {1, u, ···, un-1} forms a basis of F(u) over F, i.e., each element of F(u) can be written uniquely as c0 + c1u + ··· + cnun-1 where ci ∈ F, and [F(u):F] = n.

Example. Let p(x) = x2 -2 be an irreducible polynomial in ℚ[x], x = $\sqrt{2}$ is its root in some extension E of ℚ. This theorem says,

(i) ℚ($\sqrt{2}$) is a subfield of E generated by ℚ and $\sqrt{2}$.

(ii) The set {1, $\sqrt{2}$} is the basis set of ℚ($\sqrt{2}$) because degree(x2 -2 ) = 2. Moreover, every element in ℚ($\sqrt{2}$) can be uniquely represented as linear combination of 1 and $\sqrt{2}$, and [ℚ($\sqrt{2}$):ℚ] = 2.

Proof:

Let p(x) be any irreducible polynomial in F[x], and let u be a root of p(x) in an extension E of E.

(i) Let F(u) denote the smallest subfield of E containing F and u. F[u] = {b0 + b1u + ··· + bmum ∈ E, where b0 + b1x + ··· + bmxm ∈ F[x]}

We claim that F[u] = F(u). Consider the map Φ:F[x] → E, defined as Φ(b0 + b1x + ··· + bmxm) = b0 + b1u + ··· + bmum where b0 + b1x + ··· + bmxm ∈ F[x].

We claim that Φ is a homomorphism.

Φ ( (b0 + b1x + ··· + bmxm) + (c0 + c1x + ··· + cmxm) ) = Φ ((b0+c0) + (b1+c1)x + ··· + (bm+cm)xm) = (b0+c0) + (b1+c1)u + ··· + (bm+cm)um = (b0 + b1u + ··· + bmum) + (c0 + c1u + ··· + cmum) = Φ(b0 + b1x + ··· + bmxm) + Φ(c0 + c1x + ··· + cmxm)

Analogously, Φ ( (b0 + b1x + ··· + bmxm) · (c0 + c1x + ··· + cmxm) ) = Φ ((b0c0) + (b0c1+b1c0)x + ··· + (bmcm)x2m) = (b0c0) + (b0c1+b1c0)u + ··· + (bmcm)u2m) = (b0 + b1u + ··· + bmum) · (c0 + c1u + ··· + cmum) = Φ(b0 + b1x + ··· + bmxm) · Φ(c0 + c1x + ··· + cmxm)

ker(Φ) contains all those polynomials f(x) such that Φ(f(x)) = 0, i.e., f(u) = 0. By assumption u is a root of p(x), p(x) ∈ Ker(Φ).

As F[x] is a polynomial ring and a principal ideal domain (PID). Thus, since Ker(Φ) is an ideal, it is generated by a single element, say Ker(Φ) = ⟨g(x)⟩ ⇒[p(x) ∈ Ker(Φ)] p(x) = g(x)·h(x) for some h(x) ∈ F[x]. However, because p(x) is irreducible, h(x) is a scalar (h(x) ∈ F)

∴ ⟨g(x)⟩ = ⟨p(x)⟩ = Ker(Φ).

Using the Fundamental Theorem of Homomorphism, Image(Φ) ≋ F[x]/Ker(Φ) ↭ {b0 + b1u + ··· + bmum ∈ E, where b0 + b1x + ··· + bmxm ∈ F[x]} ≋ F[x]/Ker(Φ) ↭ F[u] ≋ F[x]/Ker(Φ) = F[x]/⟨p(x)⟩. Remember that p(x) is irreducible in F[x] ↭ the ideal generated by this polynomial ⟨p(x)⟩ is maximal and F[x]/⟨p(x)⟩ is a field. Moreover, it is the smallest subfield of E containing F and u. Thus, F[u] = F(u).

(ii) Consider the elements 1, u, u2, ···, un-1 in F(u) and let p(x) be any minimal or irreducible polynomial of u over F of degree n.

Let’s suppose by reduction to the absurd that {1, u, u2, ···, un-1} is linearly dependent. Let α0·1 + α1·u + ··· + αn-1·un-1 = 0, where αi ∈ F and not all αi’s are zero ⇒ u satisfies a polynomial of degree n-1 < degree(p(x)), and we have assumed that p(x) is the minimal (or irreducible) polynomial of u over F ⊥ ⇒ {1, u, u2, ···, un-1} is linearly independent.

Next, we need to prove that {1, u, u2, ···, un-1} is the generating set of F(u). Let f(u) ∈ F(u) ⇒[Recall F[u] = {b0 + b1u + ··· + bmum ∈ E, where b0 + b1x + ··· + bmxm ∈ F[x]}≋ F[x]/Ker(Φ) ≋ F[x]/⟨p(x)⟩ ≋ F(u) ] f(x) ∈ F[x] and p(x) ∈ F[x] ⇒ [By the division algorithm] f(x) = p(x)q(x) + r(x) where r(x) = 0 or deg(r(x)) < deg(p(x)) ⇒ f(u) = p(u)q(u) + r(u) ⇒ [u is a root of p(x)] f(u) = r(u) 🚀.

As deg(r(x)) < deg(p(x)) = n ⇒ deg(r(x)) < n. r(x) = β0 + β1x + ··· + βn-1xn-1 is a polynomial of F[x] of degree n-1

∴ f(u) =🚀 r(u) = β0 + β1u + ··· + βn-1un-1. Hence, for any arbitrary f(u) ∈ F(u) is generated from {1, u, u2, ···, un-1}. Thus, {1, u, u2, ···, un-1} is a basis of F(u) over F and this set has n-elements, so [F(u):F] = n∎

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, 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.
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