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 F^x = F -{0} is an Abelian group under multiplication.

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

We define the characteristic of a field F, char(F), as the smallest positive integer p such that p·1_F = 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·1_F + m·1_F = (m + n)·1_F
2. (n·1_F)(m·1_F) = mn·1_F

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·1_F = 1 + 1 + ··_n·· + 1 + 1 = 0 ⇒ ab·1_F = 0 ⇒ (a·1_F)(b·1_F) = 0 ⇒ [A field is a ring with no zero divisors.] a·1_F = 0 or b·1_F = 0 ⊥ (char(F) = n). Therefore, the smallest positive integer p such that p·1_F = 1 + 1 + ··· + 1 (p times) = 0 if such a p exists, it is necessarily a prime. Futhermore, if n·1_F = 0, then n is divisible by char(F) = p, and p·n = n + n + ··· + n (p times) = n·1_F + n·1_F + ··· + n·1_F (p times) =[By assumption, n·1_F = 0] 0.

Let’s define (-n)·1_F = -(n·1_F) for n > 0, 0·1_F = 0, then we have the following ring homomorphism Φ: ℤ → F, defined as Φ(n) = n·1_F 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 1_F, the prime subfield, that is, the field generated by the multiplicative identity 1_F 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.)  

Examples.

• Every field is an extension of its prime subfield.
• ℂ is an extension of both ℝ and ℚ, ℝ is an extension of ℚ
• Let ℚ($\sqrt{2}$) = {$a+\sqrt{2}b|a, b ∈ ℚ$}. If we have a field, ℚ, and we want a field where the equation x² - 2 = 0 has a solution, we extend the field of rational to ℚ($\sqrt{2}$)
• ℚ($\sqrt{2},\sqrt{3}$) = {$a+\sqrt{3}b|a, b ∈ ℚ(\sqrt{2})$} = {$c+\sqrt{2}d+\sqrt{3}e+\sqrt{6}f|c, d, e, f ∈ ℚ$} is an extension field of ℚ.
• Similarly ℂ = ℝ(i) = {a + bi|a, b ∈ ℝ, i^2 = -1}, the complex number is an extension of ℝ that contains a zero of x² + 1, namely i.

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 + ··· + a_nxⁿ = (x -α₁)(x -α₂) ··· (x -α_n)

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_ℝ(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, ···, v_m}, and the basis of E over F be {w₁, v₂, ···, w_n}.

∀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, ···, v_m}, and E is a vector space over F with basis {w₁, v₂, ···, w_n}, so a_i ∈ 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 {v_iw_j} 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 {v_iw_j} 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 b_ij = 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, ···, v_m} is a basis of K over E ⇒ c_i = 0, i = 1, 2, ···, m ⇒ $c_i=\sum_{j=1}^n (b_{ij}w_j)=0$ where b_ij ∈ F and {w₁, v₂, ···, w_n} is a basis of E over F ⇒ b_ij = 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.).

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

• x⊕y = (σ*)^-1(σ*(x) + σ*(y))
• x⊙y = (σ*)^-1(σ*(x) σ*(y))

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 + ··· + a_nxⁿ, be a polynomial in F[x], a_n ≠ 0, a_i ∈ 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)⟩ ··· + a_nxⁿ + ⟨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 + ··· + a_nxⁿ has a root in E, where a_i ∈ F, a_n ≠ 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 + ··· + a_n(x + ⟨p(x⟩)ⁿ = a₀ + a₁(x + ⟨p(x⟩) + ··· + a_n(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 + ··· + b_mu^m ∈ E | b₀ + b₁x + ··· + b_mx^m ∈ F[x]}

2. If the degree of p(x) is n, the set {1, u, ···, u^n-1} forms a basis of F(u) over F, i.e., each element of F(u) can be written uniquely as c₀ + c₁u + ··· + c_nu^n-1 where c_i ∈ 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 + ··· + b_mu^m ∈ E, where b₀ + b₁x + ··· + b_mx^m ∈ F[x]}

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

We claim that Φ is a homomorphism.

Φ ( (b₀ + b₁x + ··· + b_mx^m) + (c₀ + c₁x + ··· + c_mx^m) ) = Φ ((b₀+c₀) + (b₁+c₁)x + ··· + (b_m+c_m)x^m) = (b₀+c₀) + (b₁+c₁)u + ··· + (b_m+c_m)u^m = (b₀ + b₁u + ··· + b_mu^m) + (c₀ + c₁u + ··· + c_mu^m) = Φ(b₀ + b₁x + ··· + b_mx^m) + Φ(c₀ + c₁x + ··· + c_mx^m)

Analogously, Φ ( (b₀ + b₁x + ··· + b_mx^m) · (c₀ + c₁x + ··· + c_mx^m) ) = Φ ((b₀c₀) + (b₀c₁+b₁c₀)x + ··· + (b_mc_m)x^2m) = (b₀c₀) + (b₀c₁+b₁c₀)u + ··· + (b_mc_m)u^2m) = (b₀ + b₁u + ··· + b_mu^m) · (c₀ + c₁u + ··· + c_mu^m) = Φ(b₀ + b₁x + ··· + b_mx^m) · Φ(c₀ + c₁x + ··· + c_mx^m)

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 + ··· + b_mu^m ∈ E, where b₀ + b₁x + ··· + b_mx^m ∈ 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, ···, u^n-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, ···, u^n-1} is linearly dependent. Let α₀·1 + α₁·u + ··· + α_n-1·u^n-1 = 0, where α_i ∈ F and not all α_i’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, ···, u^n-1} is linearly independent.

Next, we need to prove that {1, u, u², ···, u^n-1} is the generating set of F(u). Let f(u) ∈ F(u) ⇒[Recall F[u] = {b₀ + b₁u + ··· + b_mu^m ∈ E, where b₀ + b₁x + ··· + b_mx^m ∈ 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 + ··· + β_n-1x^n-1 is a polynomial of F[x] of degree n-1

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

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.