Field Theory II

Recall

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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.

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)

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.

Definitions. An element α ∈ E is algebraic over F if there exists a non-constant polynomial p(x) ∈ F[x] such that p(α)=0, i.e., α is the root of p(x). Example: E = ℂ, F = ℚ, α = $\sqrt{2}$ is algebraic over ℚ, p(x) = x² -2, and p($\sqrt{2}$)=0. Similarly, i is algebraic over ℚ, p(x) = x² + 1, p(i) = 0. An element that is not algebraic, i.e., there no exists a non-constant polynomial p(x) such that p(α)=0 is called transcendental, e.g., π.

An extension E of F is called algebraic if each element of E is algebraic over F. An extension that is not algebraic, it is called transcendental, e.g, the extension ℝ over ℚ is a transcendental extension, and ℂ = ℝ(i) and ℚ($\sqrt{5}$) are algebraic extensions over ℝ and ℚ respectively (x² + 1, x² -5).

The minimal polynomial

The minimal polynomial of u over F is the monic (the leading coefficient is equal to 1) irreducible polynomial in F[x] of which u is a root.

Theorem. Let E be an extension field of F, and let u ∈ E be algebraic over F. Let p(x) ∈ F[x] be a polynomial of least degree such that p(u) = 0, we call p the minimal polynomial of u over E, Then,

(i) p(x) is irreducible over F
(ii) if g(x) ∈ F[x] is such that g(u) = 0, then p(x) | g(x)
(iii) There is exactly one monic polynomial p(x) ∈ F[x] of least degree such that p(u) = 0. This monic polynomial always exists because we can always divide the minimal polynomial by its leading coefficient, and still u be a root.

Proof.

(i) Let’s suppose for the sake of contradiction, that p(x) is reducible over F. Let p(x) = p₁(x)·p₂(x) such that deg(p₁(x)), deg(p₂(x)) < deg(p(x)).

We know that u is a root of p(x) ⇒ p(u) = 0 ⇒ p₁(u)·p₂(u) = 0 ⇒ p₁(u) = 0 or p₂(u) = 0, and deg(p₁(x)), deg(p₂(x)) < deg(p(x)), i.e., u satisfies a polynomial of degree less that p(x) ⊥ [By assumption, p(x)∈ F[x] is a polynomial of least degree such that p(u) = 0]. Therefore, p(x) is irreducible.

(ii) Let g(x) ∈ F[x] such that g(u) = 0. By the division algorithm, g(x) = p(x)q(x) + r(x) where r(x) = 0 or degree(r(x)) < deg(p(x)) ⇒

1. If r(x) = 0 ⇒ g(x) = p(x)q(x) ⇒ p(x) | g(x)∎
2. Otherwise, g(u) = p(u)q(u) + r(u) ⇒ By (ii) assumption, g(u) = 0, and p(u) = 0, so r(u) = 0, and deg(r(x)) < deg(p(x)) ⊥

(iii) Let’s assume by reduction to the absurd, that g(x) is another monic polynomial of least degree which g(u) = 0 ⇒ [By (ii)] g(x)| p(x). However, p(x)|g(x) just by interchanging the roles of p(x) and g(x) and using (ii) again, g(x)|p(x) ⇒ p = Φ·g, g = ψ·p ⇒ p = Φ·ψ·p [When we multiply polynomials, their degrees are added, deg(f·g) = deg(f) + deg(g)] deg(Φ·ψ) = 0 ⇒ Φ and ψ need to be constant ⇒ [Since both p(x) and g(x) are monic polynomials, that is, the leading coefficients are 1] p(x) = g(x) ∎

Theorem. If E is a finite extension of F, then E is an algebraic extension of F.

Proof. Let’s select an arbitrary element of E, say u ∈ E, and [E:F] = n. The set {1, u, ··· , u^n-1, uⁿ} is linearly dependent ⇒ there exists a₀, a₁, ···, a_n (not all zero) in F such that a₀ + a₁u + ··· a_nuⁿ = 0. Thus, we have found a polynomial f(x) = a₀ + a₁x + ··· a_nxⁿ ∈ F[x] such that u in E is a root. Then, u ∈ E is an algebraic element over F. Moreover, u ∈ E was selected completely as an arbitrary of E. Hence, E is an algebraic extension of F ∎

Proposition. If p(x) is the minimal polynomial of α over F and degree(p(x))= n. Then, [F(α):F] = n, and, in particular, it is finite, so by our previous result, F(α) is an algebraic extension. More formally, if E is an extension of F and α ∈ E is algebraic over F, then F(α) is an algebraic extension of F and deg(F(α):F)=deg(p(x)) where p is the minimal polynomial of α.

Proof.

Let p be the minimal polynomial of α over F, say p(x) = xⁿ + a_n-1x^n-1 + ··· +a₀ where a_i ∈ F.

We claim that the collection {α⁰=1, α, α², ···, α^n-1} is a basis for F(α).

• Lineal independence. Suppose β₀·1 + β₁α + ··· + β_n-1α^n-1 = 0, β_i∈ F. If β_i are not all zeros, consider the polynomial f = β₀ + β₁x + ··· + β_n-1x^n-1. Then, f(α)=0, f is a non-zero polynomial (f ∈ F[x]) of degree n-1 < n ⊥ (p is the minimal polynomial of α).

• Spans. F(α) = F[α] = {$\sum_{i=0}^m β_iα^i$| m ∈ ℕ, β_i ∈ F}. Obviously F[α] ⊆ F(α), since F[α] is the minimal subring containing F and α. Besides, F(α) ⊆ F[α] is obvious, too, just take all coefficients β_i to zero except β₁=1 to get α, and similarly you can get all elements of F. It is more complicated to show that F[α] is a field, that is, it is closed under all field operations.

• Finally, F(α) = F[α] = $\sum_{i=0}^m β_iα^i$| m ∈ ℕ, β_i ∈ F} =[💡{α⁰=1, α, α², ···, α^n-1} linear independence] $\sum_{i=0}^{n-1} β_iα^i$| β_i ∈ F}. Consider p = xⁿ + ax^n-1 + ··· +a₀. Then, 0 = αⁿ + a_n-1α^n-1 + ··· +a₀ ⇒ αⁿ = -a_n-1α^n-1 + ··· -a₀, so we can express αⁿ as a linear combination of {1, α, ···, α^n-1}, and we can apply the same idea with α^m, m > n, and therefore, F[α] = $\sum_{i=0}^m β_iα^i$| m ∈ ℕ, β_i ∈ F} = $\sum_{i=0}^{n-1} β_iα^i$| β_i ∈ F}.

Fact. Every finite extension is an algebraic extension, but the converse is not true, i.e., every algebraic extension is not a finite extension.

Example. E = ℚ($\sqrt{2},\sqrt{3},···,\sqrt{p}$) the smallest subfield of ℝ containing the rationals and all square root of positive primes is an algebraic extension but it is not a finite extension.

(i) E is an algebraic extension. Let α ∈ E ⇒ ∃r ∈ ℤ⁺ such that α ∈ ℚ($\sqrt{p_1},\sqrt{p_2},···,\sqrt{p_r}$). We know that ℚ($\sqrt{p_1},\sqrt{p_2},···,\sqrt{p_r}$) is a finite extension of degree 2^r ⇒ [Every finite extension is algebraic] α is algebraic over ℚ, and α was selected arbitrary, so E is an algebraic extension.

Notice that ℚ($\sqrt{2}$), its corresponding polynomial is x²-2 = 0, then the degree of the extension is 2. Analogously, ℚ($\sqrt{2}, \sqrt{3}, \sqrt{5}$) their corresponding polynomials are x² -2, x² -3, and x² + 5, so the degree of the extension is 2·2·2 = 6 = 2³.

(ii) E is an infinite extension. Claim: Each new adjunction of the square root of a prime is a proper extension, i.e. ℚ ⊂ ℚ($\sqrt{2}$) ⊂ ℚ($\sqrt{2}, \sqrt{3}$) ⊂ ···

Let F = ℚ($\sqrt{p_1},\sqrt{p_2},···, \sqrt{p_n}$) where p_i i=1, ··· n are distinct primes. Let’s show that ℚ($\sqrt{p_1},\sqrt{p_2},···, \sqrt{p_n}, \sqrt{q}$) is a proper extension of F, we must show that $\sqrt{q}$ ∉ F, and prove it by using induction on n.

1. If n = 0 ⇒ F = ℚ ⇒ $\sqrt{q}$, q prime is an irrational number and $\sqrt{q}$ ∉ F = ℚ
2. Let us assume that result is true for n-1.

Let F₀ = ℚ($\sqrt{p_1},\sqrt{p_2},···, \sqrt{p_{n-1}}$), F = F₀($\sqrt{p_n}$) is an extension F of F_o of degree 2 -a proper extension-, i.e. [F:F₀] = 2 because $\sqrt{p_n}$ is a root of x² -p_n ∈ F₀[x], and x² -p_n is irreducible in F_o[x].

We want to prove that $\sqrt{q}$ ∉ F = ℚ($\sqrt{p_1},\sqrt{p_2},···, \sqrt{p_{n}}$), so it is possible to create a proper extension ℚ($\sqrt{p_1},\sqrt{p_2},···, \sqrt{p_{n}}, \sqrt{q} $).

Let’s assume by reduction to the absurd that $\sqrt{q}$ ∈ F = F₀($\sqrt{p_n}$) ⇒ $\sqrt{q} = a·1 + b·\sqrt{p_n}$ where a, b ∈ F₀ ⇒ $q =a^2+b^2p_n+2ab\sqrt{p_n} ⇒ 0 = a^2+b^2p_n+2ab\sqrt{p_n}-q$ ⇒[This is a second equation ax²+ bx +c where the roles are a = b², b = 2ab, c = a²-q] $\sqrt{p_n}=\frac{-2ab±\sqrt{4a^2b^2-4(a^2-q)b^2}}{2b^2}=\frac{-2ab±2b\sqrt{q}}{2b^2}=\frac{-a}{b}±\frac{\sqrt{q}}{b}$ ∈[a, b ∈ F_o and by assumption, $\sqrt{q}$ ∈ F = F₀($\sqrt{p_n}$)] F₀, $\sqrt{p_n} ∈$ F₀ = ℚ($\sqrt{p_1},\sqrt{p_2},···, \sqrt{p_{n-1}}$) which is a contradiction because F = F₀($\sqrt{p_n}$) is a proper extension of F₀ ⇒ $\sqrt{q}$ ∉ F ⇒ ℚ ⊂ ℚ($\sqrt{2}$) ⊂ ℚ($\sqrt{2}, \sqrt{3}$) ⊂ ··· is an infinite properly ascending chain and E is not a finite extension of ℚ∎

Theorem. If f ∈ K[x] is monic, irreducible, (it cannot be written as a product of non constant polynomials)and f(α)= 0, then f is the minimal polynomial of α.

Proof. Suppose that f ∈ K[x] monic, irreducible, and f(α) = 0.

Suppose by reduction to the absurd, that f is not the minimal polynomial of f ↭ there is such a polynomial Φ, Φ|f ⇒ f = Φ·g ⇒ [By assumption, f is irreducible] g is constant, g ∈ K ⇒ [f is monic] g = 1 ⇒ f = Φ∎

Example. The number $\sqrt{2}$ has minimal polynomial x² -2 over ℚ, but x² -2 is not irreducible over ℝ. x² -2 is monic, f($\sqrt{2}$) = 0, and it is irreducible over ℚ, so it is the minimal polynomial of $\sqrt{2}$.

Suppose for the sake of contradiction, x²-2 is not irreducible over ℚ ↭ ∃a, b, c, d ∈ ℚ such that (ax + b)(cx +d) = x² - 2 ⇒ ac = 1, bc + ad = 0, and bd = -2 ⇒ ∃a, b, c, d ∈ ℚ such that b = ±$\sqrt{2}$a, a ≠ 0, c = 1/a, d = ±$\sqrt{2}$/a ⊥ ⇒ x² -2 is irreducible over ℚ

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
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5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
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