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] = dim_{k}L, 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's a root of, and degree of α is the degree of such polynomial, e.g., α = $\sqrt[5]{2}$ is algebraic root of x^{5} -2 = 0 and α has degree 5, 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(^{2π}⁄_{7}) transcendental? No, it is not. α is the root of 8x^{3} +4x^{2} -4x -1 with rational coefficients, and the other roots are cos(^{2π}⁄_{7}·2) and cos(^{2π}⁄_{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 (x^{2} + 1 in ℝ, x^{2} -2 in ℚ, x^{2} -2, x^{2} - 3 in ℚ), and they are called **algebraic elements**, they are the zeros or roots of some nonzero polynomial in F[x].

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

Proof.

Let α ∈ E, suppose that its minimal polynomial f(x) is not prime ⇒ if factors f(x) = g(x)h(x) where f(α) = 0 ⇒ 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 ⊥

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).

Proof.

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 has trivial kernel, Ker(Φ) = {0}, whence 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) = {$\frac{a_0+a_1α+···+a_nα^n}{b_0+b_1α+···+b_mα^m}:m,n∈ℕ, a_i, b_j ∈ F$, b_{j} 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 $\bar Φ(\frac{f(x)}{g(x)})=\frac{f(α)}{g(α)}$

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 ⇒ [A subring of a field containing 1 is an integral domain] 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 non-trivial ⇒ [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. 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(a) is a field containing F and a, p(x) is a polynomial of minimum degree such that p(a)=0.

If degree of p(x) is n, the elements of F(α) look like a_{n-1}α^{n-1} + a_{n-1}α^{n-2} + ··· + a_{0}. [F(α):F] = -finite- n = degree of the minimal polynomial- with basis {1, α, α^{2}, ···, α^{n-1}}. Conversely, if [E:F] = m, then the set {1, b, b^{2},···, b^{m-1}}, b ∈ E, is linearly dependent, c_{m-1}b^{m-1} + c_{m-2}b^{m-2} + ··· + c_{0} = 0, then b is the zero of the polynomial p(x) = c_{m-1}x^{m-1} + c_{m-2}x^{m-2} + ··· + c_{0}, so finite [E:F] = m implies E is algebraic extension of F. In other words, we can associate b with its minimal polynomial.

L/K field extension. α ∈ L, α is algebraic over K ↭ α is in a finite extension of K.

Proof.

If α ∈ L, [L : K] = n < ∞ ⇒ 1, α, α^{2}, ···, α^{n} are n+1 elements of a n-dimensional vector space ⇒ ∃linear relation: a_{0} + a_{1}α + ··· + a_{n}α^{n} = 0 ⇒ 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 extension of K.

Recall. Suppose p(x) is an irreducible polynomial in K[x], then K[x]/⟨p⟩ is a field and a ring.

The problem is the existence of inverses. Suppose q(x) ∈ K[x]/⟨p(x)⟩, q ≠ 0 ⇒ q and p are coprimes in K[x] because p is irreducible ⇒ [K[x]/⟨p(x)⟩ is an Euclidean domain] By the Euclidean algorithm, a(x)q(x) + b(x)p(x) = 1, this can be read as a(x) is the inverse of q(x) in [K[x]/⟨p⟩

α is the root of p(x) ∈ K[x], p irreducible (Figure 1)

And Image(K[x]/⟨p(x)⟩) in L is a field containing α, and it is also a finite extension of K (its dimension equals degree(p))

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 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 a_{1} (= a), a_{2}, ···, a_{n} and b_{1} (= b), b_{2}, ···, b_{n} be the distinct zeros or roots of p(x) and q(x) respectively.

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

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) that are also zeros of r(x). However, r(b_{j}) = p(c-db_{j}) = p(a + db -db_{j}) = p(a + d(b -b_{j})), and by election d is such that a_{i} ≠ a + d(b -b_{j}) for j > 1 ⇒ b = b_{1} is the only zero of s(x) in K[x] ⇒ s(x) = (x -b)^{u} ⇒ [s(x) is irreducible 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.

Example: $\frac{ℚ[x]}{x^2-2}≈ℚ(\sqrt{2}),\frac{ℚ(\sqrt{2})[x]}{x^2-3}≈ℚ(\sqrt{2}, \sqrt{3}), \frac{ℚ(\sqrt{2}, \sqrt{3})[x]}{x^2-5}≈ℚ(\sqrt{2}, \sqrt{3}, \sqrt{5})$

$ℚ → ℚ(\sqrt{2}) → ℚ(\sqrt{2}, \sqrt{3}) → ℚ(\sqrt{2}, \sqrt{3}, \sqrt{5})$

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 it belongs to a finite extension of F (*A finite extension of a field is an algebraic extension of F*), [F_{n}(a):F] < ∞

Since a is algebraic over E ⇒ a is the zero of some irreducible polynomial in E[x], say p(x) = b_{n}x^{n} + ··· + b_{0}, b_{i} ∈ E. Then, we construct a series of field extensions of F, as follows, F_{0} = F(b_{0}), F_{1} = F_{0}(b_{1}), ···, F_{n} = F_{n-1}(b_{n}) ⇒ F_{n} = F(b_{0}, b_{1}, ···, b_{n}), and p(x) ∈ F_{n}[x] ⇒ [F_{n}(a):F_{n}] = n. Each b_{i} is algebraic over F, we know that each [F_{i+1}:F_{i}] is finite, and therefore [F_{n}(a):F] = [F_{n}(a):F_{n}] [F_{n}:F_{n-1}] ··· [F_{1}:F_{0}][F_{0}:F] is finite.

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)]< ∞ and [F(b):F] < ∞] [F(a, b):F(b)][F(b):F] < ∞

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 root of p(x) with algebraic coefficients in K, then α is algebraic.

Proof.

Let α ∈ L, α^{n} + a_{n-1}α^{n-1} + ··· + a_{0} = 0.

Consider K ⊆ K(a_{0}) ⊆ K(a_{0}, a_{1}) ⊆ ··· ⊆ K(a_{0}···a_{n-1}) ⊆ K(a_{0}···a_{n-1}, **α**), all the extensions are finite because a_{i} are algebraic, and the last one is finite because we’ve got a polynomial with roots in the field K(a_{0}···a_{n-1}, **α**) ⇒ there is a finite extension of α, and 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, x^{2} - (e + π)x + eπ, its roots are e and π. If e + π and e + π are both 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, x^{2} +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.

α_{1} is a root of f(x) ⇒ f(x) = (x -α_{1})g(x), i.e., f(x) factors in F(α_{1})[x], where g(x) ∈ F(α_{1})[x]. Thus, the minimal polynomial of α_{2} working over F(α_{1}) must divide g(x) which has degree n-1, so [F(α_{1}, α_{2}):F(α)] ≤ deg(g(x)) = n-1. By inductions, the result follows.

[F(α_{1}, α_{2}):F(α_{1})]≤deg(f(x))=n

- F = ℚ, f = x
^{3}-2, K = ℚ($\sqrt[3]{2}$, w) where w is the primitive third root of unity, [ℚ($\sqrt[3]{2}$ : ℚ] = 6 = 3! - F = ℚ, f = x
^{3}-1 = (x-1)(x^{2}+ x +1), K = ℚ(w) where w is the primitive third root of unity, [ℚ(w) : ℚ] = 2 < 3!