“Gosh! Nothing can travel faster than light except gossip, bias and fake narratives, and bad news. Everything has a limit except human stupidity, unbridled greed, and untamed arrogance. They obey their own laws, and that’s why there are many types of infinites,” Apocalypse, Anawim, #justtothepoint.

Lemma. Let ϕ: E → F be an isomorphism of fields. Let K be an extension field of E and α ∈ K be algebraic over E with minimal polynomial p(x). Suppose that L is an extension field of F such that β(∈L) is root of Φ(p(x)) in F[x]. Then, ϕ extends to a unique isomorphism $\barϕ :E(α)→F(β)$ such that $\barϕ(α)=β~ and~ \barϕ$ agrees with Φ on E (Figure 1.h.)

Proof. If p(x) has degree n, then by the previous theorem, we can write any element in E(α) as a linear combination of 1, α, ···, α^{n-1}, i.e., a_{0} + a_{1}α + ··· + a_{n-1}α^{n-1}, therefore we can define $\barϕ :E(α)→F(β), \barϕ(a_0 + a_1α + ··· + a_{n-1}α^{n-1})=ϕ(a_0)+ϕ(a_1)β+···+ϕ(a_{n-1})β^{n-1}$

We can extend Φ to be an isomorphism from E[x] to F[x], which we will denote by Φ (we will abuse this notation for simplicity’s sake) defined by, $ϕ(a_0 + a_1x + ··· + a_{n-1}x^{n-1})=ϕ(a_0)+ϕ(a_1)x+···+ϕ(a_{n-1})x^{n-1}$. This extension agrees with the original isomorphism Φ: E → F because constant polynomials get mapped to constant polynomials.

Let’s call Φ(p(x)) = q(x), and Φ maps ⟨p(x)⟩ onto ⟨q(x)⟩ ⇒ ψ: E[x]/⟨p(x)⟩→F[x]/⟨q(x)⟩ is an isomorphism (it is straightforward for the reader to check it) defined by f(x) + ⟨p(x)⟩→Φ(f(x)) + ⟨Φ(p(x))⟩ = Φ(f(x)) + ⟨q(x))⟩, and we also have σ:E[x]/⟨p(x)⟩→E(α) and τ:F[x]/⟨q(x)⟩→F(β) isomorphisms defined by evaluation at α and β respectively, e.g., σ:E[x]/⟨p(x)⟩→E(α), $a_0 + a_1x + ··· + a_{n-1}x^{n-1} + ⟨p(x)⟩ → a_0 + a_1α + ··· + a_{n-1}α^{n-1}$ (where a_{0}, a_{1}, ···, a_{n-1} ∈ E and the natural isomorphism from E[x]/⟨p(x)⟩ to E(α) carries a_{k}x^{k} + ⟨p(x)⟩ to a_{k}α^{k}). Finally, $\barϕ=τψσ^{−1}$ is the desired isomorphism.

Notice that since p(x) is a minimal (irreducible) polynomial over E, then Φ(p(x)) = q(x) is a minimal (irreducible) polynomial over F, α and β are algebraic over K and L respectively, and σ:E[x]/⟨p(x)⟩ → E(α) and τ:F[x]/⟨q(x)⟩ → F(β) are evaluation isomorphisms, e.g, σ is the identity on E and carries x + ⟨p(x)⟩ to α, τ is the identity on F and carries x + ⟨q(x)⟩ to β.

Theorem. Let ϕ: E→ F be an isomorphism of fields and let p(x) be a non-constant polynomial in E[x] and q(x) the corresponding polynomial in F[x] under ϕ. If K and L are splitting field of p(x) and q(x) respectively, then ϕ extends to an isomorphism ψ: K → L (Figure 2.a) (ψ agrees with ϕ on E)

Proof.

Let use induction on degree(p(x)) and assume that p(x) is irreducible over E (Otherwise p(x)=(x-r_{1})(x-r_{2})···(x-r_{m})p_{1}(x)p_{2}(x)···p_{s}(x), and p_{i} are irreducible) ⇒ q(x) is irreducible over F. If degree(p(x))=1 ⇒ K = E and L = F, Φ itself is the desired mapping and we are done.

Assume that the theorem holds for all polynomials of degree less than n. Since K is a splitting field of p(x), all of the roots of p(x) are in K. Choose one of these roots, say α, such that E ⊂ E(α) ⊂ K, and take a root β of q(x) in L: F ⊂ F(α) ⊂ L ⇒ ∃ isomorphism $\barΦ:~E(α)→F(β):~ \barΦ(α)=β,~ and~ \barΦ$ agrees with Φ on E.

α and β are roots of p(x) and q(x) respectively ⇒ p(x) = (x -α)f(x), q(x) = (x -β)g(x), degree(f(x)) < degree(p(x)) and degree(g(x)) < degree(q(x)) ⇒ [By our induction hypothesis] **∃ an isomorphism ψ: K → L such that ψ agrees with $\barΦ$ on E(α) and therefore with Φ on E.**

- x
^{4}+ 1 = 0 over ℚ ⇒ x^{4}= -1 ⇒ x^{8}= 1. x is an 8th root of unity = e^{i2π}⁄_{8}, i = 0, 1,… 7. Therefore, x = $e^{\frac{2iπ}{4}} = ±cos\frac{π}{4}±isin\frac{π}{4}=±\frac{1}{\sqrt{2}}±i\frac{1}{\sqrt{2}}$. Let K be the splitting field of x^{4}+ 1 over ℚ. K = $\mathbb{Q}(±\frac{1}{\sqrt{2}}±i\frac{1}{\sqrt{2}}) = \mathbb{Q}(\sqrt{2}, i)$.

$[\mathbb{Q}(\sqrt{2}, i): \mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, i): \mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}): \mathbb{Q}]$ = 2·2 = 4, and that’s true because $[\mathbb{Q}(\sqrt{2}, i): \mathbb{Q}(\sqrt{2})]$ ≤ 2 because x^{2} + 1 = 0 is the irreducible polynomial of i, but $\mathbb{Q}(\sqrt{2})⊆\mathbb{R}$ and i ∉ ℝ, in other words, $\mathbb{Q}(\sqrt{2}, i) ≠ \mathbb{Q}(\sqrt{2})$ ⇒ $[\mathbb{Q}(\sqrt{2}, i): \mathbb{Q}(\sqrt{2})]$ = 2.

e^{ix} = cosx + i·sinx, where is i the imaginary unit.

- Let f(x) = x
^{8}-2. The roots for this polynomial over ℚ are $\sqrt[8]{2},ξ_8\sqrt[8]{2},ξ_8^2\sqrt[8]{2},···,ξ_8^7\sqrt[8]{2}$ where ξ_{8}is a primitive 8th root of unity. Using the same argument as before we can demonstrate that the splitting field of x^{8}-2 over ℚ is K = $ℚ(ξ_8,\sqrt[8]{2})$.

$[\mathbb{K}: ℚ] = [\mathbb{K}: ℚ(\sqrt[8]{2})]·[ℚ(\sqrt[8]{2}): ℚ] = 2 · 8 = 16$ because $[ℚ(\sqrt[8]{2}): ℚ]=8$ because x^{8} -2 is irreducible over ℚ (Eisenstein, p = 2), deg(x^{8} -2) = 8, and $\sqrt[8]{2}∉\mathbb{Q}$

Futhermore, $\sqrt[8]{2}∈ℚ(\sqrt[8]{2})⇒\sqrt{2}∈ℚ(\sqrt[8]{2})⇒$ [ξ_{8}=$\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}$, it can be described as a polynomial of i] ξ_{8} ∈ $ℚ(\sqrt[8]{2})(i) ⇒ K ⊆ ℚ(\sqrt[8]{2})(i),$ but $ℚ(\sqrt[8]{2})(i) ⊄ \mathbb{R}$ since i ∉ ℝ, hence $\mathbb{K} ⊄ \mathbb{R}⇒$It is left as an exercise, $ K=ℚ(\sqrt[8]{2})(i)⇒[K:ℚ(\sqrt[8]{2})]=[ℚ(\sqrt[8]{2})(i):ℚ(\sqrt[8]{2})] = 2, i∉ℚ(\sqrt[8]{2})$, x^{2}+1 = 0 is irreducible over $ℚ(\sqrt[8]{2})$ and deg(x^{2} +1) = 2.

- The splitting field of f(x) = x
^{n}-a over ℚ is ℚ($\sqrt[n]{a}$, w), where w is a primitive nth root of unity and each of $\sqrt[n]{a},w\sqrt[n]{a}, w^2\sqrt[n]{a}, …, w^{n-1}\sqrt[n]{a}$ is a root of x^{n}-a in ℚ($\sqrt[n]{a}, w$).Clearly the splitting field must contain $\sqrt[n]{a}$, but it is also true that $w^k\sqrt[n]{a}$, k=1, ···, n-1 are also roots of x

^{n}-a.

Besides, let L be a splitting field of f. If α and β are roots of f, then (β/α)^{n} = a/a = 1 ⇒ (β/α) = ξ ⇒ β = ξα where = ξ^{n}=1, i.e., ξ ∈ L is a n-th root of unity. **Fixing α, we have, in L, n distinct β**, and the roots of f are given by ξα as ξ runs through these n-th roots of unity (0, 1, ··· n-1).

- Prove that $ℚ(\sqrt{2}, \sqrt{3})=ℚ(\sqrt{2} + \sqrt{3})$

Solution:

- Let’s prove that $ℚ(\sqrt{2} + \sqrt{3})⊆ℚ(\sqrt{2}, \sqrt{3})$. $\sqrt{2}, \sqrt{3} ∈ ℚ(\sqrt{2}, \sqrt{3})⇒ \sqrt{2} + \sqrt{3} ∈ ℚ(\sqrt{2}, \sqrt{3})$ ∎
- We know that $\sqrt{3} + \sqrt{2} ∈ ℚ(\sqrt{2} + \sqrt{3}),~ 💣\sqrt{3} - \sqrt{2} ∈ ℚ(\sqrt{2} + \sqrt{3}) ⇒ \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} = 2\sqrt{3} ∈ ℚ(\sqrt{2} + \sqrt{3}) ⇒ \sqrt{3} ∈ ℚ(\sqrt{2} + \sqrt{3})$ because 2
^{-1}=^{1}⁄_{2}∈ ℚ. Similarly, we can prove $\sqrt{2}∈ ℚ(\sqrt{2} + \sqrt{3})~ ⇒ ℚ(\sqrt{2}, \sqrt{3})⊆ℚ(\sqrt{2} + \sqrt{3})$ ⇒ $ℚ(\sqrt{2}, \sqrt{3})=ℚ(\sqrt{2} + \sqrt{3})$ ∎💣$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})=1,~ so~ \sqrt{3} - \sqrt{2}$ is a multiplicative inverse of $(\sqrt{3} + \sqrt{2}) ∈ ℚ(\sqrt{2} + \sqrt{3})$ that is an extension field of ℚ, so it contains the multiplicative inverses of its non-negative elements.

$ℚ(\sqrt{2}, \sqrt{3})=ℚ(\sqrt{2})(\sqrt{3})=${a + b$\sqrt{3} | a, b ∈ ℚ(\sqrt{2})$} = {$(α+β\sqrt{2})+(γ+δ\sqrt{2})\sqrt{3}| α, β, γ, δ ∈ ℚ$} = {$α + β\sqrt{2}+γ\sqrt{3}+δ\sqrt{6}| α, β, γ, δ ∈ ℚ$}

Theorem. Let p(x) be a polynomial in F[x]. Then, there exists a splitting field K of p(x) that is unique up to isomorphism, i.e, any two splitting fields of p(x) over F are isomorphic.

Proof. Suppose that they are K, L splitting fields of p(x) over F. The result follows from the previous theorem by letting Φ be the identity from F to F.

Theorem. Let F be a field, let E = F(α) be an extension of F, where α is algebraic over F and let f ∈ F[x] be the irreducible polynomial of α over F. Let Φ: F → K be a homomorphism from F to a field K, and let L be an extension of K. If β ∈ L is a root of Φ(f), then there is a unique extension of Φ to a homomorphism Φ': E → L such that ϕ'(α) = β and fixes F.

Proof

Let β ∈ L be a root of Φ(f) ⇒ ∃σ isomorphism: F(α) ≋ F[x]/⟨f⟩, such that σ(a) = a + ⟨f⟩ ∀a ∈ F, σ(α) = x + ⟨f⟩, a_{0} + a_{1}α + ··· a_{n-1}α^{n-1} → a_{0} + a_{1}x + ··· + a_{n-1}x^{n-1} + ⟨f⟩.

Let ev_{β}∘Φ be the homomorphism F[x] → L defined as follows, given a polynomial g ∈ F[x], let Φ(g) be the polynomial obtained by applying Φ to the coefficients of g (Φ(a_{n}x^{n} + ··· + a_{1}x + a_{0}) = Φ(a_{n})x^{n} + ··· + Φ(a_{1})x + Φ(a_{0}) ), then ev_{β}∘Φ(g) be the evaluation of Φ(g) = Φ(a_{n})x^{n} + ··· + Φ(a_{1})x + Φ(a_{0}) at β, that is, ev_{β}∘Φ is a homomorphism from F[x] to K. **ev _{β}∘Φ(g) = Φ(a_{n})β^{n} + ··· + Φ(a_{1})β + Φ(a_{0})**

∀a ∈ F, ev_{β}∘Φ(a) = Φ(a), ev_{β}∘Φ(x) = β. By assumption, β ∈ L is a root of Φ(f) ⇒ Φ(f)(β) = 0 ⇒ f ∈ Ker(ev_{β}∘Φ) ⇒ ⟨f⟩ ⊆ Ker(ev_{β}∘Φ) ⇒ [f is irreducible ⇒ ⟨f⟩ is a maximal ideal and ev_{β}∘Φ is not the trivial homomorphism] ⟨f⟩ = Ker(ev_{β}∘Φ). Then, there is an induced homomorphism e: F[x]/⟨f⟩ → L, and **φ = e∘σ: F(α) → L such that φ(a) = Φ(a) ∀a ∈ F, φ(α) = β.**

φ is uniquely specified by the conditions φ(a) = Φ(a) ∀a ∈ F, φ(α) = β because every element of E = F(α) can be written as $\sum_{i=0}^{n-1} a_iα^i$ ({1, α, ··· α^{n-1}} is a basis of E), so $φ(\sum_{i=0}^{n-1} a_iα^i)=\sum_{i=0}^{n-1} Φ(a_i)φ(α)^i=\sum_{i=0}^{n-1} Φ(a_i)β^i$

- Recall, σ isomorphism: F(α) ≋ F[x]/⟨f⟩, a
_{0}+ a_{1}α + ··· a_{n-1}α^{n-1}→ a_{0}+ a_{1}x + ··· + a_{n-1}x^{n-1}+ ⟨f⟩ - e: F[x]/⟨f⟩ → L, a
_{0}+ a_{1}x + ··· a_{n-1}x^{n-1}+ ⟨f⟩ → Φ(a_{0}) + Φ(a_{1})β + ··· + Φ(a_{n-1})β^{n-1}

To sum up, every extension φ of Φ is uniquely specified by the conditions above.

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.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn (Abstract Algebra), and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- Andrew Misseldine: College Algebra and Abstract Algebra.