Cyclotomic extensions and polynomials

The important thing to remember about mathematics is not to be frightened, Richard Dawkins.

Definition. Let F be a field, K/F be an extension, α ∈ F. α is expressible by radicals over F if there exists a sequence of intermediate fields or tower of fields extensions: F = F₀ ⊆ F₁ ⊆ ··· ⊆ F_r such that

1. α ∈ F_r
2. Each extension F_i/F_i-1 is a radical extension, that is, F_i=F_i-1(α_i) with α_i^n_i ∈ F_i-1 or, alternatively, α_i is the root of an irreducible polynomial in F_i-1[x] of the form x^n_i -c_i.

The elements of K can be obtained from those of F by means of rational operations together with the taking of roots, e.g. F = ℚ, $(3+\sqrt{2})^{\frac{1}{5}}$ lies in a field F₂, where F₁ = ℚ(α₀), $α_0^2=2$, F₂ = F₁(α₁), $α_1^5=3+\sqrt{2}$.

If F contains a primitive nth root of unity, then a radical extension K/F is a Kummer extension. The idea is to translate radical extensions (this is the messy realm of solving polynomials by radicals) to Kummer extensions (↭ cyclic extensions) and apply group, ring, and Galois theory, a much more conceptual and friendly approach.

Let n be a positive integer, F a field such that char(F) = 0 or char(F) = p > 0, p ɫ n. Then, xⁿ -1 ∈ F[x] is separable. Let K/F be the splitting field of xⁿ -1 over F. Then, K/F is Galois and is called a Cyclotomic extension.

We are confining ourselves to fields F of characteristic zero or p, p ɫ n, and so can be sure that the splitting field K of xⁿ -1 over F is both normal and separable. f’ = nx^n-1 ≠ 0 because p ɫ n.

Definition. A finite extension K/F is Abelian if K/F is Galois and Gal(K/F) is Abelian.

K/F is not in general cyclic, e.g., F = ℚ, n = 8, Gal(K/F) ≋ ℤ/2ℤ x ℤ/2ℤ

Lemma. Let G be an Abelian group.Let a, b ∈ G be elements such that ord(a)=m, ord(b) = n. then, ∃c ∈ G: ord(c) = lcm(m, n).

Proof.

If (m, n) = 1 ⇒ lcd(m, n) = m·n. Consider the element ab ∈ G: (ab)^{lcd(m, n)} = (ab)^mn = abab··· (mn times) ···ab = [G is Abelian] = a··· (mn times) ··· b ··· (mn times) ···b = a^mbⁿ = [By assumption, ord(a)=m, ord(b)=n] ee = e ⇒ ord(ab) = k ≤ nm.

e = (ab)^k = (ab)^km = [G Abelian] a^kmb^km = [By assumption, ord(a)=m] eb^km = b^km ⇒ ord(b) = n | km ⇒ [(m, n)=1] n | k

Mutatis mutandis, e = (ab)^k = (ab)^kn = [G Abelian] a^knb^kn = [By assumption, ord(b)=n] a^kne = a^kn ⇒ ord(a) = m | kn ⇒ [(m, n)=1] m | k

n | k, m | k ⇒ nm | k ⇒ [ord(ab) = k ≤ nm] ord(ab) = k = nm.

Suppose (n, m) ≠ 1, let (n, m) = k = p₁^r₁p₂^r₂···p_s^r_s for distinct primes p_i and strictly positive powers r_i.

If we could find an element of G with order p_i^r_i, then the product of these elements would have order k because 💡prime powers are relatively prime to prime powers of different primes.

Let take an arbitrary i, 1 ≤ i ≤ s, p_i^r_i. We notice that p_i^r_i divides either m or n, say m without losing any generality ⇒ a^m/p_ir_i has order p_i^r_i ∎

Proposition. Let K be any field, let G ⊆ K^x be a finite subgroup of K^x. Then, G is a cyclic group.

Proof.

Notice that G ⊆ K^x is Abelian. Let N = max {ord(a) | a ∈ G}, so there exists b ∈ G: ord(b) = N.

Claim: ∀a ∈ G, ord(a) divides N.

ord(a) = n, ord(b) = N ⇒ [Previous Lemma] ∃c ∈ G: ord(c) = lcm(n, N) ≥ [lcm(n, N) ≥ n, lcm(n, N) ≥ N] N ⇒ [ord(c)=lcm(n, N)≥ N and N = max {ord(a) | a ∈ G}] ord(c) = lcm(n, N) = N ⇒ n | N.

ord(a) divides N ∀a ∈ G ⇒ ∀a ∈ G, a^N = 1 ⇒ every element of G is a root of the polynomial x^N -1 in K

However, a polynomial of degree n over a field has at most n zeros, counting multiplicity and G ⊆ K^x ⇒ #number of roots of x^N -1 in K ≤ N ⇒ |G| ≤ N.

On the other hand, |G| ≥ N because N = ord(b) b ∈ G ⇒ [The order of an element can never exceed the order of the group] |G| = N and G = ⟨b⟩, i.e., G is cyclic ∎

Lemma. Let K be any field. The roots of xⁿ-1 in K form a cyclic subgroup of K^x = K\{0}

Proof.

U = {roots of xⁿ-1 in K} is a subgroup of K^x because: 1 ∈ U (1ⁿ = 1); α, β ∈ U ⇒ (αβ)ⁿ = αⁿ·βⁿ = 1; (α^-1)ⁿ = (αⁿ)^-1 = 1

U is a finite group in K^x cyclic ⇒ [Previous result. Let K be any field, let G ⊆ K^x be a finite subgroup of K^x. Then, G is a cyclic group.] U is cyclic∎

Definition. A generator of U_n is called a primitive nth root of unity.

Facts

• ζ is a primitive nth root of unity ↭ ord(ζ) = |U_n| = n ↭ If ζⁱ =1, i>0 ⇒ n divides i (n | i).

• U_n = [Let K be any field. The roots of xⁿ-1 in K form a cyclic subgroup of K^x = K{0}] ⟨ξ⟩, i.e., the nth roots of unity in K are 1, ζ, ζ²,··· ζ^n-1.

• ζⁱ is primitive ↭ i and n are coprime ⇒ #number of primitive nth roots of unity = #{i | 0 < i < n, (i, n) = 1} = φ(n). This is the Euler totient function where φ(1) = φ(2) = 1, φ(3) = [1, 2] 2, φ(4) = [1, 3] 2, φ(5) = [1, 2, 3, 4] 4, and so on.

• If p prime, φ(p) = p-1, and more generally n = p₁^r₁p₂^r₂··· p_k^r_k, then φ(n) = p₁^r₁-1(p₁-1)p₂^r₂-1(p₂-1)··· p_k^r_k-1(p_k-1).

Examples. F = ℚ, U_n = {complex nth roots of unity}

• (ℤ/nℤ)* is the multiplicative group of integers coprime to n modulo n, i.e., units (elements with inverses) = {$\bar i| (i, n) = 1$}, e.g., (ℤ/2ℤ)* = {$\bar 1$}, (ℤ/3ℤ)* = {$\bar 1, \bar 2$} ≋ ℤ/2ℤ, (ℤ/4ℤ)* = {$\bar 1, \bar 3$} ≋ ℤ/2ℤ, (ℤ/8ℤ)* = {$\bar 1, \bar 3, \bar 5, \bar 7$} ≋ ℤ/2ℤ x ℤ/2ℤ.

Theorem. Let n be a positive integer, F a field such that char(F) = 0 or char(F) = p > 0, p ɫ n. Let K/F be the splitting field of xⁿ-1 over F. Then, there is an injective group homomorphism between the Galois group of the cyclotomic extension and the multiplicative group (ℤ/nℤ)*, i.e., Φ: Gal(K/F) → (ℤ/nℤ)*. In particular, Cyclotomic extensions are always Abelian.

Proof.

Let σ ∈ Gal(K/F), and ξ ∈ K be a primitive nth root of unity. Recall that all the roots of xⁿ-1 form a cyclic group, and K is the splitting field of x^n-1 over F [Roots_xⁿ-1 = {1, ξ, ξ², ···, ξ^n-1}], so K = F(ξ).

Notice that σ(ξ) is also a primitive nth root of unity! because ξⁱ = 1 ↭ σ(ξⁱ) = (σ is an automorphism) σ(ξ)ⁱ = σ(1) = 1. Therefore, σ(ξ) = ξ^i_σ where 0 < i_σ < n, (i_σ, n) = 1

Example: n = 8, σ(ξ) = ξ, ξ³, ξ⁵ or ξ⁷

Let’s define Φ: Gal(K/F) → (ℤ/nℤ)*, σ → i_σ (mod n)

We could have also defined it as Φ: (ℤ/nℤ)* → Gal(K/F), i (mod n) → σ_i where σ_i(ξ) = ξⁱ

1. Φ is a group homomorphism. σ, τ ∈ G, στ(ξ) = σ(ξ^i_τ) = $(ξ^{i_τ})^{i_σ}$ = ξ^i_τi_σ ⇒ Φ(στ) = i_τi_σ = Φ(σ)Φ(τ).
2. Φ is one-to-one. Suppose Φ(σ) = 1 ∈ (ℤ/nℤ)* ⇒ σ(ξ) = ξ¹ = ξ ⇒ [K = F(ξ), σ ∈ Gal(K/F), σ fixes both F and ξ, then it fixes everything] σ = identity ⇒ Φ(Gal(K/F)) ≋ (ℤ/nℤ)*

Recall. A finite extension K/F is Abelian if K/F is Galois and Gal(K/F) is Abelian.

Since (ℤ/nℤ)* is Abelian, the embedded subgroup Gal(K/F) is Abelian ⇒ K/F is Abelian. We have proved that Cyclotomic extensions are always Abelian.

Notice that the map Φ is not always an isomorphism, e.g., F = K = ℂ, and Gal(K/F) → (ℤ/nℤ)* cannot be an isomorphism for n≥ 3: |Gal(K/F)| = [F = K, σ needs to fix everything] |{id}| = 1.

So we have already demonstrated the following theorem: Let n be a positive integer, F a field such that char(F) = 0 or char(F) = p > 0, p ɫ n. Let K/F be the splitting field of xⁿ-1 over F. Then, there is an injective group homomorphism between the Galois group of the cyclotomic extension and the multiplicative group (ℤ/nℤ)*, i.e., Φ: Gal(K/F) → (ℤ/nℤ)*. Futhermore, if F is ℚ, this map is an isomorphism, too.

Theorem. Let n be a positive integer. Let K/ℚ be the splitting field of xⁿ-1 over ℚ. Then, the Galois group of the cyclotomic extension is isomorphic to the multiplicative group (ℤ/nℤ)*. The isomorphism is given explicitly by the map Φ: Gal(K/ℚ) → (ℤ/nℤ)*, i (mod n) → σⁱ where σ_i(ξ) = ξⁱ.

Proof. ℚ has characteristic zero, K is the splitting field of xⁿ-1, let ξ be a primitive nth root of unity, and we have already demonstrated that it forms a cyclic subgroup of K^x ⇒ K = ℚ(ξ), so Φ: Gal(ℚ(ξ)/ℚ) → (ℤ/nℤ)* is an injective group homomorphism (previous result).

Let f(x) ∈ ℚ[x] be the irreducible polynomial of ξ over ℚ (i.e. a minimal polynomial with leading coefficient 1 and rational coefficients), deg(f(x)) = [ℚ(ξ) : ℚ] = [K : ℚ].

We are going to demonstrate that (i) deg(f(x)) = φ(n). Besides, if p is a prime number that does not divide n. Then, ξ^p is a root of f. More generally, (ii) zⁿ = 1 (z is a nth root of unity), z is a root of f, too, and p does not divide n ⇒ z^p is a root of f.

ξ is a root of xⁿ -1 and f(x) ∈ ℚ[x] is the irreducible polynomial of ξ ⇒ xⁿ-1 = f·h for some h ∈ ℚ[x].

Futhermore, h(x) ∈ ℤ[x]. This follows from Gauss Lemma, f, g ∈ ℚ[x]: content(fg) = content(f)·content(g). Definition. The content of a polynomial, say a_mx^m + a_m-1x^m-1 + ··· + a₁x + a₀ with coefficients in a unique factorization domain R (e.g. ℤ) is defined as the greatest common divisor of the coefficients of the polynomial, that is, cont(f) = gcd(a_m, a_m-1, ···, a₀). If f ∈ ℚ[x], cont(f) = $\frac{cont(nf)}{n}$ where n ∈ ℕ is such that nf ∈ ℤ[x]. If cont(f)=1 (f is described as primitive) ↭ f ∈ ℤ[x].

In our case, content(xⁿ -1) = 1, f is irreducible (so it is monic, the leading coefficient is one) so content(f) = 1 ⇒ [Gauss Lemma, xⁿ-1 = f·h] content(h) = 1 ⇒ h ∈ ℤ[x]

ξ is a nth root of unity ⇒ ξ^p is an nth root of unity, too ⇒ (ξ^p)ⁿ -1 = 0, that is, ξ^p is a root of xⁿ-1 = fh ⇒ f(ξ^p) = 0 (we have proved our claim and we are done) or h(ξ^p) = 0

Suppose for the sake of contradiction h(ξ^p) = 0, where h(x) ∈ ℤ[x]. Consider h₁(x) = h(x^p)∈ ℤ[x]. Then, h₁(ξ) = h(ξ^p) = 0 ⇒ ξ is a root of h₁. Since f is the irreducible polynomial of ξ ⇒ f divides h₁ and h₁ = f·g for some g ∈ ℚ[x]. But following the argument above, as f and h₁ (h ∈ ℤ[x] ⇒ h₁ ∈ ℤ[x]) are both integer polynomials ⇒ g ∈ ℤ[x].

h₁(x) = h(x^p) ≡ h(x)^p (modulo p).

Hint: h(x) = x² +2x +1. h₁(x) = h(x^p) = $(x^p)^{2}+2(x^p)+1=x^{2p}+2x^p+1$. h(x)^p = (x² +2x +1)^p ≡ (We know reduce mod p, we write the result of reducing the coefficients mod p, ℤ[x]→ℤ[x]/pℤ[x]) (x²)^p + (2x)^p +1 (mod p) -all other coefficients will cancel- ≡ [2^p ≡ 2 (mod p)] $(x^2)^p$ +2x^p + 1 (mod p) = h₁(x) modulo p

In other words, consider the map from ℤ onto ℤ/pℤ, n → $\bar n$ where $\bar n$ is the residue class {m ∈ ℤ: m ≡ n (mod p)}. This maps extends to ℤ[x] → ℤ[x]/pℤ[x] in the obvious way, v → v^†, ${(a_0 + a_1x + ··· a_nx^n)}^† = \bar a_0 + \bar a_1 + ··· + \bar a_n x^n$.

[h₁(x)]^† = [h(x^p)]^† = [(h(x))^†]^p, where the latter equality holds from repeated applications of the result that in ℤ[x]/pℤ[x], (ax + by)^p = a^px^p + b^py^p = ax^p + by^p.

Therefore, h(x)^p ≡ h₁(x) ≡ f(x)g(x) (modulo p) ⇒ [The demonstration is in the next paragraph] f(x) and h(x) have a common factor in ℤ[x]/pℤ[x]. Notice that ℤ[x]/pℤ[x] is a UFD, i.e., every element can be written as a product of irreducible elements, and in particular, let’s consider f(x).

Take r(x) irreducible factor of f(x) ⇒ r(x)|f(x)g(x) and r(x)|h(x)^p, too. Futhermore, r(x)|h(x)^p ⇒ [r(x) is an irreducible factor] r(x) | h(x), and therefore r(x) is a common factor of f(x) and h(x)

xⁿ-1 = fh in ℤ[x] ⇒ Let’s go modulo p: xⁿ -1 =$\bar f\bar h$ in ℤ[x]/pℤ[x]. However, as we have shown before f(x) and h(x) have a common factor in ℤ[x]/pℤ[x], so their degree of their gcd(f, h) > 1 (Let f and g be polynomials over the field F.Then f and g are relatively prime (gcd(f, g) = 1) if and only if f and g have no common root in any extension of F.). So in an extension field K of ℤ[x]/pℤ[x], K ≋ $\mathbb{F_p}, \bar f~ and~ \bar h$ have a common root ⇒ xⁿ -1 = $\bar f\bar h$ has a multiple root. If a ∈ K is a root of both $\bar f$ and $\bar h$, then (x -a)² divides xⁿ-1, but xⁿ-1 has distinct roots in ℂ ⊥ ⇒ ξ^p is a root of f.

Claim: If i < n, (i, n) = 1, then ξⁱ is a root of f

Proof. Let’s write the prime factorization of i, say i = p₁···p_l, (i, n) = 1 ⇒ p_j does not divide n.

ξ is a root of unity, p₁ does not divide n ⇒ [This was our latest proven result] ξ^p₁ is a root of f ⇒ [z = $({ξ}^{p_1})$, z^p = 1, p₂ does not divide n, and z is a root of f] $({ξ}^{p_1})^{p_2}$ is a root of f ⇒ $({ξ}^{p_1p_2})^{p_3}$ is a root of f ⇒ ··· ξⁱ is a root of f ∈ (ℤ[x]/nℤ[x])* ∀i (i < n, (i, n) = 1) ⇒ f has at least φ(n) -Euler toilet function- roots.

φ(n) ≤ deg(f) = [K = ℚ(ξ) and f is the irreducible polynomial of ξ over ℚ] [K : ℚ] = [Recall: K/Q is Galois] |Gal(K/ℚ)| ≤ [By the previous theorem, Φ: Gal(K/ℚ) → (ℤ/nℤ)* is a one-to-one group homomorphism] φ(n) ⇒ deg(f) = φ(n) = |Gal(K/ℚ)| ⇒ Φ: Gal(K/ℚ) → (ℤ/nℤ)* is an isomorphism ∎

Corollary. [ℚ(ξ_p) : ℚ] = Φ(p) = p - 1 where p is prime.

We already knew this because x^p-1 + ··· + x + 1 ∈ ℚ[x] is an irreducible polynomial (by Eiseinstein criteria) and it is the irreducible polynomial of ξ_p because x^p - 1 = (x -1)(x^p-1 + ··· + x + 1).

Cyclotomic polynomials. Let n be a positive integer, ξ be a primitive nth root of unity, namely, a generator of the cyclic group K = ℚ(ξ) ≋ (ℤ/nℤ)^*, {1, ξ, ξ², ···, ξ^n-1} is a basis. The cyclotomic polynomial Φ_n(x) is defined by Φ_n(x) = $\prod_{1≤k≤φ(n)}(x-z_k)$, z_k primitive nth root of unity = $\prod_{\substack{1≤k≤n \\gcd(k, n) = 1}}(x-ξ_n^k) = \prod_{\substack{1≤k≤n \\gcd(k, n) = 1}}(x-e^{\frac{2πik}{n}})$, this product is over all primitive nth roots of unity, Φ_n(x) ∈ ℂ[x].

Recall: $ξ_n^k$ is a primitive nth root of unity if and only if gcd(k, n) = 1. Euler’s formula: e^ix = cos(x) + isin(x), so the nth roots of unity can be written as $e^{\frac{2πik}{n} }$ De Moivre’s formula: (cos(x) + isin(x))ⁿ = cos(nx) + isin(nx).

Examples:

• Φ₁(x) = x -1 because the only primitive first root of unity is 1).
• Φ₂(x) = x + 1 because the only primitive square root of unity is -1.
• Φ₃(x) = (x -w)(x - w²) = x² -(w + w²)x + w³ = [w³ -1 = 0 ⇒ (w-1)(w²+w+1)=0 ⇒ because w is not 1, w + w² = -1] x² +x +1.
• Φ₄(x) = [The two primitive fourth roots of unity are i and -i] (x -i)(x + i) = x² +1.

Let L ⊆ ℂ be the splitting field for x^p -1 (F = ℚ is already our assumption), where p is prime. Then, with the exception of the root 1, all of the roots of x^p -1 are primitive, and so Φ_p(x) = x^p +x^p-1 + ··· + x +1.

• Φ₅(x) = x⁴ + x³ + x² +x + 1.
• Φ⁶(x) = x² -x + 1.

The zeros of the polynomial xⁿ -1 are precisely the nth roots of unity, each with multiplicity 1, i.e., Xⁿ -1 = $\prod_{1≤k≤n}(x-ξ_n^k)$ product of all nth roots of unity. Every nth root of unity is a primitive dth root of unity for some d that divides n ⇒ Xⁿ -1 = $\prod_{1≤k≤n}(x-ξ_n^k)=\prod_{d|n}(Φ_d(x))$

Proposition.

1. Φ_n(x) ∈ ℤ[x] .
2. Φ_n(x) is irreducible. It is indeed the irreducible polynomial of a primitive nth root of unity in ℂ, namely ξ_n.
3. deg(Φ_n(x)) = φ(n)

Proof.

(3) Let ξ be a primitive nth root of unity. Φ_n(x) = $\prod_{\substack{1≤k≤n \\gcd(k, n) = 1}}(x-ξ_n^k)$ ⇒ [ξ would be one of its factors] Φ_n(ξ) = 0. Moreover, deg(Φ_n) = #number of primitive nth roots of unity = φ(n)

(1) Xⁿ -1 = $\prod_{\substack{1≤k≤n \\gcd(k, n) = 1}}(x-ξ_n^k) =\prod_{d|n}(Φ_d(x))$

Let’s see the idea. Φ₁(x) = (x -1) ∈ ℤ[x]. x²-1 = Φ₁(x)Φ₂(x) where x²-1 and Φ₁(x) ∈ ℤ[x] ⇒ [Gauss Lemma] Φ₂(x) ∈ ℤ[x]. x³ -1 = Φ₁(x)Φ₃(x) where x³-1 and Φ₁(x) ∈ ℤ[x] ⇒ [Gauss Lemma] Φ₃(x) ∈ ℤ[x]. x¹⁰ -1 = Φ₁(x)Φ₂(x)Φ₅(x)Φ₁₀(x) ⇒ [x¹⁰-1 and by strong induction, Φ₁(x), Φ₂(x), and Φ₅(x) ∈ ℤ[x]] Φ₁₀(x) ∈ ℤ[x]

Xⁿ -1 = ($\prod_{\substack{d < n \\d|n}}(Φ_d(x))Φ_n(x)$ ⇒ [$\prod_{\substack{d < n \\d|n}}(Φ_d(x))∈ℤ[x]$ by induction hypothesis, xⁿ-1∈ℤ[x]] ⇒ Φ_n(x) ∈ ℤ[x]

(2) Φ_n(ξ_n) = 0, deg(Φ_n(x)) = [(3)] φ(n) = [ℚ(ξ_n) : ℚ], Φ_n(x) ∈ ℤ[x] ⇒ Φ_n(x) is the irreducible polynomial of ξ_n.

Examples.

• x⁶ -1 = Φ₁(x)Φ₂(x)Φ₃(x)Φ₆(x) = (x-1)(x+1)(x²+x+1)Φ₆(x)

x⁶-1 = (x³ -1)(x³ +1) = (x -1)(x +1)(x² +x +1)(x² -x +1)

Therefore, (x -1)(x +1)(x² +x +1)(x² -x +1) = (x-1)(x+1)(x²+x+1)Φ₆(x) ⇒ Φ₆(x) = x² -x +1. Notice that deg(Φ₆(x)) = φ(6) = 2.

• Φ₇(x) = [7 is prime] x⁶ + x⁵ + ··· + 1

• x⁸ -1 = Φ₁(x)Φ₂(x)Φ₄(x)Φ₈(x) = (x -1)(x +1)(x² +1)Φ₈(x)

x⁸ -1 = (x⁴ -1)(x⁴ +1) = (x -1)(x + 1)(x² +1)(x⁴ +1)

(x -1)(x + 1)(x² +1)(x⁴ +1) = (x -1)(x +1)(x² +1)Φ₈(x) ⇒ Φ₈(x) = x⁴ +1. deg(Φ₈(x)) = φ(8) = 4.

• Φ₉(x) = x⁶ + x³ + 1.

If p is prime ℚ(ξ_p)/ℚ is a cyclic extension, the Galois group Gal(ℚ(ξ_p)/ℚ)≋(ℤ/pℤ)*. If p is not prime, ℚ(ξ_p)/ℚ is not cyclic in general, e.g., Gal(ℚ(ξ₈)/ℚ)≋ ℤ/2ℤ x ℤ/2ℤ However, ℚ(ξ_p)/ℚ is always Abelian.

Abel’s Theorem. Let F be a field of characteristic 0, let p be a prime number, and a ∈ F. If x^p -a is reducible over F, then it has a linear factor x -c in F[x].

Proof. (Based on Fields and Galois Theory, John. M. Howie, Springer)

Suppose x^p -a is reducible over F, let g ∈F[x] be monic irreducible factor of f, deg(g) = d, say g(x) = x^d -b_d-1x^d-1 + ··· + (-1)^db₀

• If d = 1, then we are done.
• Suppose 1 < d < p. Let E be a splitting field of f over F and let β be a root of f in E.

Since E is a splitting field of f over F, deg(g) = d, then g factorizes in E[x], say g = (x -w^n₁β)(x -w^n₂β)···(x -w^n_dβ) where w is a primitive pth root of unity and 0 ≤ n₁ < n₂ < ··· < n_d < p.

Then, [g = (x -w^n₁β)(x -w^n₂β)···(x -w^n_dβ) = x^d -b_d-1x^d-1 + ··· + (-1)^db₀] b₀ = w^{n₁+n₂+ ···+ n_d}β^d = wⁿβ^d where n = n₁+n₂+ ···+ n_d.

b₀^p = [Since b₀ = wⁿβ^d] w^npβ^dp = [β is a root of f = x^p-a in E ⇒ β^p = a, w is a primitive pth root of unity] a^d

By assumption, p is prime ⇒ gcd(d, p) = 1 ⇒ [Bézout’s identity] ∃s, t: sd + tp = 1 ⇒ a = a^sda^tp = [b₀^p = a^d] b₀^psa^tp = (b₀^sa^t)^p, therefore c = b₀^sa^t is a root of x^p-a ⇒ [Factor theorem. A polynomial f(x) has a factor (x -α) ↭ α is a root, i.e., f(α) = 0] f has a linear factor (x -c) where c = b₀^sa^t ∈ F∎

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).