    # 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 ⊆ ··· ⊆ Fr such that

1. α ∈ Fr
2. Each extension Fi/Fi-1 is a radical extension, that is, Fi=Fi-1i) with αini ∈ Fi-1 or, alternatively, αi is the root of an irreducible polynomial in Fi-1[x] of the form xni -ci.

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 = F11), $α_1^5=3+\sqrt{2}$.

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’ = nxn-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 = ambn = [By assumption, ord(a)=m, ord(b)=n] ee = e ⇒ ord(ab) = k ≤ nm.

e = (ab)k = (ab)km = [G Abelian] akmbkm = [By assumption, ord(a)=m] ebkm = bkm ⇒ ord(b) = n | km ⇒ [(m, n)=1] n | k

Mutatis mutandis, e = (ab)k = (ab)kn = [G Abelian] aknbkn = [By assumption, ord(b)=n] akne = akn ⇒ 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 = p1r1p2r2···psrs for distinct primes pi and strictly positive powers ri.

If we could find an element of G with order piri, 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, piri. We notice that piri divides either m or n, say m without losing any generality ⇒ am/piri has order piri

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

Proof.

Notice that G ⊆ Kx 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, aN = 1 ⇒ every element of G is a root of the polynomial xN -1 in K

However, a polynomial of degree n over a field has at most n zeros, counting multiplicity and G ⊆ Kx ⇒ #number of roots of xN -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 Kx = K\{0}

Proof.

U = {roots of xn-1 in K} is a subgroup of Kx 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 Kx cyclic ⇒ [Previous result. Let K be any field, let G ⊆ Kx be a finite subgroup of Kx. Then, G is a cyclic group.] U is cyclic∎

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

# Facts

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

• Un = [Let K be any field. The roots of xn-1 in K form a cyclic subgroup of Kx = K{0}] ⟨ξ⟩, i.e., the nth roots of unity in K are 1, ζ, ζ2,··· ζn-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 = p1r1p2r2··· pkrk, then φ(n) = p1r1-1(p1-1)p2r2-1(p2-1)··· pkrk-1(pk-1).

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

Un Primitive φ(n)
n = 1 {1} 1 1
n = 2 {1, -1} -1 1
n = 3 {1, w, w2} w, w2 2
n = 4 {1, i, -i, -1} i, -i 2

• (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 xn-1 over F [Rootsxn-1 = {1, ξ, ξ2, ···, ξn-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) = ξiσ where 0 < iσ < n, (iσ, n) = 1

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

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

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

1. Φ is a group homomorphism. σ, τ ∈ G, στ(ξ) = σ(ξiτ) = $(ξ^{i_τ})^{i_σ}$ = ξiτiσ ⇒ Φ(στ) = iτiσ = 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 σi(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 KxK = 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].

f ≠ xn -1 😉 because xn -1 is not an irreducible polynomial over ℚ, mainly 1 ∈ ℚ is a root of xn -1.

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, ξp 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 ⇒ zp 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 amxm + am-1xm-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(am, am-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 ⇒ ξp is an nth root of unity, too ⇒ (ξp)n -1 = 0, that is, ξp is a root of xn-1 = fh ⇒ f(ξp) = 0 (we have proved our claim and we are done) or h(ξp) = 0

Suppose for the sake of contradiction h(ξp) = 0, where h(x) ∈ ℤ[x]. Consider h1(x) = h(xp)∈ ℤ[x]. Then, h1(ξ) = h(ξp) = 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(xp) ≡ h(x)p (modulo p).

Hint: h(x) = x2 +2x +1. h1(x) = h(xp) = $(x^p)^{2}+2(x^p)+1=x^{2p}+2x^p+1$. 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- ≡ [2p ≡ 2 (mod p)] $(x^2)^p$ +2xp + 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(xp)] = [(h(x))]p, where the latter equality holds from repeated applications of the result that in ℤ[x]/pℤ[x], (ax + by)p = apxp + bpyp = axp + byp.

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 ⊥ ⇒ ξp 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···pl, (i, n) = 1 ⇒ pj does not divide n.

ξ is a root of unity, p1 does not divide n ⇒ [This was our latest proven result] ξp1 is a root of f ⇒ [z = $({ξ}^{p_1})$, zp = 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(ξp) : ℚ] = Φ(p) = p - 1 where p is prime.

We already knew this because xp-1 + ··· + x + 1 ∈ ℚ[x] is an irreducible polynomial (by Eiseinstein criteria) and it is the irreducible polynomial of ξp because xp - 1 = (x -1)(xp-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, ···, ξn-1} is a basis. The cyclotomic polynomial Φn(x) is defined by Φn(x) = $\prod_{1≤k≤φ(n)}(x-z_k)$, zk 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, Φn(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: eix = 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 xp -1 (F = ℚ is already our assumption), where p is prime. Then, with the exception of the root 1, all of the roots of xp -1 are primitive, and so Φp(x) = xp +xp-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. Φn(x) ∈ ℤ[x] .
2. Φn(x) is irreducible. It is indeed the irreducible polynomial of a primitive nth root of unity in ℂ, namely ξn.
3. deg(Φn(x)) = φ(n)

Proof.

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

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

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(ξp)/ℚ is a cyclic extension, the Galois group Gal(ℚ(ξp)/ℚ)≋(ℤ/pℤ)*. If p is not prime, ℚ(ξp)/ℚ is not cyclic in general, e.g., Gal(ℚ(ξ8)/ℚ)≋ ℤ/2ℤ x ℤ/2ℤ However, ℚ(ξp)/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 xp -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 xp -a is reducible over F, let g ∈F[x] be monic irreducible factor of f, deg(g) = d, say g(x) = xd -bd-1xd-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 -wn1β)(x -wn2β)···(x -wndβ) where w is a primitive pth root of unity and 0 ≤ n1 < n2 < ··· < nd < p.

Then, [g = (x -wn1β)(x -wn2β)···(x -wndβ) = xd -bd-1xd-1 + ··· + (-1)db0] b0 = wn1+n2+ ···+ ndβd = wnβd where n = n1+n2+ ···+ nd.

b0p = [Since b0 = wnβd] wnpβdp = [β is a root of f = xp-a in E ⇒ βp = a, w is a primitive pth root of unity] ad

By assumption, p is prime ⇒ gcd(d, p) = 1 ⇒ [Bézout’s identity] ∃s, t: sd + tp = 1 ⇒ a = asdatp = [b0p = ad] b0psatp = (b0sat)p, therefore c = b0sat is a root of xp-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 = b0sat ∈ F∎

# Bibliography 