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 = F_{0} ⊆ F_{1} ⊆ ··· ⊆ F_{r} such that
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 F_{2}, where F_{1} = ℚ(α_{0}), $α_0^2=2$, F_{2} = F_{1}(α_{1}), $α_1^5=3+\sqrt{2}$.
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, x^{n} -1 ∈ F[x] is separable. Let K/F be the splitting field of x^{n} -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 x^{n} -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^{m}b^{n} = [By assumption, ord(a)=m, ord(b)=n] ee = e ⇒ ord(ab) = k ≤ nm.
e = (ab)^{k} = (ab)^{km} = [G Abelian] a^{km}b^{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^{kn}b^{kn} = [By assumption, ord(b)=n] a^{kn}e = 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 = p_{1}^{r1}p_{2}^{r2}···p_{s}^{rs} for distinct primes p_{i} and strictly positive powers r_{i}.
If we could find an element of G with order p_{i}^{ri}, 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}^{ri}. We notice that p_{i}^{ri} divides either m or n, say m without losing any generality ⇒ a^{m/piri} has order p_{i}^{ri} ∎
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 x^{n}-1 in K form a cyclic subgroup of K^{x} = K\{0}
Proof.
U = {roots of x^{n}-1 in K} is a subgroup of K^{x} because: 1 ∈ U (1^{n} = 1); α, β ∈ U ⇒ (αβ)^{n} = α^{n}·β^{n} = 1; (α^{-1})^{n} = (α^{n})^{-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.
ζ is a primitive nth root of unity ↭ ord(ζ) = |U_{n}| = n ↭ If ζ^{i} =1, i>0 ⇒ n divides i (n | i).
U_{n} = [Let K be any field. The roots of x^{n}-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, ζ^{2},··· ζ^{n-1}.
ζ^{i} 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 = p_{1}^{r1}p_{2}^{r2}··· p_{k}^{rk}, then φ(n) = p_{1}^{r1-1}(p_{1}-1)p_{2}^{r2-1}(p_{2}-1)··· p_{k}^{rk-1}(p_{k}-1).
Examples. F = ℚ, U_{n} = {complex nth roots of unity}
U_{n} | Primitive | φ(n) | |
---|---|---|---|
n = 1 | {1} | 1 | 1 |
n = 2 | {1, -1} | -1 | 1 |
n = 3 | {1, w, w^{2}} | w, w^{2} | 2 |
n = 4 | {1, i, -i, -1} | i, -i | 2 |
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 x^{n}-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 x^{n}-1 form a cyclic group, and K is the splitting field of x^{n-1} over F [Roots_{xn-1} = {1, ξ, ξ^{2}, ···, ξ^{n-1}}], so K = F(ξ).
Notice that σ(ξ) is also a primitive nth root of unity! because ξ^{i} = 1 ↭ σ(ξ^{i}) = (σ is an automorphism) σ(ξ)^{i} = σ(1) = 1. Therefore, σ(ξ) = ξ^{iσ} where 0 < i_{σ} < n, (i_{σ}, n) = 1
Example: n = 8, σ(ξ) = ξ, ξ^{3}, ξ^{5} or ξ^{7}
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) = ξ^{i}
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 x^{n}-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 x^{n}-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) → σ^{i} where σ_{i}(ξ) = ξ^{i}.
Proof. ℚ has characteristic zero, K is the splitting field of x^{n}-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) z^{n} = 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 x^{n} -1 and f(x) ∈ ℚ[x] is the irreducible polynomial of ξ ⇒ x^{n}-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_{m}x^{m} + a_{m-1}x^{m-1} + ··· + a_{1}x + a_{0} 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}, ···, a_{0}). 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(x^{n} -1) = 1, f is irreducible (so it is monic, the leading coefficient is one) so content(f) = 1 ⇒ [Gauss Lemma, x^{n}-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 x^{n}-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 h_{1}(x) = h(x^{p})∈ ℤ[x]. Then, h_{1}(ξ) = h(ξ^{p}) = 0 ⇒ ξ is a root of h_{1}. Since f is the irreducible polynomial of ξ ⇒ f divides h_{1} and h_{1} = f·g for some g ∈ ℚ[x]. But following the argument above, as f and h_{1} (h ∈ ℤ[x] ⇒ h_{1} ∈ ℤ[x]) are both integer polynomials ⇒ g ∈ ℤ[x].
h_{1}(x) = h(x^{p}) ≡ h(x)^{p} (modulo p).
Hint: h(x) = x^{2} +2x +1. h_{1}(x) = h(x^{p}) = $(x^p)^{2}+2(x^p)+1=x^{2p}+2x^p+1$. h(x)^{p} = (x^{2} +2x +1)^{p} ≡ (We know reduce mod p, we write the result of reducing the coefficients mod p, ℤ[x]→ℤ[x]/pℤ[x]) (x^{2})^{p} + (2x)^{p} +1 (mod p) -all other coefficients will cancel- ≡ [2^{p} ≡ 2 (mod p)] $(x^2)^p$ +2x^{p} + 1 (mod p) = h_{1}(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^{†}, ${(a_0 + a_1x + ··· a_nx^n)}^† = \bar a_0 + \bar a_1 + ··· + \bar a_n x^n$.
[h_{1}(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^{p}x^{p} + b^{p}y^{p} = ax^{p} + by^{p}.
Therefore, h(x)^{p} ≡ h_{1}(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)
x^{n}-1 = fh in ℤ[x] ⇒ Let’s go modulo p: x^{n} -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 ⇒ x^{n} -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 x^{n}-1, but x^{n}-1 has distinct roots in ℂ ⊥ ⇒ ξ^{p} is a root of f.
Claim: If i < n, (i, n) = 1, then ξ^{i} is a root of f
Proof. Let’s write the prime factorization of i, say i = p_{1}···p_{l}, (i, n) = 1 ⇒ p_{j} does not divide n.
ξ is a root of unity, p_{1} does not divide n ⇒ [This was our latest proven result] ξ^{p1} is a root of f ⇒ [z = $({ξ}^{p_1})$, z^{p} = 1, p_{2} 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 ⇒ ··· ξ^{i} 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, ξ^{2}, ···, 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:
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.
The zeros of the polynomial x^{n} -1 are precisely the nth roots of unity, each with multiplicity 1, i.e., X^{n} -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 ⇒ X^{n} -1 = $\prod_{1≤k≤n}(x-ξ_n^k)=\prod_{d|n}(Φ_d(x))$
Proposition.
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) X^{n} -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. Φ_{1}(x) = (x -1) ∈ ℤ[x]. x^{2}-1 = Φ_{1}(x)Φ_{2}(x) where x^{2}-1 and Φ_{1}(x) ∈ ℤ[x] ⇒ [Gauss Lemma] Φ_{2}(x) ∈ ℤ[x]. x^{3} -1 = Φ_{1}(x)Φ_{3}(x) where x^{3}-1 and Φ_{1}(x) ∈ ℤ[x] ⇒ [Gauss Lemma] Φ_{3}(x) ∈ ℤ[x]. x^{10} -1 = Φ_{1}(x)Φ_{2}(x)Φ_{5}(x)Φ_{10}(x) ⇒ [x^{10}-1 and by strong induction, Φ_{1}(x), Φ_{2}(x), and Φ_{5}(x) ∈ ℤ[x]] Φ_{10}(x) ∈ ℤ[x]
X^{n} -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, x^{n}-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.
x^{6}-1 = (x^{3} -1)(x^{3} +1) = (x -1)(x +1)(x^{2} +x +1)(x^{2} -x +1)
Therefore, (x -1)(x +1)(x^{2} +x +1)(x^{2} -x +1) = (x-1)(x+1)(x^{2}+x+1)Φ_{6}(x) ⇒ Φ_{6}(x) = x^{2} -x +1. Notice that deg(Φ_{6}(x)) = φ(6) = 2.
Φ_{7}(x) = [7 is prime] x^{6} + x^{5} + ··· + 1
x^{8} -1 = Φ_{1}(x)Φ_{2}(x)Φ_{4}(x)Φ_{8}(x) = (x -1)(x +1)(x^{2} +1)Φ_{8}(x)
x^{8} -1 = (x^{4} -1)(x^{4} +1) = (x -1)(x + 1)(x^{2} +1)(x^{4} +1)
(x -1)(x + 1)(x^{2} +1)(x^{4} +1) = (x -1)(x +1)(x^{2} +1)Φ_{8}(x) ⇒ Φ_{8}(x) = x^{4} +1. deg(Φ_{8}(x)) = φ(8) = 4.
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(ξ_{8})/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-1}x^{d-1} + ··· + (-1)^{d}b_{0}
Since E is a splitting field of f over F, deg(g) = d, then g factorizes in E[x], say g = (x -w^{n1}β)(x -w^{n2}β)···(x -w^{nd}β) where w is a primitive pth root of unity and 0 ≤ n_{1} < n_{2} < ··· < n_{d} < p.
Then, [g = (x -w^{n1}β)(x -w^{n2}β)···(x -w^{nd}β) = x^{d} -b_{d-1}x^{d-1} + ··· + (-1)^{d}b_{0}] b_{0} = w^{n1+n2+ ···+ nd}β^{d} = w^{n}β^{d} where n = n_{1}+n_{2}+ ···+ n_{d}.
b_{0}^{p} = [Since b_{0} = w^{n}β^{d}] w^{np}β^{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^{sd}a^{tp} = [b_{0}^{p} = a^{d}] b_{0}^{ps}a^{tp} = (b_{0}^{s}a^{t})^{p}, therefore c = b_{0}^{s}a^{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 = b_{0}^{s}a^{t} ∈ F∎