Kummer extensions

Definition. Let ζ be an element in an extension K/F. We say that ζ is an nth root of unity if ζⁿ = 1. Futhermore, ζ is said to be a primitive nth root of unity if n is the smallest positive integer such that ζⁿ = 1. In other words, ζⁿ = 1 and ζⁱ ≠ 1 ∀1 ≤ i < n."

• {1} is the set of first roots of unity, 1 is the primitive.
• {1, -1} is the set of square roots (2nd roots) of unity, -1 is the primitive square root.
• The primitive 3rd (cube) roots of unity in ℂ are w, w². The primitive 3rd (cube) roots of unity in $\mathbb{F_7}$ are 2, 4: 2³ = 7 ≡ 1 (mod 7), 4³ = 64 ≡ 1 (mod 7).
• {1, -1, i, -i} is the set of 4th roots of unity. i and -i are the two primitives 4th roots of unity.

Definition. Let n ≥ 2 be a positive integer, F a field such that char(F) = 0 or char(F) = p > 0, p ɫ n (p does not divide n), and F contains a primitive nth root of unity, ξ_n. A Kummer extension of F is an extension K/F such that [K : F] = n and K is the splitting field of an irreducible polynomial xⁿ-a where a ∈ F, i.e., K = F($\sqrt[n]{a}$). We say that K is obtained by adjoining to the base field F a root b = $\sqrt[n]{a}$ of the equation xⁿ = a. (the symbol $\sqrt[n]{a}$ for a ∈ F will be used to denote any root of the polynomial xⁿ -a ∈ F[x])

Examples

Recall. Suppose F is a field, char(F) ≠ 2. Let K/F be a degree 2 extension. Then, K/F is Galois and K = F(α) where α² ∈ F. In other words, a degree 2 extension can be obtained by adding a square root of an element of F.

• F = ℚ, n = 2 ⇒ -1 ∈ F which is the primitive square root of unity. A Kummer extension is simply a quadratic (degree two) extension $ℚ(\sqrt{α})/ℚ$ where α ∈ ℚ is a non-square element, i.e., $\sqrt{α}∉ℚ$. Notice that x² - α is irreducible.
• $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not a Kummer extension because ℚ does not contain a primitive 3rd root of unity (w and w²). However, $\mathbb{Q}(\sqrt[3]{2}, w)/\mathbb{Q(w)}$ is indeed a Kummer extension.
• Let K be the splitting field of x³ -2 over $\mathbb{F_7}$. It is a Kummer extension, n = 3, p = 7, p ɫ n, and 2 and 4 are primitive 3rd roots of unity and they “live” in $\mathbb{F_7}$.

  Notice that 2 is a primitive 3rd root of unity: 2 ≠ 1, 2² = 4 ≠ 1, 2³ = 1.

Theorem. Let n be a positive integer, F a field such that char(F) = 0 or char(F) = p > 0, p ɫ n containing a primitive nth root of unity. Let K/F be an extension of degree n ([K : F] = n). The following statements are equivalent:

1. K/F is a Kummer extension, i.e., ∃a ∈ F such that xⁿ-a is an irreducible polynomial and K is the splitting field of xⁿ -a over F.
2. K/F is a cyclic extension, i.e., K/F is Galois and Gal(K/F) is cyclic.

Theorem. Let n be a positive integer, F a field such that char(F) = 0 or char(F) = p > 0, p ɫ n containing a primitive nth root of unity. Let 0 ≠ a ∈ F, let K be the splitting field of xⁿ-a over F. Then,

1. K/F is a cyclic extension
2. |Gal(K/F)| = n ↭ xⁿ -a is irreducible over F.

Proof. (This proof relies heavily on NPTEL-NOC IITM, Introduction to Galois Theory.)

(1) Let K be the splitting field of xⁿ -a over F. Besides, let α ∈ K be a root of xⁿ -a.

If K = F, K/F is cyclic ∎

Let’s assume K ≠ F, α ∉ F (K is the splitting field of xⁿ -a ⇒ K is generated by the roots of the polynomial xⁿ-a ⇒ there exists some roots that are not in F), and ζ ∈ F be a primitive nth root of unity.

If ξ ∈ F is a primitive nth root of unity, then the distinct elements ζⁱα are all the roots of f in K [First, notice that all these distinct elements live in K: α ∈ K, ζ ∈ F ⊆ K, ζα ∈ K] (ζα)ⁿ = ζⁿαⁿ = [αⁿ = a, ζⁿ = 1] a. So {α, ζα, ζ²α, ···, ζ^n-1α} are the n distinct roots of f in K ⇒ K is the splitting field of the set {α, ζα, ζ²α, ···, ζ^n-1α} and [K is Galois over F ↭ K is a splitting field of a separable polynomial over F] K/F is Galois.

Recall: An irreducible polynomial f ∈ F[x] is separable iff the greatest common factor of f and f' is 1, it is expressed or written as (f, f') = 1 ↭ f and f' have no common roots

Every polynomial over a field of characteristic 0 is separable. If char(F) = p > 0, p ɫ n, f’(x) = nx^n-1. 0 is the only root of f’ and has, therefore, no roots in common with f(x).

Let σ ∈ Gal(K/F) ⇒ (σ(α))ⁿ = [σ is an homomorphism] σ(αⁿ) = [α is a root of f, αⁿ = a] σ(a) = [σ ∈ Gal(K/F), σ fixes F, a ∈ F] a ⇒ σ(α) is a root of xⁿ -a. Therefore, ∀σ ∈ Gal(K/F), σ(α) = ζⁱα for some i.

Let define a map: Φ: Gal(K/F) → ℤ/nℤ, σ → i_σ (mod n), where σ(α) = ζ^i_σα.

1. Φ is well-defined. Suppose i ≠ j and ζⁱα = ζ^jα, the question is Φ(i) ≡ Φ(j) (mod n)? ⇒ [*α^-1] ζⁱ = ζ^j ⇒ ζ^i-j = 1 ⇒ [ζ is a primitive nth root of 1] i-j is divisible by n, i ≡ j (mod n). In other words, σ(α) = ζⁱα = ζ^jα ⇒ i ≡ j (mod n), that is, i ∈ [$\bar j$] or, equivalently, j ∈ [$\bar i$] in ℤ/nℤ.
2. Φ is a group homomorphism. Φ(σ₁σ₂) = [σ₁(α) = ζ^i_σ1α, σ₂(α) = ζ^i_σ2α ⇒ σ₁σ₂(α) = σ₁(ζ^i_σ2α) = ζ^i_σ2(σ₁(α)) = ζ^i_σ2(ζ^i_σ1α) = ζ^i_σ1+i_σ2α] i_σ1+i_σ2 (mod n) = Φ(σ₁) + Φ(σ₂) (mod n).
3. Φ is injective, Φ(σ) = 0 ⇒ i_σ ≡ 0 (mod n) ⇒ σ(α) = ζ⁰α = α ⇒ [K = F(α), σ fixes F and σ fixes α, then σ fixes everything in K] σ = id.

Therefore, Gal(K/F) is isomorphic to a subgroup of ℤ/nℤ. Since ℤ/nℤ is cyclic, so is Gal(K/F), and K/F is a cyclic extension (K/F is Galois and Gal(K/F) is cyclic).

(2) [K : F] = K/F is a Galois extension |Gal(K/F)| = n ↭ xⁿ -a is irreducible over F.

(2⇒) [K : F] = |Gal(K/F)| = n ⇒ [We already know that K = F(α)] ⇒ [F(α) : F] = n ⇒ deg(g) = n where g is the irreducible polynomial of α over F ⇒ [α root f, f(α) = 0] g | f ⇒ [ deg(g) = deg(f) = n] f = g, so f is irreducible over F.

(2⇐) Suppose xⁿ -a is irreducible, α is a root of f ⇒ [K : F] = [(K=)F(α) : F] = n = K/F is a Galois extension |Gal(K/F)| ∎

David S. Dummit and Richard M. Foote: Abstract Algebra. Published 1990, Prentice Hall, 625ss. has a different version: Let F be a field of characteristic not dividing n which contains the n^th roots of unity. Then, the extension F($\sqrt[n]{a}$) for a ∈ F is cyclic over F of degree dividing n.

Their proof is quite similar to the one shown above (less than a sketch): (1) K = F($\sqrt[n]{a}$) is Galois because it is the splitting field for xⁿ - a. (2) For any σ ∈ Gal(K/F), σ($\sqrt[n]{a}$) is another root of the polynomial, hence σ($\sqrt[n]{a}$) = ξ_σ($\sqrt[n]{a}$) for some n^th root of unity ξ_σ. (3) This gives a map: Gal(K/F) → μ_n (the group of n^th roots of unity), σ → ξ_σ. (4) This map is an injection of Gal(K/F) into the cyclic group μ_n∎

Theorem. Let n be a positive integer, F a field such that char(F) = 0 or char(F) = p > 0, p ɫ n containing a primitive nth root of unity. Let K/F be a cyclic extension of degree n ([K : F] = n). Then, K is the splitting field of an irreducible polynomial xⁿ -a over F, a ∈ F.

Proof.

Let G = Gal(K/F), K/F is Galois, G is cyclic. Let ζ ∈ F be a primitive nth root of unity.

[K : F] = K/F is a Galois extension |Gal(K/F)| = n. Futhermore, by assumption G is cyclic, too ⇒ ∃σ ∈ G that is a generator of G ⇒ [G cyclic, |G|=n] 1, σ, σ²,···, σ^n-1 are all distinct F-automorphism of K ⇒ [Independence of field homomorphisms. Let K, L be two fields, let σ₁, σ₂, ···, σ_n: K → L be distinct field homomorphisms (σ_i≠σ_j, ∀i, j: i≠j). If a₁σ₁(α) + a₁σ₁(α)+ ··· + a_nσ_n(α) = 0 ∀α ∈ K, then σ₁, σ₂, ···, σ_n are independent. Where independent means there is no non-trivial dependence a₁σ₁(α) + a₁σ₁(α)+ ··· + a_nσ_n(α) = 0 which holds for every α ∈ K, that is, a₁ = a₂ = ··· = a_n = 0.], i.e., 1, σ, ···, σ^n-1 are (distinct characters and therefore) independent as functions K → K over F, 1 + ζσ + ζ²σ² + ··· + ζ^n-1σ^n-1 ≇ 0, i.e., not identically zero as a function on K, ∃β ∈ K, β ≠ 0: (1 + ζσ + ζ²σ² + ··· + ζ^n-1σ^n-1)(β) ≠ 0

∃β ∈ K, β ≠ 0: β + ζσ(β) + ζ²σ²(β) + ··· + ζ^n-1σ^n-1(β) = α (∈ K) ≠ 0. Let’s apply σ to α:

σ(α) = [Abusing notation] σα = σβ + ζσ²(β) + ζ²σ³(β) ··· + ζ^n-1σⁿ(β) = [σⁿ=id, σ is the generator of the group of automorphisms G of order n] σβ + ζσ²β + ζ²σ³β ··· + ζ^n-2σ^n-1β + ζ^n-1β [ζⁿ = 1 ⇒ ζ^n-1ζ = 1 ⇒ ζ^n-1 = ζ^-1] σα = ζ^-1(ζσβ + ζ²σ²β + ζ³σ³β ··· + ζ^n-1σ^n-1β + β) = [the elements in α are just reordered] ζ^-1α ⇒ σα = ζ^-1α.

Claim: αⁿ ∈ F, and it is going to be a = αⁿ, i.e., what we are looking for.

σ(αⁿ) = [σ homomorphism] (σα)ⁿ = [σα = ζ^-1α] (ζ^-1α)ⁿ = $ζ^{-1^{n}}α^n=α^n$ ⇒ [$ζ^{-1^{n}}=ξ^{-n}=ζ^{n^{-1}}=1^{-1}=1$] σ(αⁿ) = αⁿ ⇒ [e.g. σ²(αⁿ)=σ(σ(αⁿ))=σ(αⁿ)=αⁿ] σⁱ(αⁿ) = αⁿ ∀i ⇒ αⁿ ∈ K^G(it is fixed by all automorphisms of G) = [K/F is Galois ↭ K^G = F] F, so αⁿ ∈ F.

Let a = αⁿ ∈ F. xⁿ -a splits completely over F(α) because all the n roots of xⁿ -a, namely α, ζα, ζ²α,···, ζ^n-1α, are in F(α) and F(α) is obviously generated by α over F ⇒ F(α) is the splitting field of xⁿ -a over F.

αⁿ = a, (ζα)ⁿ = ζⁿαⁿ = a, (ζ²α)ⁿ = (ζⁿ)²αⁿ = a, etc.

Futhermore, let’s demonstrate that f = xⁿ -a is irreducible over F. We know σ(α) = [Abusing notation] σα = ζ^-1α, σ²α = σ(ζ^-1α) = ζ^-1σ(α) = ζ^-2α […] σⁱα = ζ^-iα ∈ F(α) ⇒ σⁱ(F(α)) ⊆ F(α)

∀i, 0 ≤ i ≤ n: σⁱ restricts to an automorphism of F(α). Besides, σⁱ ≠ σ^j for i ≠ j, 0 ≤ i,j < n.

For the sake of contradiction, let’s suppose that σⁱ(α) = σ^j(α) ⇒ ζ^-iα = ζ^-jα ⇒ ζ^-i = ζ^-j ⇒ [0 ≤ i,j < n] i = j ⇒ Each σⁱ ∈ Gal(F(α)/F) and they are all distinct, n ≤ |Gal(F(α)/F)| = [F(α)/F is a Galois extension because it is the splitting field of a separable polynomial, namely xⁿ-a] [F(α):F] ≤ [F(α) is an intermediate field] [K : F] = [By assumption] n ⇒ [F(α):F] = [K : F] = n ⇒ F(α) = K ⇒ xⁿ -a is irreducible (because deg_Fα = [F(α):F] = n, f(α) = 0, and deg(f) = n)∎

David S. Dummit and Richard M. Foote: Abstract Algebra. Published 1990, Prentice Hall, 625ss. has a different version: Any cyclic extension of degree n over a field F of characteristic not dividing n which contains the n^th roots of unity is of the form F($\sqrt[n]{a}$) for some a ∈ F.

Their proof is quite similar, too: (less than a sketch) 1. K/F cyclic extension, G = ⟨σ⟩. 2. Define the Lagrange resolvent (α, ξ) = α + ξσ(α) + ξ²σ²(α) + ··· + ξ^n-1σ^n-1(α). 3. Apply the automorphism σ to (α, ξ): σ(α, ξ) = ξ^-1(α, ξ). 4. σ(α, ξ)ⁿ = (ξ^-1)ⁿ(α, ξ)ⁿ = (α, ξ)ⁿ, so σ(α, ξ)ⁿ is fixed by Gal(K/F), hence is an element of F for any α ∈ K. 5. Let ξ be a primitive n^th root of unity, since 1, σ,···, σ^n-1 are linear independent (different characters) ⇒ ∃α ∈ K: (α, ξ) ≠ 0 and iterating σⁱ(α, ξ) = ξ^-i(α, ξ), and therefore σⁱ does not fix (α, ξ) for any i < n, 🔑this element cannot lie in any proper intermediate field of K/F ⇒ K = F($\sqrt[n]{a}$) = F((α, ξ)).

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).
4. Fields and Galois Theory. Howie, John. M. Springer Undergraduate Mathematics Series.
5. Fields and Galois Theory. Morandi. P., Springer.
6. Fields and Galois Theory. By Evan Dummit, 2020.
7. David S. Dummit and Richard M. Foote: Abstract Algebra. Published 1990, Prentice Hall