    # 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, ξn. 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])

You may find a different version of Kummer extension in literature. A Kummer extension is a field extension K/F, where F contains n distinct nth roots of unity (i.e. roots of xn -1) and K/F has Abelian Galois group of exponent n.

# 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.
• $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ is not a Kummer extension because ℚ does not contain a primitive 3rd root of unity (w and w2). However, $\mathbb{Q}(\sqrt{2}, w)/\mathbb{Q(w)}$ 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, ···, ζn-1α} are the n distinct roots of f in K ⇒ K is the splitting field of the set {α, ζα, ζ2α, ···, ζn-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) = nxn-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 → iσ (mod n), where σ(α) = ζiσα.

1. Φ is well-defined. Suppose i ≠ j and ζiα = ζjα, the question is Φ(i) ≡ Φ(j) (mod n)? ⇒ [*α-1] ζi = ζj ⇒ ζi-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 = ζjα ⇒ 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) = ζiσ1α, σ2(α) = ζiσ2α ⇒ σ1σ2(α) = σ1iσ2α) = ζiσ21(α)) = ζiσ2iσ1α) = ζiσ1+iσ2α] iσ1+iσ2 (mod n) = Φ(σ1) + Φ(σ2) (mod n).
3. Φ is injective, Φ(σ) = 0 ⇒ iσ ≡ 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 nth 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($\sqrt[n]{a}$) = ξσ($\sqrt[n]{a}$) for some nth root of unity ξσ. (3) This gives a map: Gal(K/F) → μn (the group of nth roots of unity), σ → ξσ. (4) This map is an injection of Gal(K/F) into the cyclic group μn

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,···, σn-1 are all distinct F-automorphism of K ⇒ [Independence of field homomorphisms. Let K, L be two fields, let σ1, σ2, ···, σn: K → L be distinct field homomorphisms (σi≠σj, ∀i, j: i≠j). If a1σ1(α) + a1σ1(α)+ ··· + anσn(α) = 0 ∀α ∈ K, then σ1, σ2, ···, σn are independent. Where independent means there is no non-trivial dependence a1σ1(α) + a1σ1(α)+ ··· + anσn(α) = 0 which holds for every α ∈ K, that is, a1 = a2 = ··· = an = 0.], i.e., 1, σ, ···, σn-1 are (distinct characters and therefore) independent as functions K → K over F, 1 + ζσ + ζ2σ2 + ··· + ζn-1σn-1 ≇ 0, i.e., not identically zero as a function on K, ∃β ∈ K, β ≠ 0: (1 + ζσ + ζ2σ2 + ··· + ζn-1σn-1)(β) ≠ 0

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

σ(α) = [Abusing notation] σα = σβ + ζσ2(β) + ζ2σ3(β) ··· + ζn-1σn(β) = [σn=id, σ is the generator of the group of automorphisms G of order n] σβ + ζσ2β + ζ2σ3β ··· + ζn-2σn-1β + ζn-1β [ζn = 1 ⇒ ζn-1ζ = 1 ⇒ ζn-1 = ζ-1] σα = ζ-1(ζσβ + ζ2σ2β + ζ3σ3β ··· + ζn-1σn-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. σ2n)=σ(σ(αn))=σ(αn)=αn] σin) = αn ∀i ⇒ αn ∈ KG(it is fixed by all automorphisms of G) = [K/F is Galois ↭ KG = 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,···, ζn-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 ≠ σj for i ≠ j, 0 ≤ i,j < n.

For the sake of contradiction, let’s suppose that σi(α) = σj(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) = Kxn -a is irreducible (because degFα = [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 nth 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) + ··· + ξn-1σn-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 nth root of unity, since 1, σ,···, σn-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
Bitcoin donation 