Definition. Let ζ be an element in an extension K/F. We say that **ζ is an nth root of unity if** ζ^{n} = 1. Futhermore, ζ is said to be a primitive nth root of unity if n is the smallest positive integer such that ζ^{n} = 1. In other words, ζ^{n} = 1 and ζ^{i} ≠ 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, w
^{2}. The primitive 3rd (cube) roots of unity in $\mathbb{F_7}$ are 2, 4: 2^{3}= 7 ≡ 1 (mod 7), 4^{3}= 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 x^{n}-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 x ^{n} = a.** (the symbol $\sqrt[n]{a}$ for a ∈ F will be used to denote any root of the polynomial 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 x^{n} -1) and K/F has Abelian Galois group of exponent n.

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 α^{2} ∈ 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 x^{2}- 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 w
^{2}). However, $\mathbb{Q}(\sqrt[3]{2}, w)/\mathbb{Q(w)}$ is indeed a Kummer extension. - Let K be the splitting field of x
^{3}-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, 2

^{2}= 4 ≠ 1, 2^{3}= 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:

- K/F is a Kummer extension, i.e., ∃a ∈ F such that x
^{n}-a is an irreducible polynomial and K is the splitting field of x^{n}-a over F. - 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 x^{n}-a over F. Then,

- K/F is a cyclic extension
- |Gal(K/F)| = n ↭ x
^{n}-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 x^{n} -a over F. Besides, let α ∈ K be a root of x^{n} -a.

If K = F, K/F is cyclic ∎

Let’s assume K ≠ F, α ∉ F (K is the splitting field of x^{n} -a ⇒ K is generated by the roots of the polynomial x^{n}-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 ζ ^{i}α 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)

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(α^{n}) = [α is a root of f, α^{n} = a] σ(a) = [σ ∈ Gal(K/F), σ fixes F, a ∈ F] a ⇒ σ(α) is a root of x^{n} -a. Therefore, **∀σ ∈ Gal(K/F), σ(α) = ζ ^{i}α for some i.**

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

- Φ is well-defined. Suppose i ≠ j and ζ
^{i}α = ζ^{j}α, the question is Φ(i) ≡ Φ(j) (mod n)? ⇒ [*α^{-1}] ζ^{i}= 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) = ζ^{i}α = ζ^{j}α ⇒ i ≡ j (mod n), that is, i ∈ [$\bar j$] or, equivalently, j ∈ [$\bar i$] in ℤ/nℤ. - Φ is a group homomorphism. Φ(σ
_{1}σ_{2}) = [σ_{1}(α) = ζ^{iσ1}α, σ_{2}(α) = ζ^{iσ2}α ⇒ σ_{1}σ_{2}(α) = σ_{1}(ζ^{iσ2}α) = ζ^{iσ2}(σ_{1}(α)) = ζ^{iσ2}(ζ^{iσ1}α) = ζ^{iσ1+iσ2}α] i_{σ1}+i_{σ2}(mod n) = Φ(σ_{1}) + Φ(σ_{2}) (mod n). - Φ is injective, Φ(σ) = 0 ⇒ i
_{σ}≡ 0 (mod n) ⇒ σ(α) = ζ^{0}α = 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 ↭ x^{n} -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 x^{n} -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 x ^{n} - a*. (2) For any σ ∈ Gal(K/F), σ($\sqrt[n]{a}$) is another root of the polynomial, hence σ($\sqrt[n]{a}$) = ξ

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 x^{n} -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, σ^{2},···, σ^{n-1} are all distinct F-automorphism of K ⇒ [**Independence of field homomorphisms**. Let K, L be two fields, let σ_{1}, σ_{2}, ···, σ_{n}: K → L be distinct field homomorphisms (σ_{i}≠σ_{j}, ∀i, j: i≠j). If a_{1}σ_{1}(α) + a_{1}σ_{1}(α)+ ··· + a_{n}σ_{n}(α) = 0 ∀α ∈ K, then σ_{1}, σ_{2}, ···, σ_{n} are independent. Where independent means there is no non-trivial dependence a_{1}σ_{1}(α) + a_{1}σ_{1}(α)+ ··· + a_{n}σ_{n}(α) = 0 which holds for every α ∈ K, that is, a_{1} = a_{2} = ··· = a_{n} = 0.], i.e., 1, σ, ···, σ^{n-1} are (distinct characters and therefore) independent as functions K → K over F, 1 + ζσ + ζ^{2}σ^{2} + ··· + ζ^{n-1}σ^{n-1} ≇ 0, i.e., not identically zero as a function on K, ∃β ∈ K, β ≠ 0: (1 + ζσ + ζ^{2}σ^{2} + ··· + ζ^{n-1}σ^{n-1})(β) ≠ 0

∃β ∈ K, β ≠ 0: β + ζσ(β) + ζ^{2}σ^{2}(β) + ··· + ζ^{n-1}σ^{n-1}(β) = α (∈ K) ≠ 0. Let’s apply σ to α:

σ(α) = [Abusing notation] σα = σβ + ζσ^{2}(β) + ζ^{2}σ^{3}(β) ··· + ζ^{n-1}σ^{n}(β) = [σ^{n}=id, σ is the generator of the group of automorphisms G of order n] σβ + ζσ^{2}β + ζ^{2}σ^{3}β ··· + ζ^{n-2}σ^{n-1}β + ζ^{n-1}β [ζ^{n} = 1 ⇒ ζ^{n-1}ζ = 1 ⇒ ζ^{n-1} = ζ^{-1}] σα = ζ^{-1}(ζσβ + ζ^{2}σ^{2}β + ζ^{3}σ^{3}β ··· + ζ^{n-1}σ^{n-1}β + β) = [the elements in α are just reordered] ζ^{-1}α ⇒ **σα = ζ ^{-1}α**.

Claim: α^{n} ∈ F, and it is going to be a = α^{n}, i.e., what we are looking for.

σ(α^{n}) = [σ homomorphism] (σα)^{n} = [**σα = ζ ^{-1}α**] (ζ

Let a = α^{n} ∈ F. x^{n} -a splits completely over F(α) because all the n roots of x^{n} -a, namely α, ζα, ζ^{2}α,···, ζ^{n-1}α, are in F(α) and F(α) is obviously generated by α over F ⇒ **F(α) is the splitting field of x ^{n} -a over F.**

α^{n} = a, (ζα)^{n} = ζ^{n}α^{n} = a, (ζ^{2}α)^{n} = (ζ^{n})^{2}α^{n} = a, etc.

Futhermore, let’s demonstrate that f = x^{n} -a is irreducible over F. We know σ(α) = [Abusing notation] σα = ζ^{-1}α, σ^{2}α = σ(ζ^{-1}α) = ζ^{-1}σ(α) = ζ^{-2}α […] σ^{i}α = ζ^{-i}α ∈ F(α) ⇒ σ^{i}(F(α)) ⊆ F(α)

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

For the sake of contradiction, let’s suppose that σ^{i}(α) = σ^{j}(α) ⇒ ζ^{-i}α = ζ^{-j}α ⇒ ζ^{-i} = ζ^{-j} ⇒ [0 ≤ i,j < n] i = j ⇒ **Each σ ^{i} ∈ Gal(F(α)/F) and they are all distinct**, n ≤ |Gal(F(α)/F)| = [F(α)/F is a Galois extension because

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) + ξ^{2}σ^{2}(α) + ··· + ξ^{n-1}σ^{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 σ^{i}(α, ξ) = ξ^{-i}(α, ξ), and therefore σ^{i} 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)).

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Fields and Galois Theory. Howie, John. M. Springer Undergraduate Mathematics Series.
- Fields and Galois Theory. Morandi. P., Springer.
- Fields and Galois Theory. By Evan Dummit, 2020.
- David S. Dummit and Richard M. Foote: Abstract Algebra. Published 1990, Prentice Hall