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If $μ_{n} \subseteq F$ and $char F ∤ n$ , then a degree- $n$ extension is cyclic exactly when it is $F (α)$ with $α^{n} \in F$ and $x^{n} - α^{n}$ irreducible

Statement

Let $F$ be a field, let $n \geq 1$ , assume $char F ∤ n$ , and assume $F$ contains a primitive $n$ -th root of unity. For a finite extension $K / F$ of degree $n$ , the following are equivalent:

$K / F$ is cyclic.
There exists $α \in K$ such that $K = F (α)$ , $α^{n} \in F$ , and $x^{n} - α^{n}$ is irreducible over $F$ .

Facts & Assumptions

Given: A field $F$ , an integer $n \geq 1$ , a primitive $n$ -th root of unity $ζ \in F$ , and a finite extension $K / F$ of degree $n$ .

[F1]

The Lagrange resolvent is $R_{σ, ζ} (x) = \sum_{i = 0}^{n - 1} ζ^{- i} σ^{i} (x)$ (The Lagrange resolvent attached to a cyclic action and a root of unity).

[L1]

Distinct characters are linearly independent (Dedekind's linear independence theorem for distinct characters).

[L2]

If an extension contains one nonzero root $α$ of $x^{n} - a$ , then all roots are $ζ^{i} α$ , and the splitting field is obtained by adjoining $α$ and the relevant roots of unity (After adjoining one nonzero root $α$ of $x^{n} - a$ , all roots are $ζ α$ with $ζ^{n} = 1$ ).

[L3]

When $char F ∤ n$ , the polynomial $x^{n} - 1$ is separable and a splitting field has cyclic root-of-unity group of order $n$ ( $t^{n} - 1$ is separable over $K$ exactly when the characteristic does not divide $n$ , and then a splitting field carries $n$ distinct $n$ -th roots of unity).

[L4]

A finite extension is Galois exactly when it is the splitting field of a separable polynomial (Equivalent characterizations of a finite Galois extension).

[L5]

Embeddings of a simple algebraic extension correspond to the distinct roots of the minimal polynomial ( $F$ -embeddings of $F (α)$ into an algebraically closed field correspond to the distinct roots of $m_{α}$ ).

Proof

technique · direct

1.1

For the forward direction, assume $K / F$ is cyclic and choose a generator $σ$ of $Gal (K / F)$ . The distinct powers $1, σ, \dots, σ^{n - 1}$ are distinct characters $K^{\times} \to K^{\times}$ , so [L1] makes the resolvent operator $R_{σ, ζ}$ of [F1] nonzero. Choose $γ \in K$ with $α : = R_{σ, ζ} (γ) \neq 0$ .

F1L1choose

1.2

For the converse direction, assume $K = F (α)$ , $α^{n} = a \in F$ , and $x^{n} - a$ is irreducible. If $α = 0$ , then $a = 0$ and the irreducible polynomial is $x^{n}$ , which forces $n = 1$ . Hence $K = F$ , so the extension is cyclic. Assume from now on that $α \neq 0$ . Then $[K : F] = n$ , and [L2] shows every root of $x^{n} - a$ is $ζ^{i} α$ with $ζ^{i} \in F$ . Because [L3] makes the $n$ roots of $x^{n} - 1$ distinct, these roots $ζ^{i} α$ are distinct as well. Hence all roots of $x^{n} - a$ already lie in $K$ , and the polynomial is separable, so $K$ is its splitting field and [L4] makes $K / F$ Galois.

L2L3L4algebra

2.1

In the nonzero case of step 1.2, the root $ζ α$ of the irreducible polynomial $x^{n} - a$ determines, by [L5], an $F$ -embedding $K = F (α) \to K$ sending $α$ to $ζ α$ , hence an automorphism $τ$ of $K / F$ . Its powers send $α$ to $ζ^{i} α$ , so $τ$ has order $n$ . Because $[K : F] = n$ and a finite Galois extension has at most $[K : F]$ automorphisms, $Gal (K / F) = ⟨ τ ⟩$ is cyclic of order $n$ .

step 1.2L4L5algebra

2.2

Because $σ (α) = \sum_{i = 0}^{n - 1} ζ^{- i} σ^{i + 1} (γ) = ζ α,$ the element $α^{n}$ is fixed by $σ$ and hence by the whole cyclic Galois group, so $α^{n} \in F$ . If $0 < d < n$ and $α^{d} \in F$ , then $α^{d} = σ (α^{d}) = ζ^{d} α^{d},$ so $ζ^{d} = 1$ , contradicting primitivity of $ζ$ . Therefore no smaller positive power of $α$ lies in $F$ . The minimal polynomial of $α$ divides $x^{n} - α^{n}$ and has degree $n = [K : F]$ , so it is exactly $x^{n} - α^{n}$ and $K = F (α)$ .

step 1.1algebra

3.1

Steps 2.2 and 2.1 prove the equivalence.

step 2.2step 2.1∎

Remarks

The irreducibility clause is what forces degree exactly $n$ . Without it, adjoining one $n$ -th root can produce a proper divisor of $n$ .

Depends on

Used by

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Fields and Galois Theory, v5.10, Proposition 5.27 (standard reference, not scraped)
J. Ash, Basic Abstract Algebra, Section 6.7 (standard reference, not scraped)

If μn⊆Fμn​⊆F and char⁡F∤ncharF∤n, then a degree-nn extension is cyclic exactly when it is F(α)F(α) with αn∈Fαn∈F and xn−αnxn−αn irreducible