PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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After adjoining one nonzero root $α$ of $x^{n} - a$ , all roots are $ζ α$ with $ζ^{n} = 1$

Statement

Let $n \geq 1$ , let $F$ be a field, and let $a \in F$ . Suppose an extension $E / F$ contains a nonzero root $α$ of $x^{n} - a$ . Then $a \neq 0$ , and for $β \in E$ , $β^{n} = a ⟺ β = ζ α for some ζ \in E with ζ^{n} = 1.$ Consequently, if $μ_{n} (E) = {ζ \in E : ζ^{n} = 1}$ and $E$ contains every root of $x^{n} - a$ , its splitting field inside $E$ is $F (α, μ_{n} (E))$ .

Facts & Assumptions

Given: A positive integer $n$ , a field extension $E / F$ , and a nonzero $α \in E$ satisfying $α^{n} = a \in F$ .

[F1]

Every nonzero field element is a unit (Field).

[F2]

In a commutative group, $(x y)^{n} = x^{n} y^{n}$ and $(x^{- 1})^{n} = (x^{n})^{- 1}$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute).

[F3]

A splitting field is generated over the base by all roots of the polynomial (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

Proof

technique · direct, both directions

1.1

Since $α \neq 0$ , [F1] gives $α^{- 1}$ . Also $a = α^{n} \neq 0$ , because a product of nonzero field elements is nonzero.

1.2

Conversely, if $β = ζ α$ and $ζ^{n} = 1$ , then [F2] gives $β^{n} = ζ^{n} α^{n} = a$ .

2.1

If $β^{n} = a$ , put $ζ = β α^{- 1}$ . By [F2], $ζ^{n} = β^{n} (α^{n})^{- 1} = a a^{- 1} = 1$ , and $β = ζ α$ .

F1F2step 1.1

3.1

Thus the root set in $E$ is exactly $α μ_{n} (E)$ . The field generated by that set equals $F (α, μ_{n} (E))$ : it contains $α = 1 α$ , and from any root $ζ α$ it recovers $ζ = (ζ α) α^{- 1}$ ; the reverse containment follows because each $ζ α$ is a root. Now [F3] gives the splitting-field assertion.

F1F3step 2.1step 1.2∎

Depends on

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Dependency tree · two levels

25 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, Chapter 2 (standard reference, not scraped)

After adjoining one nonzero root αα of xn−axn−a, all roots are ζαζα with ζn=1ζn=1