ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Nonabelian frobenius conjugacy class

Example

The splitting field of $T^{3} - 2$ over $Q$ has Galois group $S_{3}$ . At p=5 its Frobenius conjugacy class consists of all three transpositions: distinct choices of prime can give distinct elements.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Frobenius cycle type and prime splitting: Let $F \in Z [T]$ be monic separable with splitting field L, and let p be a rational prime not dividing $Disc (F)$ . Then p is unramified in L and $\overset{ˉ}{F} \in F_{p} [T]$ is squarefree. The degrees of its monic irreducible factors, with each distinct factor counted once, are exactly the cycle lengths of arithmetic Frobenius on the roots of F.

[F2]

Frobenius elements above a prime are conjugate: In a finite Galois extension L/K let the nonzero prime p be unramified. If $σ P = P^{'}$ above p, then ${Frob}_{P^{'}} = σ {Frob}_{P} σ^{- 1} .$ Thus p determines one conjugacy class. If the Galois group is abelian, the element is independent of P.

[F3]

Eisenstein criterion over the integers: Let $f = a_{n} x^{n} + \dots + a_{0} \in Z [x]$ be primitive with $n \geq 1$ . If there is a prime $p$ such that $p ∤ a_{n}, p ∣ a_{i} for every i < n, p^{2} ∤ a_{0},$ then $f$ is irreducible in $Q [x]$ .

Verification

1.1

Eisenstein at 2 proves $T^{3} - 2$ irreducible. Let a be its positive real root. The real cubic field $Q (a)$ does not contain the nonreal cube root of unity $ζ_{3}$ . Its quadratic polynomial remains irreducible over that real field, so $L = Q (a, ζ_{3})$ has degree six and contains all roots. The faithful permutation action on its three roots embeds its order-six Galois group into $S_{3}$ , hence is an isomorphism.

1.2

For roots r of $T^{3} - 2$ , the derivative is $3 r^{2}$ . Their product is 2, so $\prod_{r} 3 r^{2} = 27 \cdot 4 = 108$ . Pairing differences contributes $(- 1)^{3}$ , giving discriminant -108. Thus 5 is a good prime. Modulo 5, direct multiplication gives $(T + 2) (T^{2} + 3 T + 4) = T^{3} - 2$ ; the quadratic discriminant is $9 - 16 = 3$ modulo 5, which is not among the squares 0,1,4. Its degrees are therefore 1 and 2.

algebra

2.1

The Frobenius cycle theorem gives cycle type (1,2), a transposition. The Frobenius elements above 5 constitute its entire conjugacy class. Conjugating a transposition in $S_{3}$ yields each of the three transpositions and no other permutation. Thus the class contains three distinct elements.

F1F2step 1.1step 1.2∎

Depends on

Used by

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Dependency tree · two levels

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Sources

§9.2.2, cube root of two; Milne Theorem 8.23 (standard reference, not scraped)