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ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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x4+x3+x2+x+1 has Galois group C4 over Q

Example

The polynomial x4+x3+x2+x+1 has Galois group C4 over Q.

Facts & Assumptions

Given: Eisenstein's criterion (Eisenstein criterion over the integers), the correspondence between conjugate roots and simple-extension embeddings (F-embeddings of F(α) into an algebraically closed field correspond to the distinct roots of mα), and the resolvent formula (The coefficient formula and discriminant of the quartic resolvent).

[L1]

In the unique-root resolvent branch, irreducibility over the resolvent splitting field distinguishes D4 from C4 (The five-case resolvent classification of an irreducible quartic Galois group).

Verification

technique · direct
1.1

After substituting x+1, the polynomial becomes x4+5x3+10x2+10x+5, which is Eisenstein at 5. Thus the original polynomial is irreducible.

givenalgebra
1.2

The resolvent formula gives R(y)=y3y23y+2=(y2)(y2+y1), so it has exactly one rational root.

givenalgebra
2.1

If ζ is a root, then ζ5=1 and ζ1. The roots are the distinct elements ζ,ζ2,ζ3,ζ4, all in Q(ζ), so this degree-four simple extension is the splitting field. The embedding ζζ2 is an automorphism and has order four on the exponents modulo 5; it therefore generates the full Galois group, which is C4.

step 1.1givenconstruct
3.1

Step 2.1 proves the group directly, while step 1.2 places it in the unique-root branch described by [L1]; the quadratic resolvent splitting field makes the quartic reducible there, as the C4 row requires.

step 2.1step 1.2L1

Depends on

Used by

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Sources