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$x^{4} + x^{3} + x^{2} + x + 1$ has Galois group $C_{4}$ over $Q$

Example

The polynomial $x^{4} + x^{3} + x^{2} + x + 1$ has Galois group $C_{4}$ 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 $m_{α}$ ), 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 $D_{4}$ from $C_{4}$ (The five-case resolvent classification of an irreducible quartic Galois group).

Verification

technique · direct

1.1givenalgebra

After substituting $x + 1$ , the polynomial becomes $x^{4} + 5 x^{3} + 10 x^{2} + 10 x + 5$ , which is Eisenstein at $5$ . Thus the original polynomial is irreducible.

1.2givenalgebra

The resolvent formula gives $R (y) = y^{3} - y^{2} - 3 y + 2 = (y - 2) (y^{2} + y - 1)$ , so it has exactly one rational root.

2.1step 1.1givenconstruct

If $ζ$ is a root, then $ζ^{5} = 1$ and $ζ \neq 1$ . The roots are the distinct elements $ζ, ζ^{2}, ζ^{3}, ζ^{4}$ , all in $Q (ζ)$ , so this degree-four simple extension is the splitting field. The embedding $ζ \mapsto ζ^{2}$ is an automorphism and has order four on the exponents modulo $5$ ; it therefore generates the full Galois group, which is $C_{4}$ .

3.1step 2.1step 1.2L1∎

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 $C_{4}$ row requires.

Depends on

Used by

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Sources

J. S. Milne, Fields and Galois Theory, v5.10, quartic subgroup table (standard reference, not scraped)
K. Conrad, Galois Groups of Cubics and Quartics, Section 3 (standard reference, not scraped)

x4+x3+x2+x+1x4+x3+x2+x+1 has Galois group C4C4​ over QQ