Proving x⁵ - 4x + 2 is Not Solvable

Abstract Algebra · Axiom Academy

EXAMPLE Proving x⁵ - 4x + 2 is Not Solvable Using Galois theory to show a quintic polynomial cannot be solved by radicals Excellent work! You've just proven that a specific quintic polynomial cannot be solved by radicals. Here's what we learned: Irreducibility is the foundation: Eisenstein's criterion quickly showed f(x) is irreducible over ℚ, which guarantees the Galois group acts transitively. The discriminant reveals structure: Since Disc(f) is not a perfect square, the Galois group doesn't lie entirely in A₅, so it must contain a transposition (2-cycle). Reduction modulo primes gives cycle information: Working mod 2 showed G contains a 5-cycle. The combination of transitivity, a 2-cycle, and a 5-cycle forces G = S₅. Solvability requires a solvable Galois group: The fundamental theorem of Galois theory tells us that f is solvable by radicals if and only if Gal(f/ℚ) is a solvable group. S₅ is not solvable: The symmetric group S₅ has no composition series with abelian quotients (its derived series doesn't reach e ), making it non-solvable. This exemplifies Abel-Ruffini theorem: there is no general formula (using radicals) for solving quintic equations. While some quintics are solvable (those with solvable Galois groups), most—like this one—are fundamentally unsolvable by radicals!

This is the written version of the interactive lesson above. See the full Abstract Algebra course.