A student claims that since the quintic is 'unsolvable,' the equation x⁵ - 1 = 0 has no solutions. What is wrong with this claim?
Ax⁵ - 1 = 0 actually has no real solutions, so the student is partly correct
BThe insolvability result applies only to degree-6 and higher polynomials, not quintics
Cx⁵ - 1 = 0 has roots expressible by radicals; insolvability applies only to the general quintic with Galois group S₅
DThe Fundamental Theorem of Algebra only guarantees roots for polynomials over the reals, not over the complex numbers
The insolvability result says the *general* quintic — one with Galois group S₅ — cannot be solved by radicals. Specific quintics can be solvable: x⁵ - 1 = 0 has roots that are fifth roots of unity, expressible as e^(2πik/5), and its Galois group is Z₅, which is abelian and hence solvable. Moreover, every degree-5 polynomial has exactly five complex roots by the Fundamental Theorem of Algebra — insolvability only says those roots cannot always be written using radicals, not that they don't exist.
Question 2 Multiple Choice
Why is A₅ the critical obstacle in proving that S₅ is not solvable?
AA₅ is larger than S₅, so the derived series cannot pass through it
BA₅ is abelian, so the derived series reaches {e} too quickly to represent radical extensions
CA₅ is simple and non-abelian — it equals its own commutator subgroup, so the derived series gets stuck and never reaches {e}
DA₅ contains S₅ as a normal subgroup, preventing the composition series from terminating
A group is solvable if its derived series — the sequence G, [G,G], [[G,G],[G,G]], … — eventually reaches the trivial group {e}. For S₅, the derived series goes S₅ → A₅ → A₅ → ⋯. It gets stuck at A₅ because A₅ is simple: it has no proper non-trivial normal subgroups, which means its commutator subgroup is itself. A simple non-abelian group (like A₅) cannot be solvable, because solvability requires each quotient in the derived series to be abelian. Since A₅ is the 'terminal obstacle,' the series never reaches {e}, and S₅ is not solvable.
Question 3 True / False
The insolvability of the quintic proves that degree-5 polynomials have no roots.
TTrue
FFalse
Answer: False
This is the most important misconception to dispel. The Fundamental Theorem of Algebra guarantees that every degree-5 polynomial has exactly five roots in the complex numbers. Insolvability is not about existence of roots — it is about expressibility. The roots of a general quintic exist and can be approximated numerically to arbitrary precision; what cannot be done is write them as a finite combination of field operations and radicals starting from the coefficients. Special functions like the Bring radical can express quintic roots, just not radicals.
Question 4 True / False
A polynomial is solvable by radicals if and only if its Galois group is a solvable group — one whose derived series terminates at the trivial group.
TTrue
FFalse
Answer: True
This is the Fundamental Theorem of Galois Theory applied to radical solvability. A solvable group has a composition series with abelian (specifically cyclic of prime order) quotients, which corresponds precisely to a tower of field extensions where each step adjoins an nth root. Going up such a tower produces a splitting field for the polynomial via radicals. If the Galois group is not solvable — as with S₅ — no such tower can exist, and the polynomial cannot be solved by radicals. The group-theoretic condition and the field-theoretic condition are exactly equivalent.
Question 5 Short Answer
Why is A₅ unsolvable as a group, and how does this prove that a general quintic cannot be solved by radicals?
Think about your answer, then reveal below.
Model answer: A₅ is simple and non-abelian: its only normal subgroups are {e} and A₅ itself. This means A₅ equals its own commutator subgroup [A₅, A₅] = A₅, so the derived series of A₅ never decreases — it stays at A₅ forever. Because the derived series of S₅ reaches A₅ and then stalls (since A₅ = [A₅, A₅]), the derived series of S₅ never reaches {e}, making S₅ not solvable. Since a polynomial is solvable by radicals iff its Galois group is solvable, and the general quintic has Galois group S₅ (which is not solvable), no radical expression can produce its roots.
The simplicity of A₅ is the key fact that makes A₅ — and hence S₅ — unsolvable. Simplicity means there is no way to 'factor out' A₅ into smaller pieces via normal subgroup quotients, which is exactly what solvability requires. This is why the same argument fails for degree 4: S₄ is solvable (its derived series is S₄ → A₄ → V₄ → {e} → ⋯, where V₄ is the Klein four-group), which corresponds to the existence of the quartic formula.