V.21 The Insolubility of the Quintic

Martin W. Liebeck

Every student will be familiar with the formula for the roots of a quadratic polynomial ax² + bx + c, namely (-b ± )/2a. Perhaps less familiar is the fact that there is also a formula for the roots of a cubic: write the cubic as x³ + ax²+ bx+c, and make the substitution y = x + a to rewrite it in the form y³ + 3hy + k. The roots of this are then of the form

While the quadratic formula was known to the Greeks, the cubic formula was not found until the sixteenth century. In the same century a formula for the roots of quartic (degree 4) polynomials was also found. The formulas for quadratics, cubics, and quartics all arise by applying a sequence of arithmetic operations (addition, subtraction, multiplication, division) together with extraction of roots (square roots, cube roots, and so on) to the coefficients of the original polynomial. Such a formula is called a radical expression for the roots.

The next step, naturally enough, was the quintic (i.e., polynomial of degree 5). However, several hundred years passed without anyone finding a radical formula for the roots of a general quintic polynomial.

There was a good reason for this. There is no such formula. Nor is there a formula for polynomials of degree greater than 5. This fact was first established in the early nineteenth century by ABEL [VI.33] (who died aged twenty-six), after which GALOIS [VI.41] (who died aged twenty-one) built an entirely new theory of equations that not only explained the nonexistence of formulas but laid the foundations for a whole edifice of algebra and number theory known as Galois theory, a major area of modern-day research.

One of the key ideas of Galois was to associate with any polynomial f = f(x) a GROUP [I.3 §2.1] Gal(f) (the Galois group of f), which is a finite group that permutes the roots of f. This group is defined in terms of certain FIELDS [I.3 §2.2], which for these purposes can be thought of as subsets F of the COMPLEX NUMBERS [I.3 §1.5] having the property that if a, b are any two elements of F, then all the numbers a + b, a - b, ab, and a/b also lie in F (where we assume that b ≠ 0 in the last case to avoid dividing by 0). The standard mathematical language for this property is to say that F is “closed under” the usual arithmetic operations of addition, subtraction, multiplication, and division. For example, the rationals form a field, as does () = {a + b : a, b ∈ } (this is clearly closed under addition, subtraction, and multiplication, and is also closed under division since 1/(a + b) = a/(a² - 2b²) - b/(a² - 2b²)). A polynomial f(x) of degree n with rational coefficients has n complex roots by THE FUNDAMENTAL THEOREM OF ALGEBRA [V.13]—call them α₁,. . . , α_n. The splitting field of f is defined to be the smallest field containing and all the α_i, and is written as (α₁, . . . α_n). For example, the polynomial x² - 2 has roots ± , so its splitting field is (), defined above. Less trivially, x³ - 2 has roots α, αω², αω², where α = 2^1/3, the real cube root of 2, and ω = e^2Πi/3 so its splitting field is (α,ω), which consists of all complex numbers a₁ + a₂α + a₃α² + a₄ω + a₅αω + a₆α²ω with a_i ∈ . (Notice that we do not have to include ω² in such expressions since ω³ = 1, so (ω - 1)(ω² + ω + 1) = ω³ - 1 = 0, which implies that ω² = -ω - 1.)

Let E = (α₁,. . . , α_n) be the splitting field of our polynomial f. An automorphism of E is a bijection φ : E → E that preserves addition and multiplication—in other words, φ(a + b) = φ(a) + φ(b) and φ(ab) = φ(a)φ(b) for all a, b ∈ E. Such a function necessarily also preserves subtraction and division, and fixes every rational number. Denote by Aut(E) the set of all automorphisms of E. For example, when E = y(), any automorphism φ satisfies

and therefore φ() = or -. In the first case φ(a + b) = a + b for all a, b — , while in the second φ(a + b)= a - b. Both of these are automorphisms of E; call them φ₁, φ₂, so that Aut(E)= {φ₁φ₂}.

The composition φ ° ψ of two automorphisms ψ, φ of E is again an automorphism, and so is the inverse function φ^-1, while the identity function ι defined by ι(e) = e for all e ∈ E is also an automorphism. Since composition of functions is an associative operation, it follows that Aut(E) is a group under composition. Define the Galois group Gal(f) of our polynomial f(x) with splitting field E to be this group Aut(E). Thus, for example, Gal(x² - 2) = {φ¹φ²}. Notice that φ¹ is the identity ι, while φ = φ² ° φ² = φ¹, so this is just a cyclic group of order 2. Similarly, if f(x) = x³ -2, with splitting field E = (α, ω) as above, then any φ ∈ Aut(E) satisfies φ(α)³ = φ(α³) = φ(2) = 2, and therefore φ(α) = α, αω, or αω²; likewise φ(ω) = ω or ω². Once φ(α) and φ(ω) are specified, φ is completely determined (since φ(a₁+a₂α+. . .+a₆α²ω) = a₁ + a₂φ(α)+. . .+ a₆φ(α)²φ(ω)), so there are just sixpossibilities for the automorphism φ. It turns out thateach of these is indeed an automorphism, and therefore Gal(x³ - 2) is a group of order 6. In fact, this group is isomorphic to the SYMMETRIC GROUP [III.68] S₃, ascan be seen by considering each automorphism as apermutation of the three roots of f(x).

Now that the Galois group is defined, it is possible to state some of Galois’s fundamental results that lead to the insolubility of the quintic. Each subgroup H of G = Gal(f) has a fixed field H^†, which is defined to be the set of all numbers a ∈ E such that φ(a) = a for all φ ∈ H. Galois proved that the association between H and H^† gives a one-to-one correspondence between subgroups of G and fields which lie between and E (the so-called intermediate subfields of E). The condition that f(x) has a radical formula for its roots leads to certain special kinds of intermediate subfields, and hence to certain special subgroups of G, and eventually to Galois’s most famous theorem: the polynomial f(x) has a radical formula for its roots if and only if its Galois group Gal(f) is a soluble group. (This means that G = Gal(f) has a sequence of subgroups 1 = G₀ < G₁ < . . . < G_r = G such that for each i, G_i is a NORMAL SUBGROUP [I.3 §3.3] of G_i+1 and the factor group G_i+1/G_i is Abelian.)

It follows from Galois’s theorem that to demonstrate the insolubility of the quintic, it is enough to produce a quintic f(x) such that Gal(f) is not a soluble group. An example of such a quintic is f(x) = 2x⁵ - 5x⁴ + 5: one can show first that Gal(f) is isomorphic to the symmetric group S₅; and second that S₅ is not a soluble group. Here is a brief sketch of how the argument goes. First one establishes that f(x) is an irreducible polynomial (i.e., is not the product of two rational polynomials of smaller degree) with five distinct complex roots. Thus, as observed above, Gal(f) can be regarded as a subgroup of S₅ that permutes the five roots. By sketching the graph of f(x) one can easily see that three of its roots are real and that the other two, call them α₁ and α₂, are complex conjugates of each other. Since the complex conjugation map z → always gives an automorphism in Gal(f), it follows that Gal(f) is a subgroup of S₅ that contains a 2-cycle, namely (α₁α₂). Another basic general fact is that the Galois group of an irreducible polynomial permutes the roots transitively, meaning that for any two roots α_i, α_j there exists an automorphism in Gal(f) that sends α_i to α_j. Thus, our group Gal(f) is a subgroup of S₅ that permutes the five roots transitively and contains a 2-cycle. At this point some fairly elementary group theory shows that Gal(f) must actually be the whole of S₅. Finally, the fact that S₅ is not a soluble group follows easily from the fact that the alternating group A₅ is a non-Abelian simple group (i.e., it has no normal subgroups apart from the identity subgroup and A₅ itself).

These ideas can be extended to produce polynomials of any degree n ≥ 5 that have Galois group S_n, and that are therefore not soluble by radicals. The reason this cannot be done for quartics, cubics, and quadratics is that S₄ and all its subgroups are soluble groups.