Solution of Equations in Radicals

In this section we return to a subject touched upon in the introduction, the solution of equations in radicals.

Let f = F[x] be a nonconstant polynomial

First, let us clarify what we mean when we ask for a "solution in radicals" of the equation f(x) = 0. Roughly speaking, we mean a set of formulas which express the zeros of f in terms of the coefficients a₁,...,a_n using only addition, subtraction, multiplication, division, and extraction of roots. Let us reformulate this idea in terms of field extensions.

Definition 1: Let E be a field extension of F. We say that E is a radical extension of F if E = F(₁,...,_t), where ₁^m₁ F, ₁^m₁ F(₁,...,_i-1) (i = 2,3,...,t) for some integers m₁,m₂,...,m_t.

Remark: In the definition of a radical extension, let m denote the least common multiple of m₁, m₂,...,m_t Then for each i(1 < i < t), m = k_im_i for some positive integer k_i. Therefore,

₁^m₁ F ₁^m = (₁^m₁)^k₁ F,

_i^m_i F(₁,...,_i-1)

_i^m = (_i^m_i)^k_i F(₁,...,_i-1)  (i = 2,...,t).

Thus, if we replace each m_i by m, we see that in the definition of a radical extension, we may assume that all m_i are equal.

Roughly speaking, a radical extension of F is obtained by succesively adjoining a sequence of radical to F.

Let F = Q in all the following examples:

Example 1: E = Q( is a radical extension of Q.

Example 2: E = Q(, ) is a radical extension of Q.

Example 3: E = Q(), where is a primitive nth root of unity, is a radical extension of Q since 1 = ⁿ Q.

Definition 2: We say that f is solvable in radicals if the splitting field E_f of f over F is contained in some radical extension E of F.

The reader should have no difficulty convincing himself that the definition of solvability in radicals coincides with the roughly stated notion which we previously introduced. For indeed, if E_f = F(₁,...,_n), where ₁,...,_n are the zeros of f. Therefore, to say that E_f is contained in some radical extension of F just means that the zeros _i can be expressed in terms of radicals and in addition, subtraction, multiplication, and division.

Proposition 3: Let E/F be a radical extension. Then E is contained in a radical extension E' of F such that E'/F is a Galois extension.

Proof: Let E = F(₁,...,_n), where ₁^m₁ F, ₁^m₁ F(₁,...,_i-1) ( i = 2,...,t ). Set _i^m_i = a_i and let for each i(1 < i < t), a_ij(1 < j < k_i) be the conjugates of a_i over F. Let E' be the smallest subfield of F containing F and all the zeros of the polynomials X^m_i - a_ij (1 < i < t, 1 < j < k_i). It is clear that E' is normal over F and hence E'/F is a Galois extension (since E' is a splitting field). Moreover, it is clear that E' E, since for some j, a_ij = a_i hence a_i E' (1 < i < t). Finally, E' is a radical extension of F.

Let us now investigate the properties of the Galois group of a radical extension E/F. Our main result is

Theorem 4: Let E/F be a radical Galois extension with Galois group G. Then G is a solvable group.

Proof: Let us utilize the preceding Remark and write E = F(₁,...,_t) where ₁ⁿ F, _iⁿ F(₁,...,_i-1) (i = 2,...,t). Let be a primitive nth root of unity and let E' = E(). Since E/F is a Galois extension, E is obtained from F by adjoining all roots of a finite collection of polynomials belonging to F[X]. In order to get E' from E, we adjoin all roots of the polynomial Xⁿ - 1 F[X]. Therefore, E' is obtained by adjoining to F all the roots of a finite collection of polynomials belonging to F[X], and hence E'/F is a Galois extension. Let H = Gal(E'/F), and set

E₀ = F,   E₁ = F(),

E_i = F(,₁,...,_i-1)   (i = 2,...,t+1).

Then

F = E₀ ... E_t+1 = E'.

By the fundamental theorem of Galois theory, to the chain of intermediate fields (1), there corresponds a chain of subgroups of H:

H = H₀ H₁ ... H_i+1 = {1},

where H_i = Gal(E'/E_i). Let us show that the chain (2) of subgroups of H is a normal series with abelian factors. Indeed, E₁ is the splitting field over E₀ of the polynomial Xⁿ - 1. Therefore, E₁/E₀ is a Galois extension with abelian Galois group. Therefore, by Theorem 8 of the section of the fundamental theorem of Galois theory, H₁ H₀ and Gal(E₁/E₀) = H₀/H₁ is abelian. Next, note that E_i(i > 1) contains the nth roots of unity and E_i+1 = E_i(_i), where _i is a zero of Xⁿ - a_i, where a_i = _iⁿ E_i. By Example 5 of the section on the Galois group of a polynomial, E_i+1/E_i(i > 1) is a Galois extension with abelian Galois group. Therefore, by Theorem 8 of the section on the fundamental theorem of Galois theory, H_i+1 H_i and Gal(E_i+1/E_i) = H_i/H_i+1 (i = 1,2,...,t) is abelian. Thus we have proved that (2) is a normal series for H having abelian factors. Therefore, H is a solvable group. It is no easy to show that G is solvable. We have

F E E',  G = Gal(E/F),  H = Gal(E'/F).

Let J = Gal(E'/E). By the fundamental theorem of Galois theory, J < H. But E/F is a Galois extension, so that by Theorem 8 of the section on the fundamental theorem, J H and G = Gal(E/F) = H/J. But then G is a quotient of the solvable group H and hence is solvable by Corollary 9 of the section on solvable groups.

Corollary 5: Let f F[X] be a nonconstant polynomial. If f is solvable by radicals, then Gal_F( f ) is solvable.

Proof: If f is solvable in radicals, then there exists a radical extension E of F such that

F E_f E.

Without loss of generality, by Proposition 3, we may assume that E/F is a Galois extension. Let H = Gal(E/F). Since E_f /F is a normal extension, E_f /F is a Galois extension. Le G = Gal(E/E_f ). Then by Theorem 8 of the section on the fundamental theorem, we have G H and

But H is solvable by Theorem 4, so that H/G is solvable. Thus, Gal_F(f ) is solvable.

Corollary 6: There exist fifth-degree polynomials in Q[X] which are not solvable in radicals. Thus, the quintic (fifth-degree) equation has no general solution in radicals.

Proof: We saw in the section on the Galois group of a polynomial that f = X⁵ - 6X + 3 has the property that Gal_Q( f ) = S₅. If f is solvable in radicals, then, by Corollary 5, S₅ is solvable. But Since A₅ is a simple group (Theorem 9 of symmetric groups) a composition series for S₅ is given by

S₅ A₅ {1},

and the factors are

S₅/A₅ Z₂,    A₅/{1} A₅.

But A₅ is nonabelian, so that S₅ is not solvable. Therefore, f is not solvable in radicals.

We have seen that if a polynomial f is solvable in radicals, then Gal_F(f ) is solvable. In the remaining part of this section we will prove the converse. Actually we will accomplish considerably more. If Gal_F(f ) is solvable, we will actually give a method for constructing a radical extension containing E_f . In particular, this will lead us to an expression of the zeros of f in terms of radicals. When the procedure of this section is applied to polynomials of degrees 2,3,or 4 we will get the formulas which were mentioned in the introduction. In order to carry out our program, it will be necessary to study in some detail the structure of a Galois extension E/F of prime degree p where F contains the pth roots of unity. Our main result will be

Theorem 7: Let F be a field and let E/F be a Galois extension of prime degree p. Assume that F contains the pth roots of unity. Then there exists a F such that E = F(), where is a zero of X^p - a.

The proof of Theorem 7 is accomplished via a trick going back to Lagrange. Let be a primitive pth root of unity, an arbitrary pth root of unity. Then is of the form ^a (a = 0,1,...,p - 1). Let E. Since E/F is a Galois extension of prime order p, Gal(E/F) is cyclic of degree p. Let be a generator of Gal(E/F). Then

Gal(E/F) = {1, , ², ..., ^p-1}.

Let us define the Lagrange resolvent <,> by

<,> = + + ... + ^p-1p-1.

It is clear that <,> E. We may restate Theorem 7 as follows:

Theorem 8: Let F be a field and let E/F be a Galois extension of prime degree p. Assume that F contains the pth roots of unity, and let E - F. Then there exists a pth root of unity such that

(1) <,>^p F,

(2) E = F(<,>).

It is clear that Theorem 8 implies Theorem 7, upon setting a = <,>^p.

Proof: First, let us show that a least one of the Lagrange resolvents <^a,> (a = 1,...,p - 1) is nonzero. Let us assume the contrary. Then

<1,> = ^-a<^a,>

= ^-aavv

= ^v(v-1)a.

If is a primitive pth root of unity, then

0 = ^p - 1 = ( - 1)(^p-1 + ^p-2 + ... + 1)

^p-1 + ^p-2 + ... + 1 = 0.

Let 0 < v < p - 1, v1 and set = ^v-1. Then is a primitive pth root of unity, so that by (4),

^a = ^a(v-1) = 0.

Therefore, by (3),

<1,> = p · ,

= p^-1<1,>.

Note that

<1,> = + + ² + ... + ^p-1.

Therefore,

<1,> = + ² + ... +

= <1,>,

and thus ^a<1,> = <1,> (a = 1,...,p - 1), so that <1,> is left fixed by every element of Gal(E/F). We therefore deduce from the fundamental theorem of Galois theory, that <1,> F. But, by (5), this implies that F. And every element of F is invariant under ^-1, so that = F. But this contradicts the hypothesis that E - F. Finally, we can conclude that at least one of the Lagrange resolvents <^a,> (a = 1,...,p - 1) is nonzero. Let = ^a, where 1 < a < p - 1 is chosen so that <,> 0.

Let us now calculate the effect of on <,>. From the definition of <,>, we have

<,> = ( + + ²² + ... + ^p-1p-1)

= + ² + ²³ + ... + ^p-1p

= ^-1( + ²² + ³³ + ... + ^p)

= ^-1<,>,

where we have used the fact that F and therefore () = . Therefore, since ^p = 1,

(<,>^p) = <,>^p,

so that <,>^p is left fixed by all elements of Gal(E/F) = {1,,...,^p-1}. But this implies that <,>^p F, hence (1) holds.

Since 1, <,> = ^-1<,> <,>. Therefore, <,> is not left fixed by , and <,> F by the fundamental theorem of Galois theory. Thus, F(<,>) F. But, since deg(E/F = p, a prime, and F F(<,>) E, we see that deg(F(<,>)/F) = 1 or p. In the former case, F(<,>) = F, which is a contradiction. Therefore, deg(F(<,>)/F) = p, from which it follows that deg(E/F(<,>)) = 1, and thus E = F(<,>).

Let us apply Theorem 7 to prove

Theorem 9: Let f F[X] be a nonconstant polynomial such that Gal_F(f ) is solvable. Then f is solvable in radicals.

Proof: Let deg(E_f /F) = n and let be a primitive nth root of unity. Set E_f() = E', F() = F'. The relationship between the fields E_f, F, E', and F' is illustrated in figure 2, where a line connecting the two fields denotes a containment relation. It suffices to show that E'/F is a radical extension. For then, since F E_f E', we see that f is solvable in radicals. Note that E' is obtained from F by adjoining all zeros of f and Xⁿ - 1, and thus E'/F is a Galois extension. Let G = Gal(E'/F) and let H be the subgroup of G corresponding to E_f under the Galois correspondence. Then Gal(E'/E_f ) = H. Moreover, since E_f /F is a Galois extension, HG and Gal(E_f /F) = G/H. But Gal(E_f /F) = Gal_F(f ) is solvable. Therefore, G/H is solvable. Moreover, by Example 5 of the section the Galois group of a polynomial, H = Gal(E_f()/E_f ) is abelian and hence solvable. Thus, since H and G/H are solvable, G is solvable. Let J be the subgroup of G corresponding to F' under the Galois correspondence. Then J = Gal(E'/F') and J is a subgroup of a solvable group and hence is solvable. For the relationship between the various Galois groups we have defined, see Figure 2.

Since J is solvable, there exists a composition series

J = J₀ J₁ ... J_t = {1}

such that J_i/J_i+1 is a cyclic group of prime order p_i (0 < i < t - 1). Let F_i be the fixed field of J_i. Then J_i = Gal(E'/F_i) and

F' = F₀ F₁ ... F_t = E'.

Moreover, since J_iJ_i+1, we see that F_i+1/F_i is a Galois extension with Galois group J_i/J_i+1. Thus, F_i+1/F_i is a Galois extension of prime degree p_i. Let us now show that F_i contains the p_ith roots of unity. This will allow us to apply Theorem 7 to the extension F_i+1/F_i.

If Gal(E'/F'), then the restriction of to E is an F-automorphism of E, that is, an element of Gal(E/F) [because E' = E(), F' = F()]. Therefore, let us define the function

:Gal(E'/F') Gal(E/F),

by () = the restriction of to E ( Gal(E'/F')). It is trivial to check that is an isomorphism. Therefore, J = Gal(E'/F') is isomorphic to a subgroup of G/H = Gal(E_f /F). In particular, since p_i divides the order of J and since G/H has order n, we see that p_i|n (0 < i < t - 1). If is a p_ith root of unity, then ^p_i = 1, so that ⁿ = 1 (since p_i|n). Therefore, every p_ith root of unity is an nth root of unity. But F' = F() F_i (i = 0,...,t) and is a primitive nth root of unity. Therefore, F_i (i = 1,...,t) contains the p_ith roots of unity, as asserted.

We may now apply Theorem 7 to each of the extensions F_i+1/F_i (0 < i < t - 1). We see that there exists _i+1 F_i+1 such that (1) F_i+1 = F_i(_i+1) and (2) _i+1 is a zero of a polynomial of the form X^p_i - a_i+1 (a_i+1 F_i). Thus, we derive that

E' = F_t = F(,₁,...,_t)

and

ⁿ = 1 F,   ₁^p_i F(),

= a_i+1 F_i = F(,₁,...,_i)

Thus, E' is a radical extension of F.

Theorem 10: Let f F[X] be a nonconstant polynomial of degree at most 4. Then f is solvable in radicals.

Proof: By Corollary 5 of the section on the Galois group of a polynomial, Gal_F(f ) is a subgroup of S_n (n < 4). Since a subgroup of a solvable subgroup is solvable, Theorem 9 implies that it suffices to show that S₁, S₂, S₃ and S₄ are solvable groups. This is obvious for S₁ and S₂ since these groups are abelian. A composition series for S₃ is

S₃A₃{1}

and the composition factors are S₃/A₃ Z₂, A₃/{1} Z₃. Thus, S₃ is solvable. A composition series for S₄ is given by

S₄A₄{(1),(12)(34),(13)(24),(14)(23)}{(1),(12)(34)}{1}.

The composition factors are isomorphic to

Z₂, Z₃, Z₂, Z₂.

Thus, S₄ is solvable.