### 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

f = Xn + aXn-1 + ... + an,

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,...,an 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 1m1 F, 1m1 F(1,...,i-1) (i = 2,3,...,t) for some integers m1,m2,...,mt.

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

1m1 F 1m = (1m1)k1 F,
imi F(1,...,i-1)
im = (imi)ki F(1,...,i-1)  (i = 2,...,t).

Thus, if we replace each mi by m, we see that in the definition of a radical extension, we may assume that all mi 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 Ef 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 Ef = F(1,...,n), where 1,...,n are the zeros of f. Therefore, to say that Ef 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 1m1 F, 1m1 F(1,...,i-1) ( i = 2,...,t ). Set imi = ai and let for each i(1 < i < t), aij(1 < j < ki) be the conjugates of ai over F. Let E' be the smallest subfield of F containing F and all the zeros of the polynomials Xmi - aij (1 < i < t, 1 < j < ki). 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, aij = ai hence ai 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(),
Ei = F(,1,...,i-1)   (i = 2,...,t+1).

Then

(1)
F = E0 ... Et+1 = E'.

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

(2)
H = H0 H1 ... Hi+1 = {1},

where Hi = Gal(E'/Ei). 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 Ei(i > 1) contains the nth roots of unity and Ei+1 = Ei(i), where i is a zero of Xn - ai, where ai = in Ei. By Example 5 of the section on the Galois group of a polynomial, Ei+1/Ei(i > 1) is a Galois extension with abelian Galois group. Therefore, by Theorem 8 of the section on the fundamental theorem of Galois theory, Hi+1 Hi and Gal(Ei+1/Ei) = Hi/Hi+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 GalF( f ) is solvable.

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

F Ef E.

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

GalF(f ) = Gal(Ef /F) = H/G.

But H is solvable by Theorem 4, so that H/G is solvable. Thus, GalF(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 GalQ( 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 GalF(f ) is solvable. In the remaining part of this section we will prove the converse. Actually we will accomplish considerably more. If GalF(f ) is solvable, we will actually give a method for constructing a radical extension containing Ef . 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 Xp - 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,>
(3)
= -aavv
= v(v-1)a.

If is a primitive pth root of unity, then

0 = p - 1 = ( - 1)(p-1 + p-2 + ... + 1)
(4)
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 · ,
(5)
= 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 GalF(f ) is solvable. Then f is solvable in radicals.

Proof: Let deg(Ef /F) = n and let be a primitive nth root of unity. Set Ef() = E', F() = F'. The relationship between the fields Ef, 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 Ef 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 Ef under the Galois correspondence. Then Gal(E'/Ef ) = H. Moreover, since Ef /F is a Galois extension, HG and Gal(Ef /F) = G/H. But Gal(Ef /F) = GalF(f ) is solvable. Therefore, G/H is solvable. Moreover, by Example 5 of the section the Galois group of a polynomial, H = Gal(Ef()/Ef ) 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.

Figure 2: The subfields of E' and their corresponding Galois Groups.

Since J is solvable, there exists a composition series

J = J0 J1 ... Jt = {1}

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

F' = F0 F1 ... Ft = E'.

Moreover, since JiJi+1, we see that Fi+1/Fi is a Galois extension with Galois group Ji/Ji+1. Thus, Fi+1/Fi is a Galois extension of prime degree pi. Let us now show that Fi contains the pith roots of unity. This will allow us to apply Theorem 7 to the extension Fi+1/Fi.

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(Ef /F). In particular, since pi divides the order of J and since G/H has order n, we see that pi|n (0 < i < t - 1). If is a pith root of unity, then pi = 1, so that n = 1 (since pi|n). Therefore, every pith root of unity is an nth root of unity. But F' = F() Fi (i = 0,...,t) and is a primitive nth root of unity. Therefore, Fi (i = 1,...,t) contains the pith roots of unity, as asserted.

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

E' = Ft = F(,1,...,t)

and

n = 1 F,   1pi F(),
= ai+1 Fi = F(,1,...,i)
(1 < i < t - 1).

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, GalF(f ) is a subgroup of Sn (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.