
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 = X^{n} + aX^{n1} + ... + a_{n},
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 a_{1},...,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},...,_{i1}) (i = 2,3,...,t) for some integers m_{1},m_{2},...,m_{t}.
Remark: In the definition of a radical extension, let m denote the least common multiple of m_{1}, m_{2},...,m_{t} Then for each i(1 < i < t), m = k_{i}m_{i} for some positive integer k_{i}. Therefore,
_{1}^{m1} F _{1}^{m} = ( _{1}^{m1}) ^{k1} F,
_{i}^{m} = ( _{i}^{mi}) ^{ki} F( _{1},..., _{i1}) (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},...,_{i1}) ( i = 2,...,t ). Set _{i}^{mi} = 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^{mi}  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 _{1}^{n} F, _{i}^{n} F(_{1},...,_{i1}) (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 X^{n}  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
E _{0} = F, E _{1} = F( ),
E _{i} = F( , _{1},..., _{i1}) (i = 2,...,t+1).
Then
(1)
F = E _{0} ... 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:
(2)
H = H _{0} H _{1} ... 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, E_{1} is the splitting field over E_{0} of the polynomial X^{n}  1. Therefore, E_{1}/E_{0} is a Galois extension with abelian Galois group. Therefore, by Theorem 8 of the section of the fundamental theorem of Galois theory, H_{1} H_{0} and Gal(E_{1}/E_{0}) = H_{0}/H_{1} 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 X^{n}  a_{i}, where a_{i} = _{i}^{n} 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
Gal_{F}(f ) = Gal(E_{f }/F) = H/G.
But H is solvable by Theorem 4, so that H/G is solvable. Thus, Gal_{F}(f ) is solvable.
Corollary 6: There exist fifthdegree polynomials in Q[X] which are not solvable in radicals. Thus, the quintic (fifthdegree) equation has no general solution in radicals.
Proof: We saw in the section on the Galois group of a polynomial that f = X^{5}  6X + 3 has the property that Gal_{Q}( f ) = S_{5}. If f is solvable in radicals, then, by Corollary 5, S_{5} is solvable. But Since A_{5} is a simple group (Theorem 9 of symmetric groups) a composition series for S_{5} is given by
S _{5} A _{5} {1},
and the factors are
S _{5}/A _{5} Z_{2}, A _{5}/{1} A _{5}.
But A_{5} is nonabelian, so that S_{5} 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}, ..., ^{p1}}.
Let us define the Lagrange resolvent <,> by
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
(3)
If is a primitive pth root of unity, then
0 = ^{p}  1 = (  1)( ^{p1} + ^{p2} + ... + 1)
(4)
^{p1} + ^{p2} + ... + 1 = 0.
Let 0 < v < p  1, v1 and set = ^{v1}. Then is a primitive pth root of unity, so that by (4),
Therefore, by (3),
(5)
Note that
<1, > = + + ^{2} + ... + ^{p1}.
Therefore,
= <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
where we have used the fact that F and therefore () = . Therefore, since ^{p} = 1,
so that <,>^{p} is left fixed by all elements of Gal(E/F) = {1,,...,^{p1}}. 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 X^{n}  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.
Figure 2: The subfields of E' and their corresponding Galois Groups.
Since J is solvable, there exists a composition series
J = J _{0} J _{1} ... 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' = F _{0} F _{1} ... F _{t} = E'.
Moreover, since J_{i}J_{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_{i}th 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 Fautomorphism 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_{i}th root of unity, then ^{pi} = 1, so that ^{n} = 1 (since p_{i}n). Therefore, every p_{i}th 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_{i}th 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^{pi}  a_{i+1} (a_{i+1} F_{i}). Thus, we derive that
E' = F _{t} = F( , _{1},..., _{t})
and
^{n} = 1 F, _{1}^{pi} F( ),
= a _{i+1} F _{i} = 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, 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 S_{1}, S_{2}, S_{3} and S_{4} are solvable groups. This is obvious for S_{1} and S_{2} since these groups are abelian. A composition series for S_{3} is
S _{3}A _{3}{1}
and the composition factors are S_{3}/A_{3} Z_{2}, A_{3}/{1} Z_{3}. Thus, S_{3} is solvable. A composition series for S_{4} is given by
S _{4}A _{4}{(1),(12)(34),(13)(24),(14)(23)} {(1),(12)(34)} {1}.
The composition factors are isomorphic to
Z_{2}, Z_{3}, Z_{2}, Z_{2}.
Thus, S_{4} is solvable.
