Solvability by Radicals
Given the specific polynomial over the field of rational numbers , from the quadratic formula for its roots we know that its roots are ; thus the field is
the splitting field of over . Consequently there is an element in such that the extension field where is such that
it contains all the roots of
From a slightly different point of view, given the general quadratic polynomial over , we can consider it as a particular polynomial over the field of rational functions in the two variables and over ; in the extension obtained by adjoining to where , we find all the roots of . There is a formula which expresses the roots of in terms of and square roots of rational functions of these.
For a cubic equation the situation is very similar; given the general cubic equation an explicit formula can be given, involving combinations of square roots and cube roots of rational functions in . While somewhat messy, they are explicitly given by Cardan's formulas: Let
(with cube roots chosen properly); then the roots of are
where is a cube root of 1. The above formulas only serve to illustrate for us that by adjoining a certain square root and then a cube root to we reach a field in which has its roots.
For fourth-degree polynomials, which we shall not give explicitly, by using rational operations and square roots, we can reduce the problem to that of solving a certain cubic, so here too a formula can be given expressing the roots in terms of combinations of radicals (surds) of rational functions of the coefficients.
For polynomials of degree five and higher, no such universal radical formula can be given, for we shall prove that it is impossible to express their roots, in general, in this way.
Given a field and a polynomial we say that is solvable by radicals over if we can find a finite sequence of fields
such that the roots of all lie in .
If is the splitting field of over then is solvable by radicals over if we can find a sequence of fields as above such that . An important remark, and one we shall use later, in the proof of Theorem 5.7.2, is that if such an can be found, we can, without loss of generality, assume it to be a normal extension of ; we leave its proof as a problem (Problem 1).
By the general polynomial of degree over , we mean the following: Let be the field of rational functions, in the variables over , and consider the particular polynomial over the field . We say that it is solvable by radicals if it is solvable by radicals over . This really expresses the intuitive idea of ``finding a formula'' for the roots of involving combinations of th roots, for various 's, of rational functions in . For , and 4, we pointed out that this can always be done. For Abel proved that this cannot be done. However, this does not exclude the possibility that a given polynomial over may be solvable by radicals. In we shall give a criterion for this in terms of the Galois group of the polynomial. But first we must develop a few purely group-theoretical results. Some of these occurred as problems at the end of Chapter 2, but we nevertheless do them now officially.
DEFINITION: A group is said to be solvable if we can find a finite chain of subgroups
where each is a normal subgroup of and such that every factor group is abelian.
Every abelian group is solvable, for merely take and to satisfy the above definition. The symmetric group of degree 3, , is solvable for take ; is a normal subgroup of and and are both abelian being of orders 2 and 3, respectively. It can be shown that is solvable (Problem 3). For we show in Theorem 5.7.1 below that is not solvable.
We seek an alternate description for solvability. Given the group and elements in , then the commutator of and is the element . The commutator subgroup, , of is the subgroup of generated by all the commutators in . (It is not necessarily true that the set of commutators itself forms a subgroup of ) It was an exercise before that is a normal subgroup of . Moreover, the group is abelian, for, given any two elements in it, with , then
On the other hand, if is a normal subgroup of such that is abelian, then , for, given then
from which we deduce whence and so . Since contains all commutators, it contains the group these generate, namely .
is a group in its own right, so we can speak of its commutator subgroup . This is the subgroup of generated by all elements where . It is easy to prove that not only is a normal subgroup of but it is also a normal subgroup of (Problem 4). We continue this way and define the higher commutator subgroups by . Each is a normal subgroup of (Problem 4) and is an abelian group.
In terms of these higher commutator subgroups of , we have a very succinct criterion for solvability, namely,
LEMMA 5.7.1: is solvable if and only if for some integer .
Proof: If let
each being normal in is certainly normal in . Finally,
hence is abelian. Thus by the definition of solvability is a solvable group.
Conversely, if is a solvable group, there is a chain
where each is normal in and where is abelian. But then the commutator subgroup of must be contained in . Thus
We therefore obtain that .
COROLLARY: If is a solvable group and if is a homomorphic image of then is solvable.
Proof: Since is a homomorphic image of it is immediate that is the image of . Since for some for the same , whence by the lemma is solvable.
The next lemma is the key step in proving that the infinite family of groups , with , is not solvable; here is the symmetric group of degree .
LEMMA 5.7.2: Let where ; then for contains every -cycle of .
Proof: We first remark that for an arbitrary group , if is a normal subgroup of then must also be a normal subgroup of (Problem 5).
We claim that if is a normal subgroup of , where , which contains every 3-cycle in , then must also contain every 3-cycle. For suppose are in (we are using here that ); then
as a commutator of elements of must be in . Since is a normal subgroup of , for any must also be in . Choose a in such that , and where are any three distinct integers in the range from 1 to ; then
is in . Thus contains all 3-cycles.
Letting , which is certainly normal in and contains all 3-cycles, we get that contains all 3-cycles; since is normal in contains all 3-cycles; since is normal in contains all 3-cycles. Continuing this way we obtain that contains all 3-cycles for arbitrary .
A direct consequence of this lemma is the interesting group-theoretic result
THEOREM 5.7.1: is not solvable for .
Proof: If , by Lemma 5.7.2, contains all 3-cycles in for every . Therefore, for any , whence by Lemma 5.7.1, cannot be solvable.
We now interrelate the solvability by radicals of with the solvability, as a group, of the Galois group of . The very terminology is highly suggestive that such a relation exists. But first we need a result about the Galois group of a certain type of polynomial.
LEMMA 5.7.3: Suppose that the field has all th roots of unity (for some particular ) and suppose that is in . Let and let be its splitting field over . Then:
1. where is any root of .
2. The Galois group of over is abelian.
Proof: Since contains all th roots of unity, it contains ; note that but for .
If is any root of , then
are all the roots of . That they are roots is clear; that they are distinct follows from: with then since , and , we must have , which is impossible since , with . Since all of are in thus splits ; since no proper subfield of which contains also contains , no proper subfield of can split . Thus is the splitting field of , and we have proved that .
If are any two elements in the Galois group of , that is, if are automorphisms of leaving every element of fixed, then since both and are roots of and for some and . Thus
similarly, . Therefore, and agree on and on hence all of . But then whence the Galois group in abelian.
Note that the lemma says that when has all th roots of unity, then adjoining one root of to , where , gives us the whole splitting field of ; thus this must be a normal extension of .
We assume for the rest of the section that is a field which contain all th roots of unity for every integer . We have
THEOREM 5.7.2: If is solvable by radicals over then the Galois group over of is a solvable group.
Proof: Let be the splitting field of over ; the Galois group of over is . Since is solvable by radicals, there exists a sequence of fields
where and where . As we pointed out, without loss of generality we may assume that is a normal extension of . As a normal extension of is also a normal extension of any intermediate field, hence is a normal extension of each .
By Lemma 5.7.3 each is a normal extension of and since is normal over , by Theorem 5.6.6, is a normal subgroup in . Consider the chain:
As we just remarked, each subgroup in this chain is a normal subgroup in the one preceding it. Since is a normal extension of , by the fundamental theorem of Galois theory (Theorem 5.6.6) the group of over is isomorphic to However, by Lemma 5.7.3, is an abelian group. Thus each quotient group of the chain (1) is abelian.
Thus the group is solvable! Since and is a normal extension of (being a splitting field), by Theorem 5.6.6, is a normal subgroup of and is isomorphic to . Thus is a homomorphic image of which is a solvable group; by the corollary to Lemma 5.7.1, itself must then be a solvable group. Since is the Galois group of over the theorem has been proved.
We make two remarks without proof.
1. The converse of Theorem 5.7.2 is also true; that is, if the Galois group of over is solvable then is solvable by radicals over
2. Theorem 5.7.2 and its converse are true even if does not contain roots of unity.
Recalling what is meant by the general polynomial of degree over , and what is meant by solvable by radicals, we close the chapter with the great, classic theorem of Abel:
THEOREM 5.7.3: The general polynomial of degree is not solvable by radicals.
Proof: In Theorem 5.6.3 we saw that if is the field of rational functions in the variables then the Galois group of the polynomial over was , the symmetric group of degree . By Theorem 5.7.1, is not a solvable group when , thus by Theorem 5.7.2, is not solvable by radicals over when .