From: Beachy/Blair,

Excerpted from Sections 8.1 and 8.4

If your viewer can't handle subscripts, try the old version.

Historically, the problem of solving polynomial equations by radicals
has motivated a great deal of study in algebra.
The quadratic formula gives a solution of the equation
ax^{2} + bx + c = 0, where a is nonzero,
expressed in terms of its coefficients, and using a square root.
More generally, we say that an equation

a_{n}x^{n} + ... + a_{1}x + a_{0} = 0

The solution by radicals of cubic and quartic equations was discovered in the sixteenth century, and from then until the beginning of the nineteenth century, some of the best mathematicians of the period (such as Euler and Lagrange attempted to find a similar solution by radicals for equations of degree five. It is generally agreed that the first correct proof of the insolvability of the quintic was published by Abel in 1826.

Abel attacked the general problem of when a polynomial equation could be solved by radicals. His papers inspired Galois to formulate a theory of solvability of equations involving the structures we now know as groups and fields. Galois worked with fields to which all roots of the given equation had been adjoined. He then considered the set of all permutations of these roots that leave the coefficient field unchanged. The permutations form a group, called the Galois group of the equation. From the modern point of view, the permutations of the roots can be extended to automorphisms of the field, and form a group under composition of functions. Then an equation is solvable by radicals if and only if its Galois group is ``solvable''.

An automorphism f of a field F is a one-to-one correspondence f:F->F such that

f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b),

for all a,b in F. That is, f is an automorphism of the additive group of F, and since f(a) is nonzero whenever a in nonzero, it is also an automorphism when restricted to the multiplicative group of F. We use the notation Aut(F) for the group of all automorphisms of F. The identity element 1 must be left fixed by f, since by elementary results on group homomorphisms we have f(1)=1.
To study solvability by radicals of a polynomial
equation f(x)=0, we let K be the field generated by the
coefficients of f(x).
We let F be an extension field of K
that is generated by the roots of f(x) over K.
This is called the *splitting field* for f(x) over K,
and is unique up to isomorphism.

We know, by Kronecker's theorem, that we can always find roots of a polynomial in some extension field. The question is whether or not the roots have a particular form. Galois considered permutations of the roots that leave the coefficient field fixed. The modern approach is to consider the automorphisms determined by these permutations.

A part of the development of Galois theory
is introduced below, via some of the propositions and theorems
from Chapter 8 of **Abstract Algebra**.

**
Proposition 8.1.1.
**
Let F be an extension field of K.
The set of all automorphisms f:F->F such
that f(a)=a for all a in K is a group under
composition of functions.

**
Definition 8.1.2.
**
Let F be an extension field of K. The set

{ f in Aut(F) | f(a) = a for all a in K }

is called the
**
Definition 8.1.3.
**
Let K be a field, let p(x) belong to K[x],
and let F be a splitting field for p(x) over K.
Then Gal(F/K) is called the
*Galois group* of p(x) over K,
or the *Galois group of the equation p(x) = 0* over K.

**
Proposition 8.1.4.
**
Let F be an extension field of K,
and let p(x) belong to K[x].
Then any element of Gal(F/K) defines a permutation of the
roots of p(x) that lie in F.

**
Theorem 8.1.6.
**
Let K be a field, let p(x) be a polynomial in K[x],
and let F be a splitting field for p(x) over K.
If p(x) has no repeated roots, then
|Gal(F/K)|=[F:K].

The next goal is a deeper study of splitting fields. It can be proved that the following conditions are equivalent for an extension field F of a field K:

- (1) F is the splitting field over K of a polynomial with no repeated roots;
- (2) there is a finite group G of automorphisms of F such that a is in K if and only if f(a)=a for all f in G;
- (3) F is a finite extension of K; the minimal polynomial over K of any element in F has no repeated roots; and if p(x) is an irreducible polynomial in K[x] and has a root in F, then it splits over F.

**
Proposition 8.4.2.
**
Let p(x) be the polynomial
x^{n} - 1,
and let F be the splitting field of p(x) over
a field K of characteristic zero.
Then Gal(F/K) is an abelian group.

*Sketch of the proof:*
Since K has characteristic zero,
the polynomial p(x) has n distinct roots,
and they form a cyclic subgroup C
of the multiplicative group of F.
Every element of Gal(F/K) defines an automorphism of C,
so Gal(F/K) is isomorphic to a subgroup of
the automorphism group of C, which is abelian.

**
Theorem 8.4.3.
**
Let K be a field of characteristic zero that contains
all *n*th roots of unity, let a be in K,
and let F be the splitting field of
x^{n} - a,
over K.
Then Gal(F/K) is a cyclic group whose order is a divisor of n.

Solving a polynomial equation by radicals
corresponds to adjoining *n*th roots,
as in Theorem 8.4.3.
Doing this gives an abelian factor group of the galois group,
which leads to the following definition.

**
Definition 7.6.1.
**
The group G is said to be
* solvable *
if there exists a finite chain of subgroups

G = N_{0} > N_{1} > ... > N_{k} = {e}

- (i) each subgroup in the chain is normal in the one above it, and
- (ii) each factor group determined by successive subgroups in the chain is abelian.

**
Theorem 8.4.6.
**
Let p(x) be a polynomial over a field K of characteristic zero.
The equation p(x)=0 is solvable by radicals if and only if
the Galois group of p(x) over K is solvable.

**
Theorem 8.4.8.
**
There exists a polynomial of degree 5 with rational coefficients
that is not solvable by radicals.

*Sketch of the proof*:
Let
p(x) = x^{5} - 2x^{3} - 8x - 2.
It is easy to check that the derivative of p(x) has two real roots,
corresponding to one relative maximum and one relative minimum.
Since the values of p(x) change sign between -2 and -1,
between -1 and 0, and between 2 and 3, the polynomial has
precisely three real roots.

There exists a splitting field F for p(x)
that is contained in the field * C* of complex numbers.
The polynomial p(x) is irreducible over the rational numbers
by Eisenstein's criterion,
and so adjoining a root of p(x) gives an extension of degree 5.

It follows from theorems in Galois theory and group theory that the Galois group of p(x) over the rationals must contain an element of order 5. Every element of the Galois group of p(x) gives a permutation of the roots, and so the Galois group turns out to be isomorphic to a subgroup of the symmetric group on 5 elements. This subgroup must contain an element of order 5, and it must also contain the transposition that corresponds to the element of the Galois group defined by complex conjugation. It can be shown that any subgroup contains both a transposition and a cycle of length 5 must be equal to all of the symmetric group. Therefore the Galois group of p(x) over the rationals must be isomorphic to the symmetric group on 5 elements.

The proof is completed by showing that this group is not a solvable group. With some hard work it can be shown that in the group the only candidate for a chain of normal subgroups as required in the definition of a solvable group is S

Back to the 420 / 421 home page