How Galois did it,
but first a little about Galois theory.
Modern
algebra deals with all the numbers that can be obtained from some given
initial collection using addition, subtraction, multiplication and division.
The resulting collection is called a field. A field is a set , which
may be finite or infinite, that has two distinct but closely related group
structures on it. Another way of saying this is that a field is a nontrivial
commutative ring in which all the elements, apart from zero, have an
inverse. The
most common examples are the rational numbers Q, the real numbers R,
and the complex numbers C. Each of these have group structures
corresponding to the operations of addition and multiplication. The two
operations are related in that multiplication is required to be
"distributive" with respect to addition, i.e. a(a+b)=aa+ab. Galois
theory is essentially the study of fields. In Galois theory, the
primary object of interest is the polynomial equation in one variable: x +an-1xn-1
+...+ a1x + a0 = 0
where the coefficients {ak }
are Given
this, not all permutations of the roots of a polynomial may be reasonable,
because they don't induce an automorphism of the extension field which leaves
the base field fixed. This may happen if there are polynomial relationships
among the roots with coefficients in the base field. For
instance, in the polynomial
f(x) =
(x-i)(x+i)(x-2i)(x+2i) = x4 + 5x2 + 4 the roots are x1 = i, x2 =
-i, x3 = 2i, x4= -2i. We have the relations x3
= 2 x1 and x4 = 2 x2. We can
allow permutations that exchange x1 and x2,
or x3 and x4, or both. But we can't allow
a permutation that exchanges x1 and x3.
Because the resulting field automorphism T would require T( x1)
= x3 = 2 x1 = 2T( x3) = T(2 x3) =
T(4 x1) = 4T( x1). The
set of "reasonable" permutations thus generates a set of
automorphisms of the extension field that leaves the base field fixed. This
set of automorphisms is actually a group, and it is called the Galois
group of the extension. The Galois group is a way of encoding all
available information about the relationships of the roots of polynomials
with coefficients in the base field that factor completely in the extension
field. So in order to study all roots of a given polynomial, it is sufficient
to find an extension field that contains all of the roots and examine the
Galois group. Notice
that we have managed to express one kind of mathematical problem -
description of the roots of a polynomial equation - in terms of a symmetry
group, where the symmetry in question involves permutations among the roots.
Here again, symmetry operations can be used to express a concept of
"similarity" or "likeness". In this case, certain roots
of an equation are "like" others because they satisfy the same
algebraic relations, even though they are not numerically the same. But for
all algebraic purposes they are interchangeable. For
future reference, we will simply state the fundamental facts of Galois
theory. We say that a (finite) field extension E In
1824 Abel showed that there is no formula for the roots of a general
quintic in terms of radical expressions. Abel died in 1829 working on this
problem. Galois´ solution of the cubic equation. Lets
consider the cubic equation f(x) = x3 +ax2 + bx + c
= 0. f(x - a/3) =
x3 + px + q where p =b - 2/3a and q = c + (2a3 -9ab)/27. We shall, in order to avoid
trivialities assume that x3 + px + q is irreducible in F(x).
Let the zeros of x3
+ px + q = 0 be w1, w2, and w3:
we have + (w1w2+w1w3+w2w3)x - w1w2w3 Thus S wi=
0, S (wiwj)
= p and w1w2w3 = - q. If E is a
splitting field of x3 +
px + q over F, E = F(w1,w2,w3),
then [E:F] =3 or 6 and
so G = G (E/F) isomorphic to a
subgroup of isomorphic to a subgroup of S 3 ; hence In any event G has a subgroup A = isomorphic to A3 which is
a normal subgroup of G . Since A is a normal subgroup of G there is
To find u such that K
= F(u) we seek an element of E which is left fixed by A3 ={i ,(
w1,w2,w3),(w1,w2,w3)}, but not necessarily by all of G. One such element is Note that for any permutation s of { w1,
w2, w3} we have
that s(D) = + D D2 is
called he discriminant of a cubic whose zeroes are (w1,
w2, w3). Check to see that if s Î A then s(D) = D,
whereas if s is any
other permutation of To evaluate D we should naturally evaluate D2. To do this we shall f´(x)= = Pi<j (x –
wi) (x – wj), and so D2 = - f´( w1)f´( w2)f´(
w3) = - (3w12+ p) (3w22+ p) (3w32+ p) = The terms in this expression are
clearly left invariant by every permutation of {w1,
w2, w3} and
hence are fixed by G
and so belong to F. It will be shown in the next section that such expressions
always can be written as combinations of coefficients of the cubic of which
they are zeros. Here the calculations We have w1w2w3
= - q whereas Sw12
= S(w1)2 - 2(Sw1w2)
= - 2p: and finally, Sw12w22
= (Sw1w2)2
- 2(Sw12
w2w3) = p2 - 2[(Sw1)(
w1 w2w3)] = p2. Hence Hence
D2 = -[27q2+9p3-6p3+p3] = - [27q2+4p3]. Thus
K = F( Thus
K = F(D) is an
extension of degree at most 2, and E is a cyclic extension over
K. In particular if
D Î F then A = G and then [E:F]
=3. In any event we want to study the cyclic extension E
over K. Let s = (w1,w2,w3). To
solve the equation x3
+ px + q = 0 by radicals, its Galois group has to be solvable.
We therefore adjoin to K the roots of unity x,x2,x3 = 1;
x is a
zero of x2 - x - 1 which we chose to
be x = ( -1
+ Ö3i)/2. Let K*
= K(x) = K(Ö3i). In
general K* ¹ K, and this is
certainly the case of F = Ra. We now form the splitting field E* of x3 + px
+ q over K*. We now know this is a cyclic extension of degree 3
and that the cube roots of unity belongs to K*. Now we form the Lagrange resolvents in E*: v1 = l(1,w1)
= w1 + w2 + w3 =
0 We know that v23 and v33
belongs to K* and that w1 = (v1 + v2
+v3)/3 In order to get the formulas for the zeros of the
cubic we now only v23 = (w1 + x w2 + x2w3 ) 3 = w13 + w23 + w33 + 3x( w12w2
+ w22w3+
w1w32) + +3x2(w1w22
+ w2w32+ w12w3)
+ 6 w1w2w3
= = Sw13
- 3/2S(w12w2
+ w12w3)
+ + 3/2Ö3i (w12w2+
w22w3+ w1w32 - w1w22
- w2w32 - w12w
3) + + 6 w1w2w3 = Sw13
- 3/2Sw12
(w2 + w3) +3/2Ö3iD + 6 w1w2w3 Now Sw13
= (Sw1)
3 - 3Sw12
(w2 + w3) - 6 w1w2w3 and Sw12
(w2 + w3)
= (Sw1)(
(Sw1w2)
- 3w1w2w3 Since Sw1
= 0 we have v23
= - 9/2Sw12
(w2 + w3)+
3/2Ö3iD= 27/2w1w2w3
+ 3/2Ö3iD = - 27/2q + 3/2Ö3iD Similarly by replacing x with x2 , we
obtain v33 =
we obtain v33
= - 27/2q - 3/2Ö3iD Hence we may write w1= 1/3[ (- 27/2q
+ 3/2Ö3iD)1/3
+ (- 27/2q - 3/2Ö3iD)1/3]. w2 = 1/2 (xv3 + x2 v2) t1/3 is a multivalued
function and which of the cube roots v23
and v33 are to be used? They have to be
chosen so that v23v33
=(w1 +x w2+
x2w3)(w1
+x2w2
+ x w3)
= (In general it is true that l(x,w) .
l(x-1,w) Î F). As shown this is a
necessary relation but it is also sufficient as can be verified by
testing the solution w1=(v1 + v2)/3 under the condition that v2v3 = - 3p. w13 + pw1 +
q = 1/27 (v23
+ 3v22v3 + 3v2v32 + v33) +
p/3(v2+v3) + q = =1/27 (v23
+ v33) + q = -q + q = 0 And these formulas for w1, w2 , v1,
v2, v3 and
D, together
with the
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F is Galois if E is the field
obtained by adjoining to F all roots of some irreducible polynomial
with coefficients in F. The Galois group of E over F,
Gal(E/F), is the group of automorphisms of E that leave F
fixed (i. e., that map all elements of F to themselves). The
fundamental theorem says that there is a 1:1 correspondence of intermediate
fields E' such that E