Theory of Galois

You all know that the roots of ax2 + bx + c (a,b,c complex numbers), can be found by
x = (-b + w2(b2 - 4ac))/(2a) (where w2(D) denotes any complex number whose square is D).

Since medieval times, European mathematicians have been searching for likewise solutions by radicals of
anxn + an-1xn-1 + ... + a1x + a0 = 0 (ak real or complex, n>2),
ie expressions using ak (k=0,1,..,n-1,n) and rootsymbols wm (where wm(D) denotes any complex number whose m-th power is D).

In the sixteenth century, Italian mathematicians (Tartaglia, Cardano, Ferrari, del Ferro) solved by radicals both
ax3 + bx2 + cx + d = 0, and
ax4 + bx3 + cx2 + dx + e = 0.
Almost three centuries later, Niels Hendrik Abel found that a general solution by radicals is impossible if n=5.

Shortly afterwards, Evariste Galois invented his famous theory, and published it before he died in a fight, 21 years old.
He proved that the general solution by radicals is not possible if n>4.
We will try to have a view on today's shape of his ideas that founded group theory.

We shall first introduce some concepts and ideas of Galois-theory with an example.
It must be emphasized that this lecture does not embark on any existence problem.

Example: x3 - 2

Let w3(2) be the cubic root of 2, and eps=(-1+w2(3)*i)/2, so 1 + eps + eps2 = 0.
Let f(x) = x3 - 2 = (x - w3(2))(x - eps*w3(2))(x - eps2*w3(2)).

Let K be the field generated by the rationals and the coefficients of  f, so K = Q.
Let L be the field generated by K and the roots of  f, so L = Q(w3(2),eps*w3(2),eps2*w3(2)) = Q(w3(2),eps). L is called the splitting field of  f.
It is a six-dimensional vectorspace over Q, with base 1, eps, w3(2), eps*w3(2), w3(4), eps*w3(4).
Furthermore, the group G of all K-automorphisms of L (i.e. leaving K pointwise invariant), called the Galois group of  f, contains six elements as well:

e, mapping w3(2) to w3(2) and eps to eps,
s, mapping w3(2) to eps*w3(2) and eps to eps,
s2, mapping w3(2) to eps2w3(2) and eps to eps,
t, mapping w3(2) to w3(2) and eps to eps2,
t*s = s2*t, mapping w3(2) to eps2*w3(2) and eps to eps2,
t*s2 = s*t, mapping w3(2) to eps*w3(2) and eps to eps2.

This group coincides with the group S3 of permutations of w3(2), eps*w3(2) and eps2*w3(2), the roots of  f.

Notice that {e, s, s2} is a normal subgroup of G, leaving Q(eps) pointwise invariant. It is the group of Q(eps)-automorphisms of L.
Q(eps) is called the field of invariants of {e, s, s2}.
Furthermore, G/{e, s, s2} = {e, t} is the group of Q-automorphisms of Q(eps).

There are also three subgroups of G of order 2:
i) {e, t}, with field of invariants Q(w3(2));
ii) {e, s2*t}, with field of invariants Q(eps*w3(2));
iii) {e, s*t}, with field of invariants Q(eps2*w3(2)).

The following remark is very important:
There exists a bijective mapping from the set of subgroups of G onto the set of subfields of L, reversing the inclusion-relation. For instance:
Q < Q(w3(2) < Q(eps,w3(2)) = L,
G > {e,t} > {e}.

Another example: (x5 - 1)/(x - 1)

Let f(x) = (x5 - 1)/(x - 1) = 1 + x + x2 + x3 + x4 = (x - c)(x - c2)(x - c3)(x - c4),
where c = cos(2*pi/5) + i * sin(2*pi/5), a primitive 5-th root of unity.
Let K:=Q and L:=K(c,c2,c3,c4) = Q(c).
L is the splitting field of  f. It is a vectorspace over Q, with base 1,c,c2, c3.
Its Galois group contains also 4 elements:
e, mapping c onto c;
s, mapping c onto c2;
s2, mapping c onto c4;
s3, mapping c onto c3.
In this case, the Galois-group G is a (cyclic) subgroup of S4.
{e,s2} is the only proper subgroup of it.
It is a normal subgroup, leaving Q(d) pointwise invariant, where d:=c + c4 = cos(2*pi/5).
We have Q < Q(d) < Q(c) and G > {e,s2} > {e}.
Furthermore, G/{e,s2} is the group of Q-automorphisms of Q(d).

Speaking more generally:
Let f(x) be any polynomial with complex coefficients, of degree n and without multiple roots.
Let K be the field generated by Q and the coefficients of  f; let L be the splitting field of  f.
Then there exists a group of K-automorphisms of L, of which K is the field of invariants.
It is a subgroup of the group Sn of permutations of the roots of   f, containing as many elements as a base of L over K.
Furthermore, there exists a bijective mapping from the set of subfields of L containing K onto the set of subgroups of G, reversing the inclusion-relation and mapping each subgroup onto its field of invariants.

Outline of Galois' proof

Now suppose that every zero z of such a polynomial can be expressed in its coefficient by radicals.
This means that there exists a row of field extensions K=K0 < K1 < K2 < ... < Kr=L, where Ki = Ki-1(oi) for some oi in Ki\Ki-1, with oipi belonging to Ki-1 for some prime number pi
(so oi = o'i-11/pi for some o'i-1 in Ki-1, notice that o'0 can be expressed rationally in the coefficients of   f).

Then the Galois group G of f(x) must be solvable:
there is a chain G=G0 > G1 > G2 > ... > Gr={e},
where Gi is a normal subgroup of Gi-1, and Gi/Gi-1 is cyclic of prime order pi.

Since Sn is not solvable if n>4 and for any n there exist polynomials of degree n with Galois group Sn (this is true for example if the zeroes are algebraically independent over Q), we find the famous result of Abel and Galois.

Example: the general polynomial equation of degree 4

We embark to the problem of solving by radicals x4 + p*x2 + q*x + r = 0.
Let t1, t2, t3, t4 be its complex roots.
Since x4 + p*x2 + q*x + r = (x - t1)(x - t2)(x - t3)(x - t4), we find:
0 = t1 + t2 + t3 + t4
p = t1t2 + t1t3 + t1t4 + t2t3 + t2t4 + t3t4
-q = t1t2t3 + t1t2t4 + t1t3t4 + t2t3t4
r = t1t2t3t4.

Note that these expressions, called the elementary symmetric expressions in t1, t2, t3, t4, are invariant under whatever permutation of t1, t2, t3, t4 is applied to them.
Furthermore, any expression that is invariant under every permutation of t1, t2, t3, t4, belongs to the field K=Q(p,q,r) of invariants of the Galois-group of the equation.
Therefore, such an expression can be written as a rational function of the elementary symmetric expressions that equal 0, p, -q and r.
For instance, (t1t2 + t3t4)(t2t3 + t1t4)(t1t3 + t2t4) = q2 - 4pr.

Let L:=K(t1,t2,t3,t4) be the splitting field of the polynomial.
Suppose that p,q,r are such that the Galois group, being the group of K-automorphisms of L, coincides with the group S4 of permutations of t1,t2, t3,t4.

We shall, to begin with, take a look at S4.
Its elements can be written as products of cycles. For instance (123) denotes the permutation which maps t1 to t2, t2 to t3, t3 to t1, and t4 to itself.
Again, (12)(34) maps t1 to t2 and vice versa, t3 to t4 and vice versa.

Notice that S4 has a beautiful chain of subgroups: S4 > A4 > V4 > C2 > {e}.
Here A4 is the subgroup of those permutations that can be written as a product of an even number of cycles of length 2, V4 = {e,(12)(34),(13)(24),(14)(23)} is the group of Klein, and C2 = {e,(12)(34)}.
A4 is a normal subgroup of S4, and S4/A4 is cyclic of prime order 2.
V4 is a normal subgroup of A4, and A4/V4 is cyclic of prime order 3.
C2 is a normal subgroup of V4, and V4/C2 is cyclic of prime order 2.
Finally, {e} is a normal subgroup of C2, and C2/{e} is cyclic of prime order 2.
So S4 is solvable.

Let A be the field of invariants of A4, V of V4, and Z of C2. Then we have:
K < A < V < Z < L and S4 > A4 > V4 > C2 > {e}.
So C2 is the group of Z-automorphisms of L, V4/C2 the group of V-automorphisms of Z, A4/V4 the group of A-automorphisms of V, and S4/A4 the group of K-automorphisms of A.

Now we proceed step by step as follows:

1) By our assumptions, t1, t2, t3, t4 belong to L\Z, but t1 + t2, t1t2, t3 + t4, t3t4 are invariant under (12)(34), so under C2, and therefore belong to Z. When we have found a1:=t1 + t2, a2:= t1t2, a3:=t3 + t4, a4:= t3t4, then we find t1 and t2 from x2 - a1x + a2 = 0, and t3 and t4 from x2 - a3x + a4 = 0, expressed by square roots in a1, a2, a3, a4 of Z.

2) In the same way, a1 + a3, a1a3, a2 + a4 and a2a4 are invariant under (13)(24), so under V4/C2, and hence belong to V. When we have found them, we find a1, a2, a3 and a4, being roots of x2 - (a1+a3)x + a1a3 = 0 and x2 - (a2+a4)x + a2a4 = 0.

3) Now, for instance, let us see how we can find a2 + a4 = t1t2 + t3t4.
Let b1 := t1t2 + t3t4, b2 := t2t3 + t1t4, b3 := t1t3 + t2t4;
let c1 := b1 + b2 + b3, c2 := b1b2 + b1b3 + b2b3, c3 := b1b2b3.
Then c1, c2, c3 are invariant under (123)=(12)(23), so under A4/V4, and thus belong to A. Hence b1, b2, b3 can be expressed by cubic and square roots in c1, c2 and c3 of A, being roots of x3 - c1x2 + c2x - c3 = 0, by the formulas of Cardano for the cubic equation.

4) Fortunately, c1, c2, c3 themselves belong not only to A, but even to K (this happens because the coefficient of x3 in our quartic polynomial is 0 and thus not general). You can find: c1=p, c2=-4r, c3 = q2 - 4pr.

Example: regular polygons

A regular polygon with n sides can be constructed by circles and straight limes if n is prime and n-1 a power of two. This can be seen as follows:

In this case we find that z:=exp(2*pi*i/n) is a root of the irreducible polynomial 1 + x + x2 + ... + xn-2 + xn-1.
L=Q(z) and G is cyclic of order 2m with m=n-1.
So there exists a chain of groups {e]=G0 < G1 < ... < Gm=G and a chain of fields Q=K0 < K1 < ... Km=L where Ki+1 has dimension 2 over Ki.
Since square roots can be constructed by circles and straight lines, we have finished the proof. (Draw a circle with diameter a+1; draw a perpendicular on the diameter, at distance (a-1)/2 from the center; the intersection points with the circle are 2*w2(a) apart.)
For example, we can construct z with n=5 if we can construct z+z4. Since (x - (z+z4))(x - (z2+z3)) = x2 + x + 1, we find z+z4 = (-1+w2(5))/2.

I seriously hope that you don't believe everything. Study Galois theory.