You all know that the roots of ax^{2} + bx + c (a,b,c complex numbers), can be found
by

x = (-b + w2(b^{2} - 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

a_{n}x^{n} + a_{n-1}x^{n-1} + ... +
a_{1}x + a_{0} = 0 (a_{k} real or complex, n>2),

ie expressions using a_{k} (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

ax^{3} + bx^{2} + cx + d = 0, and

ax^{4} + bx^{3} + cx^{2} + 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: x ^{3} - 2 **

Let w3(2) be the cubic root of 2, and eps=(-1+w2(3)*i)/2, so 1 + eps + eps^{2} = 0.

Let f(x) = x^{3} - 2 = (x - w3(2))(x - eps*w3(2))(x - eps^{2}*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),eps^{2}*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,

s^{2}, mapping w3(2) to eps^{2}w3(2) and eps to eps,

t, mapping w3(2) to w3(2) and eps to eps^{2},

t*s = s^{2}*t, mapping w3(2) to eps^{2}*w3(2) and eps to eps^{2},

t*s^{2} = s*t, mapping w3(2) to eps*w3(2) and eps to eps^{2}.

This group coincides with the group S_{3} of *permutations* of w3(2), eps*w3(2) and
eps^{2}*w3(2), the roots of f.

Notice that {e, s, s^{2}} 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, s^{2}}.

Furthermore, G/{e, s, s^{2}} = {__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, s^{2}*t}, with field of invariants Q(eps*w3(2));

iii) {e, s*t}, with field of invariants Q(eps^{2}*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: (x ^{5} - 1)/(x - 1)**

Let f(x) = (x^{5} - 1)/(x - 1) = 1 + x + x^{2} + x^{3} + x^{4} =
(x - c)(x - c^{2})(x - c^{3})(x - c^{4}),

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,c^{2},c^{3},c^{4}) = Q(c).

L is the splitting field of f. It is a vectorspace over Q, with base 1,c,c^{2},
c^{3}.

Its Galois group contains also 4 elements:

e, mapping c onto c;

s, mapping c onto c^{2};

s^{2}, mapping c onto c^{4};

s^{3}, mapping c onto c^{3}.

In this case, the Galois-group G is a (cyclic) *subgroup* of S_{4}.

{e,s^{2}} is the only proper subgroup of it.

It is a normal subgroup, leaving Q(d) pointwise invariant, where d:=c + c^{4} =
cos(2*pi/5).

We have Q < Q(d) < Q(c) and G > {e,s^{2}} > {e}.

Furthermore, G/{e,s^{2}} 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 S_{n} 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=K_{0} < K_{1} <
K_{2} < ... < K_{r}=L, where K_{i} = K_{i-1}(o_{i}) for
some o_{i} in K_{i}\K_{i-1}, with o_{i}^{pi}
belonging to K_{i-1} for some prime number p_{i}

(so o_{i} = o'_{i-1}^{1/pi} for some o'_{i-1} in K_{i-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=G_{0} > G_{1} > G_{2} > ... > G_{r}={e},

where G_{i} is a normal subgroup of G_{i-1}, and
G_{i}/G_{i-1} is cyclic of prime order p_{i}.

Since *S _{n} is not solvable* if n>4 and for any n there exist polynomials of
degree n with Galois group S

** Example: the general polynomial equation of degree 4**

We embark to the problem of solving by radicals x^{4} + p*x^{2} + q*x + r = 0.

Let t_{1}, t_{2}, t_{3}, t_{4} be its complex roots.

Since x^{4} + p*x^{2} + q*x + r =
(x - t_{1})(x - t_{2})(x - t_{3})(x - t_{4}), we find:

0 = t_{1} + t_{2} + t_{3} + t_{4}

p = t_{1}t_{2} + t_{1}t_{3} + t_{1}t_{4}
+ t_{2}t_{3} + t_{2}t_{4} + t_{3}t_{4}

-q = t_{1}t_{2}t_{3} + t_{1}t_{2}t_{4} +
t_{1}t_{3}t_{4} + t_{2}t_{3}t_{4}

r = t_{1}t_{2}t_{3}t_{4}.

Note that these expressions, called the *elementary symmetric expressions* in
t_{1}, t_{2}, t_{3}, t_{4}, are invariant under whatever permutation
of t_{1}, t_{2}, t_{3}, t_{4} is applied to them.

Furthermore, any expression that is invariant under every permutation of t_{1},
t_{2}, t_{3}, t_{4}, 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, (t_{1}t_{2} + t_{3}t_{4})(t_{2}t_{3}
+ t_{1}t_{4})(t_{1}t_{3} + t_{2}t_{4}) =
q^{2} - 4pr.

Let L:=K(t_{1},t_{2},t_{3},t_{4}) 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 S_{4} of permutations of t_{1},t_{2},
t_{3},t_{4}.

We shall, *to begin with*, take a look at S_{4}.

Its elements can be written as products of cycles. For instance (123) denotes the
permutation which maps t_{1} to t_{2}, t_{2} to t_{3},
t_{3} to t_{1}, and t_{4} to itself.

Again, (12)(34) maps t_{1} to t_{2} and vice versa, t_{3} to
t_{4} and vice versa.

Notice that S_{4} has a beautiful chain of subgroups: S_{4} > A_{4} >
V_{4} > C_{2} > {e}.

Here A_{4} is the subgroup of those permutations that can be written as a product of
an even number of cycles of length 2, V_{4} = {e,(12)(34),(13)(24),(14)(23)} is the
group of Klein, and C_{2} = {e,(12)(34)}.

A_{4} is a normal subgroup of S_{4}, and S_{4}/A_{4} is
cyclic of prime order 2.

V_{4} is a normal subgroup of A_{4}, and A_{4}/V_{4} is
cyclic of prime order 3.

C_{2} is a normal subgroup of V_{4}, and V_{4}/C_{2} is
cyclic of prime order 2.

Finally, {e} is a normal subgroup of C_{2}, and C_{2}/{e} is
cyclic of prime order 2.

So *S _{4} is solvable*.

Let A be the field of invariants of A_{4}, V of V_{4}, and Z of C_{2}.
Then we have:

K < A < V < Z < L and S_{4} > A_{4} > V_{4} > C_{2} > {e}.

So C_{2} is the group of Z-automorphisms of L, V_{4}/C_{2} the
group of V-automorphisms of Z, A_{4}/V_{4} the group of A-automorphisms of V,
and S_{4}/A_{4} the group of K-automorphisms of A.

Now we proceed *step by step* as follows:

1) By our assumptions, t_{1}, t_{2}, t_{3}, t_{4} belong to
L\Z, but t_{1} + t_{2}, t_{1}t_{2}, t_{3} + t_{4},
t_{3}t_{4} are invariant under (12)(34), so under C_{2}, and therefore
belong to Z. When we have found a_{1}:=t_{1} + t_{2}, a_{2}:=
t_{1}t_{2}, a_{3}:=t_{3} + t_{4}, a_{4}:=
t_{3}t_{4}, then we find t_{1} and t_{2} from x^{2} -
a_{1}x + a_{2} = 0, and t_{3} and t_{4} from x^{2} -
a_{3}x + a_{4} = 0, expressed by square roots in a_{1}, a_{2},
a_{3}, a_{4} of Z.

2) In the same way, a_{1} + a_{3}, a_{1}a_{3}, a_{2} +
a_{4} and a_{2}a_{4} are invariant under (13)(24), so under
V_{4}/C_{2}, and hence belong to V. When we have found them, we find
a_{1}, a_{2}, a_{3} and a_{4}, being roots of
x^{2} - (a_{1}+a_{3})x + a_{1}a_{3} = 0 and
x^{2} - (a_{2}+a_{4})x + a_{2}a_{4} = 0.

3) Now, for instance, let us see how we can find a_{2} + a_{4} =
t_{1}t_{2} + t_{3}t_{4}.

Let b_{1} := t_{1}t_{2} + t_{3}t_{4},
b_{2} := t_{2}t_{3} + t_{1}t_{4},
b_{3} := t_{1}t_{3} + t_{2}t_{4};

let c_{1} := b_{1} + b_{2} + b_{3},
c_{2} := b_{1}b_{2} + b_{1}b_{3} +
b_{2}b_{3}, c_{3} := b_{1}b_{2}b_{3}.

Then c_{1}, c_{2}, c_{3} are invariant under (123)=(12)(23), so
under A_{4}/V_{4}, and thus belong to A. Hence b_{1}, b_{2},
b_{3} can be expressed by cubic and square roots in c_{1}, c_{2} and
c_{3} of A, being roots of x^{3} - c_{1}x^{2} + c_{2}x
- c_{3} = 0, by the formulas of Cardano for the cubic equation.

4) Fortunately, c_{1}, c_{2}, c_{3} themselves belong not only to
A, but even to K (this happens because the coefficient of x^{3} in our quartic
polynomial is 0 and thus not general). You can find: c_{1}=p,
c_{2}=-4r, c_{3} = q^{2} - 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 +
x^{2} + ... + x^{n-2} + x^{n-1}.

L=Q(z) and G is cyclic of order 2^{m} with m=n-1.

So there exists a chain of groups {e]=G_{0} < G_{1} < ... < G_{m}=G and
a chain of fields Q=K_{0} < K_{1} < ... K_{m}=L where K_{i+1}
has dimension 2 over K_{i}.

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+z^{4}.
Since (x - (z+z^{4}))(x - (z^{2}+z^{3})) = x^{2} + x + 1, we
find z+z^{4} = (-1+w2(5))/2.

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

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