Evariste Galois 18111832

1
Evariste Galois (1811-1832)

• 1827 developed interest for mathematics
• 1828 failed the entrance exam to the Ecole
  Polytechnique, but continues to work on his own
• 1829 first mathematics paper published on
  continued fractions in the Annales de
  mathématiques.
• 25 May and 1 June he submitted articles on the
  algebraic solution of equations to the Académie
  des Sciences. Cauchy was appointed as referee of
  Galois' paper.
• 1829 entered the Ecole Normale.
• 1830 learned of a posthumous article by Abel
  which overlapped with a part of his work and
  submitted a new article On the condition that an
  equation be soluble by radicals in February. The
  paper was sent to Fourier, the secretary of the
  Academy, to be considered for the Grand Prize in
  mathematics. Fourier died in April 1830 and
  Galois' paper was never subsequently found and so
  never considered for the prize which went to Abel
  and Jacobi
• 1830 He published three papers in Bulletin de
  Férussac.
• 1830 Galois was invited by Poisson to submit a
  third version of his memoir on an equation
• 1830 For writing a political letter Galois was
  expelled and he joined On 31 December 1830 the
  Artillery of the National Guard which was
  subsequently was abolished by Royal Decree since
  the new King Louis-Phillipe felt it was a threat
  to the throne.
• 1830 In and out of prison
• 1832 Galois contracted cholera during the Paris
  epidemic. He apparently fell in love with
  Stephanie-Felice du Motel, the daughter of the
  his physician
• 1832 Galois fought a duel with Perscheux
  d'Herbinville on 30 May probably about Stephaine
  and subsequently died in Cochin hospital on 31
  May.
• 1846 Liouville published the papers of Galois in
  his Journal.

2
The basic ideas of Galois Theory

• Regard a polynomial equation f(x)0 with fixed
  the field K of coefficients!
• In a field k one can add, subtract, multiply and
  divide just like in R, Q or C.
• If f(x) has irreducible non-linear factors g(x),
  adjoin to K a root r of g(x) to obtain K(r),
  which is the field of rational functions in r
  with coefficients in K.
• Check if f(x) still has irreducible non-linear
  factors. If this is so repeat 2.
• End up with the splitting field L, which contains
  all roots of f and in which f factors completely.
• Examples
• f(x)x2-2 KQ.
• This has no roots in Q.
• Fix the root v2 and consider KQ(v2).
• Now f(x)(x- v2)(x v2) and thus we are done KL
• f(x)xp-1, p prime KQ.
• This has the root 1 in Q. xp-1(x-1)g(x).
  Here g(x)1xx2xp-1 has no root in Q.
• Fix a root z of g(x)(i.e. z s.t. zp1) and
  consider KQ(z).
• Now f(x)(x-z) (x-z2) (x-zp-1)and thus we are
  done KL

3
The basic ideas of Galois Theory

• Other examples
• f(x)x3-2, KQ
• There is no root in Q, so adjoin
  then in Q( ) f(x)(x- )(x2 x )
  i.e. f(x)(x- )g(x) with g(x)
  irreducible.
• Adjoin a root z of g(x), z31. Then
  g(x)(x- )(x- )
  Now f(x) splits in LQ( , ) Q( , z)
• f(x)x21, kR
• No root in R, so adjoin a root i, to obtain
  x21(x-i)(xi) and LR(i)C.
• Notice that if one adjoins a transcendental
  number like p then Q(p)Q(t), the rational
  functions in one variable.

4
The Galois resolvent

• Galois showed that one can describe the splitting
  field L by adjoining only one root t of a
  different equation. LK(t).
• To define a Galois resolvent of f(x) let a,b,c,
  be the roots of f(x) and LK(a,b,c,) the
  splitting field. Say f(x) has degree n, so that
  there are n roots. Consider TAUBVCW
  with integers
  A,B,C, and U,V,W, variables.
• T is called a Galois resolvent if all the n!
  elements obtained by substituting the roots
  a,b,c, for the variables U,V,W, are distinct.
• If t1, t2, t3, are the values of a Galois
  resolvent T then set F(X)(x- t1) (x- t2) (x-
  tn!)
• Let t be a root of any irreducible factor G(X)
  over K of F(x), then LK(t).
• Notice that if G(X) is Lagranges reduced
  equation and his program can be carried out if it
  is solvable.

5
Groups and Galois Theory

• Consider the example 4 (f(x)(x21),KR, LC). We
  can interchange the solutions i and i. This is
  just complex conjugation, which is a field
  isomorphism, i.e. it respects addition and
  multiplication. Also the fixed elements of the
  complex conjugation are exactly KR.
• In general there is a group which acts on L by
  field isomorphisms permuting the roots and fixing
  K. This is the Galois group.
• To see this let t1,,tr be the roots of an
  irreducible factor G(x) of F(x) over K and let
  a1,..an be the roots of f(x) which are assumed to
  be all distinct.
• ai lies in L, so aihi(t)
• For any j h1(tj), , hn(tj) are also roots of
  f(x), so they are some permutation of the ai.
• Now we get permutations by setting sj(ai)hi(tj)
• These permutations give the Galois group.

6
Groups and solvability

• In the step by step process, the group gets
  smaller in each step. Vice versa, the full group
  can be obtained in a step by step process from
  smaller groups.
• The groups which one obtains by adjoining
  radicals are of a special type.
• From these two observations Galois could give a
  criterion to say when an equation f(x) is soluble
  by radicals in terms of the Galois group.
  Technically if the group is solvable so is the
  equation.
• Examples
• For the general equation xnY1xn-1Y2xn-2
  Yn considered over K(Y1,,Yn) the
  Galois group is the group of all permutations of
  the numbers 1,,n and is not solvable for ngt4,
  hence there is no general formula for higher
  order equations
• For the equation f(x)xn-1/(x-1), the group is
  the cyclic group of rotations of order n, which
  is solvable. So there is a formula in terms of
  radicals as Gauß had previously found. The group
  is the group of orientation preserving symmetries
  of a regular n-gon.