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Evariste Galois (1811-1832) 1827 developed interest for mathematics 1828 failed the entrance exam to the Ecole Polytechnique, but continues to work on.

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Presentation on theme: "Evariste Galois (1811-1832) 1827 developed interest for mathematics 1828 failed the entrance exam to the Ecole Polytechnique, but continues to work on."— Presentation transcript:

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 1.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. 2.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. 3.Check if f(x) still has irreducible non-linear factors. If this is so repeat 2. 4.End up with the splitting field L, which contains all roots of f and in which f factors completely. Examples: 1.f(x)=x 2 -2 K=Q. 1.This has no roots in Q. 2.Fix the root √2 and consider K’=Q(√2). 3.Now f(x)=(x- √2)(x+ √2) and thus we are done K’=L 2.f(x)=x p -1, p prime K=Q. 1.This has the root 1 in Q. x p -1=(x-1)g(x). Here g(x)=1+x+x 2 +…+x p-1 has no root in Q. 2.Fix a root  of g(x)(i.e.  s.t.  p =1) and consider K’=Q(  ). 3.Now f(x)=(x-  (x-   )… (x-  p-1 )and thus we are done K’=L

3 The basic ideas of Galois Theory Other examples 3.f(x)=x 3 -2, K=Q 1.There is no root in Q, so adjoin then in Q( ): f(x)=(x- )(x 2 + x+ ) i.e. f(x)=(x- )g(x) with g(x) irreducible. 2.Adjoin a root  of g(x),   =1. Then g(x)=(x- )(x- ) Now f(x) splits in L=Q(,  Q(,  ) 4.f(x)=x 2 +1, k=R 1.No root in R, so adjoin a root i, to obtain x 2 +1=(x-i)(x+i) and L=R(i)=C. Notice that if one adjoins a transcendental number like  then Q(  )=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. L=K(t). To define a Galois resolvent of f(x) let a,b,c,… be the roots of f(x) and L=K(a,b,c,…) the splitting field. Say f(x) has degree n, so that there are n roots. Consider T=AU+BV+CW+… 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 t 1, t 2, t 3,… are the values of a Galois resolvent T then set F(X)=(x- t 1 ) (x- t 2 ) …(x- t n! ) Let t be a root of any irreducible factor G(X) over K of F(x), then L=K(t). Notice that if G(X) is Lagrange’s reduced equation and his program can be carried out if it is solvable.

5 Groups and Galois Theory Consider the example 4 (f(x)=(x 2 +1),K=R, L=C). 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 K=R. 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 t 1,…,t r be the roots of an irreducible factor G(x) of F(x) over K and let a 1,..a n be the roots of f(x) which are assumed to be all distinct. –a i lies in L, so; a i =h i (t) –For any j: h 1 (t j ), …, h n (t j ) are also roots of f(x), so they are some permutation of the a i. –Now we get permutations by setting  j (a i )=h i (t j ) –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 x n +Y 1 x n-1 +Y 2 x n-2 +…+Y n considered over K(Y 1,…,Y n ) the Galois group is the group of all permutations of the numbers 1,…,n and is not solvable for n>4, hence there is no general formula for higher order equations –For the equation f(x)=x n -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.

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