The first Memoir of Galois. II 15 th April 2015
The Fundamental Theorem The connection with the modern definition of Galois group is given by the following theorem. 1 Theorem (Fundamental Theorem) The group G of permutations defined by the polynomial F (t) is such that every rational function of the roots which is left unchanged by the permutations of G is rationally known and vice versa, every function of the roots which assumes a value in K is left unchanged by all the permutations of the group G. 1 I give a free paraphrase of Galois s original statement. See [4, Prop. I, p. 51].
Proof. Suppose that g(x 1,..., x n ) is a function of the roots. We have proved that all the roots are rational functions of V 1, then g(x 1,..., x n ) = ḡ(v 1 ). If the function ḡ is left unchanged by the substitutions σ, τ,... then ḡ(v 1 ) = ḡ(v σ ) = ḡ(v τ ) =... The number of these substitutions (which equals the degree of F(t)) is r, and ḡ(v 1 ) = ḡ(v σ ) = ḡ(v τ ) = = 1 r (ḡ(v 1) + ḡ(v σ ) + ḡ(v τ ) +... ).
Proof (continued). It follows that ḡ(v 1 ) is a symmetric function of the roots of F (t) and may be expressed by means of its coefficients, which are elements of K. Vice versa, if ḡ(v 1 ) = k, then V 1 is a root of the equation ḡ(v ) k = 0. It follows that this equation admits all the other roots of F (t) (an irreducible polynomial) and its value is left unchanged by the substitutions of G.
Remark Note that this proof does not use the fact that the substitutions identified by the roots of F (t) constitute a group. In effect, what is used is the symmetry of the roots themselves. Remark The proof given by Jordan of the fact that the roots of F (t) identify the Galois Group is the following. 2 To give it in the simplest way, it is convenient to change a little the notations: V 1 is still the value that the identity substitution 1 gives to V ; and V σ, V τ,... will denote the values that V assumes under the effect of the permutations σ, τ,... that identify the other roots of F (t). Hence F (t) = (t V 1 )(t V σ )(t V τ ). The Fundamental Theorem grants the fact that this polynomial is left unchanged by the substitutions 1, σ, τ,.... 2 See [5, p, 258]. The proof sketched by Galois is given in [6, pp. 114-116].
Remark (continued) If we apply the substitution σ, the polynomial F (t) remains unchanged, but on the other hand it becomes F (t) = (t V σ )(t V σ 2)(t V τσ ). This is possible if and only if the roots are still the same, which means that σ, σ 2, στ,... are the original permutations 1, σ, τ.... simply listed in a different order. Of course the argument used for σ may be used for every other permutation corresponding to the roots of F (t) which implies the closure under multiplication. This property is sufficient, because we have to identity a finite group.
Remark Note that this result permits us to associate to every permutation of G an automorphism of E = K(x 1,..., x n ) which fixes the elements of K. But every such automorphism gives rise to a permutation of the roots and it appears that the Galois group is exactly the group of all the permutations that are automorphisms of E which have K as fixed field. This idea will be brought out in the text of Dedekind edited by Scharlau in [7]. How influential the lectures of Dedekind were in the second half of nineteenth century is not easy to assess. We may also observe 3 that the Galois group is the largest group under which every rational function of the roots which has a rational value remains numerically unaltered. Alternatively it is the smallest group such that every rational function of the roots which remains under it numerically unaltered has a rational value. Hence the Galois group is uniquely defined by its fundamental properties. 3 See [3, pp. 147-48] who gives this formulation.
Remark A resolvent of the form previously described gives the possibility of constructing the Galois Group of an equation but other convenient functions of the roots may be used. Consider the polynomial x 4 10x 2 + 1. (1) If we denote one of its roots by α, another root is given by α, and, since (1) is a reciprocal polynomial, 1 α and 1 α are the remaining roots. Of course α is a root of the irreducible polynomial (1) of Q[x] and since all its roots are function of α we have the following presentation of the Galois Group.
Remark (continued) α α α 1 α α α α 1 α 1 α 1 α 1 α 1 α 1 α α α 1 1 1 α α α α α Note that this presentation is obtained without any need of complicated calculations.
The first example given by Galois He states as obvious that the general equation of degree m has as its group S m because in this case the symmetric functions are the only ones that can be determined rationally. 4 The result is quite evident if the general equation is considered as an equations having the roots x 1, x 2,..., x m to be conceived as indeterminates. But it is not so obvious if the general equation of degree m is an equation of the form x m a 1 x m 1 + a 2 x m 2 + + ( 1) m a m in which the coefficients are indeterminates. The modern proof consists in comparing the splitting field of this equation with K(x 1, x 2,..., x m ), where x 1,..., x m are indeterminates, but is not so straightforward. 5 Anyway, till now we have no clues about Galois mind about this result. 6 4 [4, p. 51]. 5 See [2, pp. 77-78]. 6 The results stated in [4, pp. 75-76] are without proofs.
The second example given by Galois Consider the equation x p 1 x 1 = xp 1 + x p 2 + + x + 1 = 0. (p prime) (2) In this case Galois affirms that the roots x 1, x 2,..., x p 1 can be named in such a way that the group simply consists of the p 1 powers of the cyclic permutation (x 1 x 2... x p 1 ). We have a hint of the possible proof devised by Galois by a text 7 where he considers the more general case of an irreducible equation whose roots are functions of one of them. 8 7 See [4, pp. 89-91]. 8 See the example given in the previous Remark which shows a simple calculation of the Galois Group of an equation in the case in which all the roots are functions of one of them.
We need two preliminary facts. Theorem An equation f(x) = 0 is irreducible if and only if its Galois group G is transitive. Proof. Suppose that G be not transitive and that the root x 1 may be sent by the substitution of G only into x 1, x 2,..., x m, with m < f. Then the substitutions of G do not modify the symmetric functions of x 1, x 2,..., x m, and the polynomial (x x 1 )(x x 2 ) (x x m ) has its coefficients in the ground field and is a divisor of f(x). Then f(x) must be reducible. Conversely, suppose that G be transitive and f(x) be divisible by (x x 1 ) (x x m ) with m < f. A substitution of G exists that sends x 1 into x m+1. By the effect of this substitution the polynomial (x x 1 ) (x x m ) cannot remain unchanged, because it acquires the new root x m+1. Hence it cannot have all its coefficients in the ground field. 9 9 [5, p. 259].
The second fact we need is that the order of a transitive group over n letters is divisible by n, as it is easily seen. 10 Theorem Let p be a prime number and consider the equation x p 1 x 1 = xp 1 + x p 2 + + x + 1 = 0. (3) The group of this equation is cyclic of order p 1. Proof. The polynomial at the left side of (3) is irreducible, as was proved by Gauss. Gauss also proved that Z p has a primitive root (a generator of the cyclic group of the elements different from 0). Let γ be such a primitive root and, having singled out a root r of the equation (3), denote all its roots by x 1 = r, x 2 = r γ, x 3 = r γ2,..., x p 1 = r γp 2. 10 [5, p. 29].
Proof (continued). The Galois group has been defined by a Galois Resolvent V 1 = a 1 r + a 2 r γ + a 3 r γ2 + + a p 1 r γp 1 = w(r). (4) The minimum polynomial F (t) of V 1 verifies F (w(r)) = 0. From the irreducibility of the polynomial we have also that x p 1 + x p 2 + + x + 1 0 = F (w(r γ )) = F ( ( w r γ2)) =... It follows that all the values r γj are possible candidates to produce the substitutions of the Galois group.
Proof (continued). The substitution of r γ for r produces the arrangement x p 1, x 1, x 2,..., x p 2 and proceeding in the same way we obtain all the cyclic permutations of (x 1 x 2... x p 1 ). It follows that the Galois group is contained in the cyclic group generated by the circular permutation (x 1 x 2... x p 1 ). But since its order must be a multiple of p 1, it coincides with this group. 11 11 See also [5, pp. 291-297].
Let us consider the equation x 5 1 = 0. It has the root 1 which is fixed by the substitutions of its Galois Group, and the roots r = e 2π i 5, r 2, r 3, r 4 Since 2 is a primitive root modulo 5, it is easily checked that the Galois Group is given by the permutations r r 2 r 4 r 3 r 2 r 4 r 3 r r 4 r 3 r r 2 r 3 r r 2 r 4 Note that a similar result may be obtained for every prime p. For example for p = 17, a primitive root is 3 and we may describe the corresponding cyclic group of order 16 without the necessity of complicated calculations. We have simply to list the exponent as 1, 3, 9, 10, 13, 5, 15, 11, 16, 14, 8, 7, 4, 12, 2, 6
E. Artin. Galois Theory. Notre Dame Lectures, University of Notre Dame Press, 1944. Second revised edition. E. Artin. Galois Theory. Dover, New York, 1998. Unabridged and unaltered republication of [1]. E. Dehn. Algebraic equations. Columbia University Press, 1930. E. Galois. Écrits et mémoires mathématiques d Évariste Galois. Gauthier-Villars, Paris, 1976. Publiés par R. Bourgne et J. P. Azra. Préface de J. Dieudonné. Deuxième édition revue et augmenté. Réimpression autorisée, Editions J. Gabay, 1997.
C. Jordan. Traité des substitutions et des équations algébriques. Gauthier-Villars, Paris, 1870. Éditions Jacques Gabay, 1989. P. Neumann. The mathematical writings of Évariste Galois. The European Mathematical Society, 2011. Édition critique intégrale du texte avec traduction anglaise. W. Scharlau, editor. Richard Dedekind 1831-1981. Eine Würdigung zu seinem 150. Geburtstag, Braunschweig/Wiesbaden, 1981. F. Vieweg und Sohn.