Tuesday, October 06, 2009

Galois' Memoir: Proposition 1 (Galois Group)

The following is taken from the translation of Galois' Memoir by Harold M. Edwards found in his book Galois Theory. The proof itself is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Definition 1: Galois Group Gal(P/F)

Let ri, ..., rn be the roots of a polynomial P with coefficients in F

Let V1, ..., Vm be the roots of the minimum polynomial for a given Galois Resolvent where the Galois Resolvent V = V1

Let σj be a map: σj: ri → fi(Vj) for i=1, ..., n

The set of all possible permutations { σ1, ..., σm } is called the Galois group of P(X)=0 over F and is denoted Gal(P/F).

Theorem: Galois' Proposition I

Let an equation be given whose m roots are a, b, c, ...

These will always be a group of permutations of the letters a,b,c,... which will have the following property:

1. That each function invariant under the substitution of this group will be known rationally

2. Conversely, that every function of the roots which can be determined rationally will be invariant under these substitutions

In other words:

Let f(x1, ..., xn) be a rational fraction in n indeterminates x1, ..., xn with coefficients in F.


f(r1, ..., rn) ∈ F if and only if for all σ ∈ Gal(P/F),

f(r1, ..., rn) = f(σ(r1), ..., σ(rn))


(1) Let f = φ/ψ where φ, ψ ∈ F[x1, ..., xn]

(2) First, we need to prove that f is a rational function which comes down to showing that:

ψ(x1, ..., xn) ≠ 0 → ψ(σ(x1), ..., σ(xn)) ≠ 0 where σ ∈ Gal(P/F)

(3) We can show this since:

(a) Using Lemma 3 before (see Lemma 3, here), we know that there exists functions f1, ..., fn such that:

ψ(r1, ..., rn) = ψ(f1(V), ..., fn(V))

where V is the Galois Resolvent (see Definition 1, here) and fi ∈ F[X]

(b) But since each fi ∈ F[X] and ψ ∈ F[X], it follows that there exists a function g ∈ F[X] such that:

g(V) = ψ(f1(V), ..., fn(V))

(c) Let σ be a permutation in Gal(P/F)

(d) From Lemma 4 before (see Lemma 4, here), we know that each permutation corresponds to a root of the minimal irreducible polynomial of the Galois Resolvent so that there exists V' such that:

V' is a root of the minimal irreducible polynomial of the Galois Resolvent


σ: ri → fi(V')

(e) So that we have:

ψ(σ(r1), ..., σ(rn)) = ψ(f1(V'), ..., fn(V')) = g(V')

(f) It is clear that if ψ(σ(r1), ..., σ(rn)) = 0, then g(V')= 0

(g) So that when ψ is 0, V' is a root.

(h) Using Theorem 1, here, it clear that:

g(V') =0 if and only if g(V) is 0.

(i) So further, ψ(σ(r1), ..., σ(rn)) = 0 if and only if ψ(r1, ..., rn) = 0. [from step #3a above]

(4) This shows that f(σ(r1), ..., σ(rn)) is defined when f(r1, ..., rn) is defined.

(5) Assume that f(r1, ..., rn) ∈ F

(6) From step #3a above, we have:

f(r1, ..., rn) = f(f1(V), ..., fn(V))

(7) But since each fi ∈ F[X] and f ∈ F[X], it follows that there exists a function h ∈ F[X] such that:

h(X) = f(f1(X), ..., fn(X))

(8) So that:

h(V) = f(f1(V), ..., fn(V))


h(X) - f(f1(X), ..., fn(X)) = 0

(9) If we define a function H such that:

H(X) = h(X) - f(f1(X), ..., fn(X))

It is clear that H(V)=0

(10) Using Theorem 1, here, it follows that all m roots of the minimal irreducible polynomial of the Galois Resolvent are also roots of H so that it's roots include: V1, ...., Vm

(11) So it follows for all j = 1, ..., m that:

h(Vj) = f(f1(Vj), ..., fn(Vj))

(12) Now since f(r1, ..., rn) ∈ F [by our assumption in step #5], it follows that we can define a function G such that:

G(X) = h(X) - f(r1, ..., rn)

(13) From step #6 above, we note that V is a root of G(X) and by the same logic as step #10, we conclude that all V1, ..., Vm are also roots of G(X).

(14) So that we have for all j= 1, ..., m:

h(Vj) = f(r1, ..., rn)

(15) But if this is true for all Vj, then it follows that it is true for all σ ∈ Gal(F/G) [by Lemma 4, here] so that we have:

f(σj(r1, ..., σj(rn)) = f(r1, ..., rn) for j = 1, ..., m

(16) Assume that f(σj(r1, ..., σj(rn)) = f(r1, ..., rn) for j = 1, ..., m

(17) From step #3a above, we know that:

f(r1, ..., rn) = f(f1(V), f2(V), ..., fn(V)) where f, fi ∈ F[X]

(18) So, we can define a function g(x) such that:

g(x) = f(f1(x),f2(x), ..., fn(x))

so that we have:

g(V) = f(r1, ..., rn)

(19) Since for all j, we have f(σj(r1, ..., σj(rn)) = f(r1, ..., rn), it follows that:

for all j = 1..m, g(Vj) = f(r1, ..., rn)

(20) Further we note that:

m*f(r1, ..., rn) = g(V1) + g(V2) + ... + g(Vm)

so that we have:

f(r1, ..., rn) = (1/m)*(g(V1) + ... + g(Vm))

(21) We can define the following function H such that:

H(x1, ..., xm) = (1/m)*g(x1) + ... + g(xm)

(22) H is clearly a symmetric function so it can be expressed as a rational fraction in the elementary symmetric polynomials s1, ..., sm [see Theorem 6, here]

(23) Now the elementary symmetric polynomials correspond to the coefficients of a polynomial. [see Theorem 1, here]

(24) This shows that if we substitute V1, ..., Vm for x1, ..., xm the elementary symmetric polynomial will correspond to the coefficients of the minimal irreducible polynomial for the Galois Resolvent whose coefficients are in F.

(25) Therefore H(V1, ..., Vm) ∈ F

(26) But H(V1, ..., Vm) = (1/m)*[g(V1) + g(V2) +... + g(Vm)] = f(r1, ..., rn)

(26) So we have shown that f(r1, ..., rn) ∈ F



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