Definition 1: Galois Group Gal(P/F)

Let r

_{i}, ..., r

_{n}be the roots of a polynomial P with coefficients in F

Let V

_{1}, ..., V

_{m}be the roots of the minimum polynomial for a given Galois Resolvent where the Galois Resolvent V = V

_{1}

Let σ

_{j}be a map: σ

_{j}: r

_{i}→ f

_{i}(V

_{j}) for i=1, ..., n

The set of all possible permutations { σ

_{1}, ..., σ

_{m}} 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(x

_{1}, ..., x

_{n}) be a rational fraction in n indeterminates x

_{1}, ..., x

_{n}with coefficients in F.

Then:

f(r

_{1}, ..., r

_{n}) ∈ F if and only if for all σ ∈ Gal(P/F),

f(r

_{1}, ..., r

_{n}) = f(σ(r

_{1}), ..., σ(r

_{n}))

Proof:

(1) Let f = φ/ψ where φ, ψ ∈ F[x

_{1}, ..., x

_{n}]

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

ψ(x

_{1}, ..., x

_{n}) ≠ 0 → ψ(σ(x

_{1}), ..., σ(x

_{n})) ≠ 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 f

_{1}, ..., f

_{n}such that:

ψ(r

_{1}, ..., r

_{n}) = ψ(f

_{1}(V), ..., f

_{n}(V))

where V is the Galois Resolvent (see Definition 1, here) and f

_{i}∈ F[X]

(b) But since each f

_{i}∈ F[X] and ψ ∈ F[X], it follows that there exists a function g ∈ F[X] such that:

g(V) = ψ(f

_{1}(V), ..., f

_{n}(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

and

σ: r

_{i}→ f

_{i}(V')

(e) So that we have:

ψ(σ(r

_{1}), ..., σ(r

_{n})) = ψ(f

_{1}(V'), ..., f

_{n}(V')) = g(V')

(f) It is clear that if ψ(σ(r

_{1}), ..., σ(r

_{n})) = 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(r

_{1}), ..., σ(r

_{n})) = 0 if and only if ψ(r

_{1}, ..., r

_{n}) = 0. [from step #3a above]

(4) This shows that f(σ(r

_{1}), ..., σ(r

_{n})) is defined when f(r

_{1}, ..., r

_{n}) is defined.

(5) Assume that f(r

_{1}, ..., r

_{n}) ∈ F

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

f(r

_{1}, ..., r

_{n}) = f(f

_{1}(V), ..., f

_{n}(V))

(7) But since each f

_{i}∈ F[X] and f ∈ F[X], it follows that there exists a function h ∈ F[X] such that:

h(X) = f(f

_{1}(X), ..., f

_{n}(X))

(8) So that:

h(V) = f(f

_{1}(V), ..., f

_{n}(V))

and:

h(X) - f(f

_{1}(X), ..., f

_{n}(X)) = 0

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

H(X) = h(X) - f(f

_{1}(X), ..., f

_{n}(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: V

_{1}, ...., V

_{m}

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

h(V

_{j}) = f(f

_{1}(V

_{j}), ..., f

_{n}(V

_{j}))

(12) Now since f(r

_{1}, ..., r

_{n}) ∈ F [by our assumption in step #5], it follows that we can define a function G such that:

G(X) = h(X) - f(r

_{1}, ..., r

_{n})

(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 V

_{1}, ..., V

_{m}are also roots of G(X).

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

h(V

_{j}) = f(r

_{1}, ..., r

_{n})

(15) But if this is true for all V

_{j}, then it follows that it is true for all σ ∈ Gal(F/G) [by Lemma 4, here] so that we have:

f(σ

_{j}(r

_{1}, ..., σ

_{j}(r

_{n})) = f(r

_{1}, ..., r

_{n}) for j = 1, ..., m

(16) Assume that f(σ

_{j}(r

_{1}, ..., σ

_{j}(r

_{n})) = f(r

_{1}, ..., r

_{n}) for j = 1, ..., m

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

f(r

_{1}, ..., r

_{n}) = f(f

_{1}(V), f

_{2}(V), ..., f

_{n}(V)) where f, f

_{i}∈ F[X]

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

g(x) = f(f

_{1}(x),f

_{2}(x), ..., f

_{n}(x))

so that we have:

g(V) = f(r

_{1}, ..., r

_{n})

(19) Since for all j, we have f(σ

_{j}(r

_{1}, ..., σ

_{j}(r

_{n})) = f(r

_{1}, ..., r

_{n}), it follows that:

for all j = 1..m, g(V

_{j}) = f(r

_{1}, ..., r

_{n})

(20) Further we note that:

m*f(r

_{1}, ..., r

_{n}) = g(V

_{1}) + g(V

_{2}) + ... + g(V

_{m})

so that we have:

f(r

_{1}, ..., r

_{n}) = (1/m)*(g(V

_{1}) + ... + g(V

_{m}))

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

H(x

_{1}, ..., x

_{m}) = (1/m)*g(x

_{1}) + ... + g(x

_{m})

(22) H is clearly a symmetric function so it can be expressed as a rational fraction in the elementary symmetric polynomials s

_{1}, ..., s

_{m}[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 V

_{1}, ..., V

_{m}for x

_{1}, ..., x

_{m}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(V

_{1}, ..., V

_{m}) ∈ F

(26) But H(V

_{1}, ..., V

_{m}) = (1/m)*[g(V

_{1}) + g(V

_{2}) +... + g(V

_{m})] = f(r

_{1}, ..., r

_{n})

(26) So we have shown that f(r

_{1}, ..., r

_{n}) ∈ F

QED

References

- Harold M. Edwards, Galois Theory, Springer, 1984
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001