Galois' Memoir: Corollaries to Proposition I

Although Evariste Galois does not state it directly, there are important implications to his Proposition I which I will present.

The content that follows is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equation.

Here is the statement of Proposition I from Galois' Memoir without proof. For a proof, see here.

Proposition I:

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

For σ ∈ Gal(P/F):

f(σ(r₁), ..., σ(r_n)) is defined whenever f(r₁, ..., r_n) is defined.

and

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

Following from this proposition are the following corollaries.

Definition 1: Automorphism

An automorphism is a bijective homomorphism from a group to itself.

For details on the meaning of this definition (including the definition of bijective and the definition of a homomorphism, see here.)

Corollary 1.1:

Each permutation σ ∈ Gal(P/F) can be extended to an automorphism of F(r₁, ..., r_n) which leaves every element in F invariant by setting

σ(f(r₁, ..., r_n)) = f(σ(r₁), ..., σ(r_n)) for any rational fraction f(x₁, ..., x_n) for which
f(r₁, ..., r_n) is defined.

Proof:

(1) Assume that f(r₁, ..., r_n) is defined.

(2) Using Proposition I (see here), we know that:

f(σ(r₁), ..., σ(r_n)) ∈ F(r₁, ..., r_n) is defined.

(3) Let σ(f(r₁,..., r_n)) = f(σ(r₁), ..., σ(r_n))

(4) First, we need to show that σ(f(r₁, ..., r_n)) leaves every element in F invariant.

(5) Assume that f(r₁, ...., r_n) = g(r₁, ..., r_n) for two rational fractions f,g ∈ F(x₁, ..., x_n)

(6) Let us define a function H such that H(x₁, ..., x_n) = f(x₁, ..., x_n) - g(x₁, ..., x_n)

(7) By step #5 above, we know that H(r₁, ..., r_n) = 0

(8) So it follows that H(r₁, ..., r_n) ∈ F since 0 ∈ F.

(9) Using Proposition I (see here), we have:

H(r₁, ..., r_n) = H(σ(r₁), ..., σ(r_n))

(10) So:

f(σ(r₁), ... σ(r_n) ) - g(σ(r₁), ..., σ(r_n)) = 0

and further:

f(σ(r₁), ..., σ(r_n)) = g(σ(r₁), ..., σ(r_n))

(11) This proves that σ(f)=σ(g) = f(σ(r₁), ..., σ(r_n)) = g(σ(r₁), ..., σ(r_n)) if and only if
f(r₁, ..., r_n) = g(r₁, ..., r_n)

(12) To complete the proof, we show that σ is an automorphism [see Definition 1 above].

(13) We know it is bijective because σ is bijective.

(14) We know that it is a homomorphism since:

σ(f + g) = f(σ(r₁), ..., σ(r_n)) + g(σ(r₁), ..., σ(r_n)) = σ(f) + σ(g)

σ(fg) = f(σ(r₁), ..., σ(r_n))*g(σ(r₁), ..., σ(r_n)) = σ(f)*σ(g)

QED


Corollary 1.2:

The set Gal(P/F) does not depend on the choice of the Galois resolvent V.

Proof:

(1) Let V' ∈ F(r₁, ..., r_n) be another Galois resolvent (see Definition 2, here) for p(x)=0,

(2) Let π' be its minimum polynomial over F.

(3) Let f_i' ∈ F(x) for i=1,...,n be a rational fraction such that:

r_i = f_i'(V') for i = 1, ..., n

(4) Using Corollary 1.1 above and using step #3 above, we have:

σ(r_i) = f_i'(σ(V'))

since σ leaves every element in F invariant and since it maps each r_i to a unique value.

(5) Likewise, we can apply σ to both sides of:

π'(V') = 0

to get:

σ(π(V')) = π'(σ(V')) = σ(0) = 0

which gives us that:

π(σ(V')) = 0

(6) This shows that σ(V') is a root of π'.

(7) Thus we have shown that every element of Gal(P/F) as defined earlier with V is also an element of Gal(P/G) as defined with V'.

(8) We can also use the same argument to show the reverse which demonstrates that Gal(P/F) for V = Gal(P/F) for V'.

QED


Corollary 1.3:

Gal(P/F) is a subgroup of the group of all permutations of r₁, ..., r_n

Proof:

(1) Gal(P/F) is by definition a subset of all the permutations of r₁, ..., r_n. [see Definition 1, here, for definition of Gal(P/F)]

(2) So, to complete this proof (see Definition 6, here and Definition 2, here), I need to show that Gal(P/F) has closure, associativity, inversion, and identity on the operation of permutations.

(3) It has closure since for any element τ ∈ Gal(P/F), the composition τ * σ ∈ Gal(P/F) since:

(a) Let τ be the map: f_i(V_j) → f_i(V_k) for some k=1,..., m

(b) Let σ be the map: r_i → f_i(V_j)

(b) It follows that τ * σ : r_i → f_i(V_k)

(c) Therefore, τ * σ ∈ Gal(P/F)

(4) It has identity since the identity map is in Gal(P/F) is clear since the map is σ₁ in the definition of the Galois Group since r₁ = f₁(V₁), r₂ = f₂(V₁), ..., and so on.

(5) It has inversion since:

(a) Let σ ∈ Gal(P/F)

(b) By definition of Gal(P/F), σ(r_i) = f_i(V_j)

(c) We know that V_j is also a Resolvent of P(X)=0. [see Lemma 4, here]

(d) So, we can define Gal(P/F) according to V_j and it will be the same. [see Corollary 1.2 above]

(e) So, it follows that the map:

f_i(V_j) → f_i(V₁)

is an element of Gal(P/F)

(6) It has associativity since:

(a) Let's assume that we have three elements of Gal(P/F) so that we have:

σ, τ, φ

(b) Let's define the following maps:

φ: f_i(V_k) → f_i(V_l)

τ: f_i(V_j) → f_i(V_k)

σ: r_i → f_i(V_j)

(c) (φ * τ ) : f_i(V_j) → f_i(V_l)

(d) (φ * τ ) * σ: r_i → f_i(V_l)

(e) (τ * σ) : r_i → f_i(V_k)

(f) φ*(τ * σ): r_i → f_i(V_l)

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001