Monday, October 12, 2009

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, ..., xn) be a rational fraction in n indeterminates x1, ..., xn with coefficients in a field F.

For σ ∈ Gal(P/F):

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


f(r1, ..., rn) ∈ F if and only if f(σ(r1),..., σ(rn)) = f(r1, ..., rn) 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, ..., rn) which leaves every element in F invariant by setting

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


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

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

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

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

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

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

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

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

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

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

H(r1, ..., rn) = H(σ(r1), ..., σ(rn))

(10) So:

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

and further:

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

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

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

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


Corollary 1.2:

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


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

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

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

ri = fi'(V') for i = 1, ..., n

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

σ(ri) = fi'(σ(V'))

since σ leaves every element in F invariant and since it maps each ri 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'.


Corollary 1.3:

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


(1) Gal(P/F) is by definition a subset of all the permutations of r1, ..., rn. [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: fi(Vj) → fi(Vk) for some k=1,..., m

(b) Let σ be the map: ri → fi(Vj)

(b) It follows that τ * σ : ri → fi(Vk)

(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(ri) = fi(Vj)

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

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

(e) So, it follows that the map:

fi(Vj) → fi(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: fi(Vk) → fi(Vl)

τ: fi(Vj) → fi(Vk)

σ: ri → fi(Vj)

(c) (φ * τ ) : fi(Vj) → fi(Vl)

(d) (φ * τ ) * σ: ri → fi(Vl)

(e) (τ * σ) : ri → fi(Vk)

(f) φ*(τ * σ): ri → fi(Vl)