Although Evariste Galois does not state it directly, there are important implications to hisProposition 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, seehere.
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.
and
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.
Proof:
(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)
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, ..., 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 thatGal(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, ..., rn
Proof:
(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)
QED
References
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, seehere.
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.
and
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.
Proof:
(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)
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, ..., 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 thatGal(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, ..., rn
Proof:
(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)
QED
References
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations
, World Scientific, 2001
