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 r_i, ..., r_n be the roots of a polynomial P with coefficients in F

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

Let σ_j be a map: σ_j: r_i → f_i(V_j) for i=1, ..., n

The set of all possible permutations { σ₁, ..., 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(x₁, ..., x_n) be a rational fraction in n indeterminates x₁, ..., x_n with coefficients in F.

Then:

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

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

Proof:

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

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

ψ(x₁, ..., x_n) ≠ 0 → ψ(σ(x₁), ..., 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 f₁, ..., f_n such that:

ψ(r₁, ..., r_n) = ψ(f₁(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(f₁(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(r₁), ..., σ(r_n)) = ψ(f₁(V'), ..., f_n(V')) = g(V')

(f) It is clear that if ψ(σ(r₁), ..., 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(r₁), ..., σ(r_n)) = 0 if and only if ψ(r₁, ..., r_n) = 0. [from step #3a above]

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

(5) Assume that f(r₁, ..., r_n) ∈ F

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

f(r₁, ..., r_n) = f(f₁(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(f₁(X), ..., f_n(X))

(8) So that:

h(V) = f(f₁(V), ..., f_n(V))

and:

h(X) - f(f₁(X), ..., f_n(X)) = 0

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

H(X) = h(X) - f(f₁(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: V₁, ...., V_m

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

h(V_j) = f(f₁(V_j), ..., f_n(V_j))

(12) Now since f(r₁, ..., 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(r₁, ..., 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 V₁, ..., V_m are also roots of G(X).

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

h(V_j) = f(r₁, ..., 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(r₁, ..., σ_j(r_n)) = f(r₁, ..., r_n) for j = 1, ..., m

(16) Assume that f(σ_j(r₁, ..., σ_j(r_n)) = f(r₁, ..., r_n) for j = 1, ..., m

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

f(r₁, ..., r_n) = f(f₁(V), f₂(V), ..., f_n(V)) where f, f_i ∈ F[X]

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

g(x) = f(f₁(x),f₂(x), ..., f_n(x))

so that we have:

g(V) = f(r₁, ..., r_n)

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

for all j = 1..m, g(V_j) = f(r₁, ..., r_n)

(20) Further we note that:

m*f(r₁, ..., r_n) = g(V₁) + g(V₂) + ... + g(V_m)

so that we have:

f(r₁, ..., r_n) = (1/m)*(g(V₁) + ... + g(V_m))

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

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

(22) H is clearly a symmetric function so it can be expressed as a rational fraction in the elementary symmetric polynomials s₁, ..., 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 V₁, ..., V_m for x₁, ..., 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(V₁, ..., V_m) ∈ F

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

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

QED

References

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

Galois' Memoir: Lemma 4

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.

Lemma 4:

Suppose one has formed the equation for V and that one has taken one of its irreducible factors so that V is the root of an irreducible equation.

Let V, V', V'', ... be the roots of this irreducible equation .

If a = f(V) is one of the roots of the given equation, f(V') will also be a root of the given equation.

In fact, we can show that:

r₁ = f₁(V), r₂ = f₂(V), ..., r_n = f_n(V)

Then it follows that for any root V' or V'' or ... the complete set of distinct roots is:

f₁(V'), f₂(V'), ..., f_n(V')

Proof:

(1) Let V be a Galois Resolvent of P(X) = 0. [see Lemma 2, here]

(2) Let r₁, ..., r_n be the roots of P(X).

(3) We know that for each r_i, there exists f_i [see Lemma 3, here] such that:

r_i = f_i(V)

with f_i(X) ∈ F(X)

(4) Let V₁ = V, V₂, ..., V_m be the roots of the minimum polynomial of V over F [see Theorem 1, here] which are in F(r₁, ..., r_n) [see Corollary 1.1, here]

(5) Since r_i = f_i(V), we have P(f_i(V)) = 0.

(6) If we view P(f_i(X)) as a polynomial, then, we have P(f_i(V_j)) = 0 for j = 1, ..., m since P(f_i(X)) and the minimal polynomial of V over F share at least one root V. [see Theorem, here]

(7) This shows that each f_i(V_j) must correspond to a root r_i since P(X)=0 if and only if X is a root.

(8) Now, I will end this proof by showing that for any root V_i, that the n roots are:

f₁(V_i), f₂(V_i), ..., f_n(V_i)

(9) Now we know that for V₁, f₁(V₁) ≠ f₂(V₁) ≠ ... ≠ f_n(V₁) [from step #3 above]

(10) Assume that f_u(V_i) = f_v(V_i)

where i ≠ 1

(11) Then V_j is a root of the polynomial f_u - f_v.

(12) But then all V₁, ..., V_m must likewise be roots of f_u - f_v. [see see Theorem, here]

(13) But this also means that V₁ is a root of f_u - f_v

(14) So it follows that f_u(V₁) = f_v(V₁)

(15) But from step #9 above this is only possible if u = v.

(16) Hence, we have shown that for any V_i,

f₁(V_i), ..., f_n(V_i) must be the n distinct roots [since they cannot be equal and each one is equal to a root.]

QED

References

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

Galois' Memoir: Lemma 4: Preliminary Results (Existence of Minimum Polynomial)

Before jumping into Lemma 4 by Evariste Galois, I will prove the existence of minimum polynomials.

The theorem in today's blog comes from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Definition 1: Minimum Polynomial of u over F

For a nonzero polynomials in F[X] which have u as a root, the minimum polynomial is the polynomial of least degree that divides all other nonzero polynomials and is unique, monic, and irreducible.

Theorem 1: Existence of Minimum Polynomials

(a) Every element u ∈ F(r₁, ..., r_n) has a polynomial expression in r₁, ..., r_n such that:

u = φ(r₁, ..., r_n)

for some polynomial φ ∈ F[x₁, ..., x_n]

(b) For every element u ∈ F(r₁, ..., r_n), there is a unique monic irreducible polynomial π ∈ F[X] such that π(u) = 0.

This polynomial π splits into a product of linear factors over F(r₁, ..., r_n)

Proof:

(1) Let u ∈ F(r₁, ..., r_n) be such that:

u = φ(r₁, ..., r_n)

for some polynomial φ ∈ F[x₁, ..., x_n]

(2) Let Θ(X,x₁, ..., x_n) = ∏(σ) [X - φ(σ(x₁), ..., σ(x_n))]

where σ runs over the set of permutations of x₁, ..., x_n

(3) Since Θ is symmetric in x₁, ..., x_n, we can write Θ as a polynomial in X and the elementary symmetric polynomials s₁, ..., s_n in x₁, ..., x_n [see Theorem 4, here]

(4) Let Θ(X,x₁, ..., x_n) = Ψ(X,s₁, ..., s_n)

for some polynomial Ψ with coefficients in F.

(5) Substituting r₁, ..., r_n into Θ, we get:

Θ(X,r₁, ..., r_n)= Ψ(X,a₁, ..., a_n) ∈ F[X]

where a_i are the coefficients of the polynomial where r₁, ..., r_n are the roots. [see Theorem 1, here]

(6) From the definition of Θ in step #2 above, we have:

Θ(u,r₁, ..., r_n) = 0

since in the permutation σ that leaves x_i unchanged, the result will be 0.

(7) It follows that Ψ(X,a₁, ..., a_n) is a polynomial in F[X] which has u as a root. [see step #5 above]

(8) We also note that Θ(X,r₁, ..., r_n) is a product of linear factors. [see step #2 above]

(9) Thus, we have shown that Ψ(X,a₁,...,a_n) splits into a product of linear factors over F(r₁, ..., r_n) [see step #4 above]

(10) We know that we can break down Ψ(X,a₁, ..., a_n) into a product of monic irreducible factors of Ψ [see Theorem 3, here] so that:

Ψ = cP₁*...*P_r

(11) Since u is a root, it follows that u must be the root of one of these monic irreducible factors.

(12) So that there exists i such that:

P_i is a monic, irreducible polynomial over F[X] and u is a root of P_i and P_i divides Ψ.

(13) Since Ψ splits into a product of linear factors over F(r₁, ..., r_n), it follows that P_i must also split into a product of linear factors over F(r₁, ..., r_n).

(14) Finally, we note that there is only one monic irreducible polynomial P_i which has u as a root since:

(a) Assume that Q ∈ F[X] is another polynomial with the same properties.

(b) First, we note that P must divide Q. [see Lemma 2, here]

(c) But, by the same argument Q must divide P.

(d) So, then it follows that P=Q.

(15) This completes part(b) of the proof.

(16) Let V be the Galois Resolvent (see Definition 1, here) such that:

V ∈ F(r₁, ..., r_n)

(17) We know that r₁, ..., r_n are rational fractions in V. [see Lemma 3, here]

(18) Since u is a rational fraction of F(r₁, ..., r_n), it follows that u is also rational fraction in V and further u ∈ F(V).

(19) So, u can be expressed as a polynomial in V and:

u = Q(V)

for some polynomial Q ∈ F[X].

(20) Substituting f(r₁, ..., r_n) for V, we obtain:

u = Q(f(r₁, ..., r_n))

(21) This is a polynomial expression in r₁, ..., r_n since Q and f are polynomials.

(22) This completes part(a) of the proof.

QED


Corollary 1.1:

Let V be any Galois Resolvent of P(X) = 0 over F and let V₁, ..., V_m be the roots of its minimum polynomial over F (among which V lies)

Then:

F(r₁, ..., r_n) = F(V) = F(V₁, ..., V_m)

Proof:

(1) r₁, ..., r_n are rational fractions in V [see Lemma 3, here]

(2) So, we have:

F(r₁, ..., r_n) ⊂ F(V)

(3) We know that the roots V₁, ..., V_m of the minimum polynomial of V are in F(r₁, ..., r_n) since:

(a) V is a polynomial g(x₁, ..., x_n) ∈ F[X] [see Definition 1, here]

(b) Applying part(b) of Theorem 1 above, we note that each V_i has g(V_i)=0 and can be expressed over linear factors of F(r₁, ..., r_n)

(4) So, we have:

F(V₁, ..., V_m) ⊂ F(r₁, ..., r_n)

(5) Since V₁, ..., V_m are roots of V, it follows that:

F(V) ⊂ F(V₁, ..., V_m)

(6) Therefore we have:

F(r₁, ..., r_n) = F(V) = F(V₁, ..., V_m)

QED

References

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