## Tuesday, October 06, 2009

### 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 ri, ..., rn be the roots of a polynomial P with coefficients in F

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

Let σj be a map: σj: ri → fi(Vj) for i=1, ..., n

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

Then:

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

f(r1, ..., rn) = f(σ(r1), ..., σ(rn))

Proof:

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

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

ψ(x1, ..., xn) ≠ 0 → ψ(σ(x1), ..., σ(xn)) ≠ 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, ..., fn such that:

ψ(r1, ..., rn) = ψ(f1(V), ..., fn(V))

where V is the Galois Resolvent (see Definition 1, here) and fi ∈ F[X]

(b) But since each fi ∈ F[X] and ψ ∈ F[X], it follows that there exists a function g ∈ F[X] such that:

g(V) = ψ(f1(V), ..., fn(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: ri → fi(V')

(e) So that we have:

ψ(σ(r1), ..., σ(rn)) = ψ(f1(V'), ..., fn(V')) = g(V')

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

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

(5) Assume that f(r1, ..., rn) ∈ F

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

f(r1, ..., rn) = f(f1(V), ..., fn(V))

(7) But since each fi ∈ F[X] and f ∈ F[X], it follows that there exists a function h ∈ F[X] such that:

h(X) = f(f1(X), ..., fn(X))

(8) So that:

h(V) = f(f1(V), ..., fn(V))

and:

h(X) - f(f1(X), ..., fn(X)) = 0

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

H(X) = h(X) - f(f1(X), ..., fn(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, ...., Vm

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

h(Vj) = f(f1(Vj), ..., fn(Vj))

(12) Now since f(r1, ..., rn) ∈ F [by our assumption in step #5], it follows that we can define a function G such that:

G(X) = h(X) - f(r1, ..., rn)

(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, ..., Vm are also roots of G(X).

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

h(Vj) = f(r1, ..., rn)

(15) But if this is true for all Vj, then it follows that it is true for all σ ∈ Gal(F/G) [by Lemma 4, here] so that we have:

f(σj(r1, ..., σj(rn)) = f(r1, ..., rn) for j = 1, ..., m

(16) Assume that f(σj(r1, ..., σj(rn)) = f(r1, ..., rn) for j = 1, ..., m

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

f(r1, ..., rn) = f(f1(V), f2(V), ..., fn(V)) where f, fi ∈ F[X]

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

g(x) = f(f1(x),f2(x), ..., fn(x))

so that we have:

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

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

for all j = 1..m, g(Vj) = f(r1, ..., rn)

(20) Further we note that:

m*f(r1, ..., rn) = g(V1) + g(V2) + ... + g(Vm)

so that we have:

f(r1, ..., rn) = (1/m)*(g(V1) + ... + g(Vm))

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

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

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

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

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

QED

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

## Monday, October 05, 2009

### 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), ..., rn = fn(V)

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

f1(V'), f2(V'), ..., fn(V')

Proof:

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

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

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

ri = fi(V)

with fi(X) ∈ F(X)

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

(5) Since ri = fi(V), we have P(fi(V)) = 0.

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

(7) This shows that each fi(Vj) must correspond to a root ri 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 Vi, that the n roots are:

f1(Vi), f2(Vi), ..., fn(Vi)

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

(10) Assume that fu(Vi) = fv(Vi)

where i ≠ 1

(11) Then Vj is a root of the polynomial fu - fv.

(12) But then all V1, ..., Vm must likewise be roots of fu - fv. [see see Theorem, here]

(13) But this also means that V1 is a root of fu - fv

(14) So it follows that fu(V1) = fv(V1)

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

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

f1(Vi), ..., fn(Vi) 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

## Sunday, October 04, 2009

### 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

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, ..., rn) has a polynomial expression in r1, ..., rn such that:

u = φ(r1, ..., rn)

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

(b) For every element u ∈ F(r1, ..., rn), 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, ..., rn)

Proof:

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

u = φ(r1, ..., rn)

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

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

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

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

(4) Let Θ(X,x1, ..., xn) = Ψ(X,s1, ..., sn)

for some polynomial Ψ with coefficients in F.

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

Θ(X,r1, ..., rn)= Ψ(X,a1, ..., an) ∈ F[X]

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

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

Θ(u,r1, ..., rn) = 0

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

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

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

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

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

Ψ = cP1*...*Pr

(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:

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

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

(14) Finally, we note that there is only one monic irreducible polynomial Pi 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, ..., rn)

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

(18) Since u is a rational fraction of F(r1, ..., rn), 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, ..., rn) for V, we obtain:

u = Q(f(r1, ..., rn))

(21) This is a polynomial expression in r1, ..., rn 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, ..., Vm be the roots of its minimum polynomial over F (among which V lies)

Then:

F(r1, ..., rn) = F(V) = F(V1, ..., Vm)

Proof:

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

(2) So, we have:

F(r1, ..., rn) ⊂ F(V)

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

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

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

(4) So, we have:

F(V1, ..., Vm) ⊂ F(r1, ..., rn)

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

F(V) ⊂ F(V1, ..., Vm)

(6) Therefore we have:

F(r1, ..., rn) = F(V) = F(V1, ..., Vm)

QED

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