## 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, , World Scientific, 2001