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')


(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.]



No comments: