Friday, October 02, 2009

Galois' Memoir: Lemma 3

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 3:

When a function V is chosen as satisfies Lemma 2 (see here), it will have the property that all the roots of the given equation can be expressed as rational functions of V.

That is:

Let P be a function of degree n with the roots: r1, ..., rn where each root is distinct.

Let V(r1, ..., rn) be the Galois resolvent where each distinct permutation has a distinct value.

Then for all roots ri ∈ F(V).


(1) Let V be a Galois Resolvent [see Definition 2, here]

(2) Let r1 be any root of an equation P

Since r1 is any root, the proof is done if we can show that ri ∈ F(V).

(3) Let:

g(x1, ..., xn) = ∏ (for each permutation σ) [V - f(x1, σ(x2), ..., σ(xn)) ] ∈ F(V)[x1, ..., xn]

where σ runs over all permutations of x2, ..., xn

(4) Since g is symmetric in x2, ..., xn, g can be written as a polynomial in x1 and the elementary polynomials s1, ..., sn-1. (see Lemma, here)

(5) Let:

g(x1, x2, ..., xn) = h(x1, ..., sn-1)

for some polynomial h with coefficients in F(V).

(6) Substituting in various ways the roots r1, ..., rn of P for the indeterminates x1, ..., xn which has the effect of substituting a1, ..., an-1 ∈ F for s1, ..., sn-1 [see Theorem 1, here, for details on the mapping between elementary symmetric polynomials and the coefficents of a polynomial], we obtain:

g(r1, r2, ..., rn) = h(r1, a1, ..., an-1)

and generalizing this, we get:

g(ri, r1, ...., ri-1, ri+1, ..., rn) = h(ri, a1, ..., an-1)

(7) Since V is a Galois Resolvent for a function f such that V = f(r1, ..., rn), we have:

V ≠ f(ri, σ(r1), σ(r2), ..., σ(ri-1), σ(ri+1), ..., σ(rn))

for i ≠ 1 and for any permutation σ of { r1, ..., ri-1, ri+1, ..., rn }. [see Lemma 2, here]

(8) Therefore:

g(ri, r1, r2, ..., ri-1, ri+1, ...., rn) ≠ 0 for i ≠ 1.

(9) On the other hand, the definition of g and V show that [see definition of g in step #3 above]:

g(r1, ..., rn) = 0

(10) From step #9 above and step #6 above, we know that:

h(X,a1, ..., an-1) ∈ F(V)[X]

vanishes for X = r1 but not for X = ri with i ≠ 1.

(11) Therefore, it is divisible by X - r1 but not by X - ri for i ≠ 1. [This follows from Girard's Theorem, see here]

(12) We then consider the monic greatest common divisor D(X) of P(X) and h(X,a1, ..., an-1) in F(V)[X]. [see Theorem 1, here for proof of the existence of a greatest common divisor for polynomials]

(13) Since P(X) = (X - r1)*...*(X - rn), we know that X - r1 divides P(X).

(14) From step #11 above, we know that X - r1 divides h.

(15) Since x - r1 divides both P(X) and h(X,a1, ..., an-1), it divides D.

(15) On the other hand, h(X,a1, ..., an-1) is not divisible by X - ri for i ≠ 1. [see step #10 above]

(16) Hence, D has no other factor than X - r1.

(17) Thus, D = X - r1 whence r1 ∈ F(V) since D ∈ F(V)[X].



1 comment:

checkdrummer said...

Hi, why are please ai from F? Although they are the values of elementary symetric polynomials - but only on variables r2,..,rn (and not r1 also) - and so I think they cannot be expressed (necessarily) by coeffitients of P.
E.g. having cubic equation with rational coeffitients and roots x1,x2,x3, then elementary symmetric polynomials on x2,x3 are also not necessarily rational...
Thank you very much for your explanation