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).
Proof:
(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].
QED
References
- Harold M. Edwards, Galois Theory
, Springer, 1984
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations