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: r

_{1}, ..., r

_{n}where each root is distinct.

Let V(r

_{1}, ..., r

_{n}) be the Galois resolvent where each distinct permutation has a distinct value.

Then for all roots r

_{i}∈ F(V).

Proof:

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

(2) Let r

_{1}be any root of an equation P

Since r

_{1}is any root, the proof is done if we can show that r

_{i}∈ F(V).

(3) Let:

g(x

_{1}, ..., x

_{n}) = ∏ (for each permutation σ) [V - f(x

_{1}, σ(x

_{2}), ..., σ(x

_{n})) ] ∈ F(V)[x

_{1}, ..., x

_{n}]

where σ runs over all permutations of x

_{2}, ..., x

_{n}

(4) Since g is symmetric in x

_{2}, ..., x

_{n}, g can be written as a polynomial in x

_{1}and the elementary polynomials s

_{1}, ..., s

_{n-1}. (see Lemma, here)

(5) Let:

g(x

_{1}, x

_{2}, ..., x

_{n}) = h(x

_{1}, ..., s

_{n-1})

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

(6) Substituting in various ways the roots r

_{1}, ..., r

_{n}of P for the indeterminates x

_{1}, ..., x

_{n}which has the effect of substituting a

_{1}, ..., a

_{n-1}∈ F for s

_{1}, ..., s

_{n-1}[see Theorem 1, here, for details on the mapping between elementary symmetric polynomials and the coefficents of a polynomial], we obtain:

g(r

_{1}, r

_{2}, ..., r

_{n}) = h(r

_{1}, a

_{1}, ..., a

_{n-1})

and generalizing this, we get:

g(r

_{i}, r

_{1}, ...., r

_{i-1}, r

_{i+1}, ..., r

_{n}) = h(r

_{i}, a

_{1}, ..., a

_{n-1})

(7) Since V is a Galois Resolvent for a function f such that V = f(r

_{1}, ..., r

_{n}), we have:

V ≠ f(r

_{i}, σ(r

_{1}), σ(r

_{2}), ..., σ(r

_{i-1}), σ(r

_{i+1}), ..., σ(r

_{n}))

for i ≠ 1 and for any permutation σ of { r

_{1}, ..., r

_{i-1}, r

_{i+1}, ..., r

_{n}}. [see Lemma 2, here]

(8) Therefore:

g(r

_{i}, r

_{1}, r

_{2}, ..., r

_{i-1}, r

_{i+1}, ...., r

_{n}) ≠ 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(r

_{1}, ..., r

_{n}) = 0

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

h(X,a

_{1}, ..., a

_{n-1}) ∈ F(V)[X]

vanishes for X = r

_{1}but not for X = r

_{i}with i ≠ 1.

(11) Therefore, it is divisible by X - r

_{1}but not by X - r

_{i}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,a

_{1}, ..., a

_{n-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 - r

_{1})*...*(X - r

_{n}), we know that X - r

_{1}divides P(X).

(14) From step #11 above, we know that X - r

_{1}divides h.

(15) Since x - r

_{1}divides both P(X) and h(X,a

_{1}, ..., a

_{n-1}), it divides D.

(15) On the other hand, h(X,a

_{1}, ..., a

_{n-1}) is not divisible by X - r

_{i}for i ≠ 1. [see step #10 above]

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

_{1}.

(17) Thus, D = X - r

_{1}whence r

_{1}∈ F(V) since D ∈ F(V)[X].

QED

References

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