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, ..., r_n where each root is distinct.

Let V(r₁, ..., 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 r₁ be any root of an equation P

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

(3) Let:

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

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

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

(5) Let:

g(x₁, x₂, ..., x_n) = h(x₁, ..., s_n-1)

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

(6) Substituting in various ways the roots r₁, ..., r_n of P for the indeterminates x₁, ..., x_n which has the effect of substituting a₁, ..., a_n-1 ∈ F for s₁, ..., 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(r₁, r₂, ..., r_n) = h(r₁, a₁, ..., a_n-1)

and generalizing this, we get:

g(r_i, r₁, ...., r_i-1, r_i+1, ..., r_n) = h(r_i, a₁, ..., a_n-1)

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

V ≠ f(r_i, σ(r₁), σ(r₂), ..., σ(r_i-1), σ(r_i+1), ..., σ(r_n))

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

(8) Therefore:

g(r_i, r₁, r₂, ..., 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(r₁, ..., r_n) = 0

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

h(X,a₁, ..., a_n-1) ∈ F(V)[X]

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

(11) Therefore, it is divisible by X - r₁ 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,a₁, ..., 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 - r₁)*...*(X - r_n), 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, ..., a_n-1), it divides D.

(15) On the other hand, h(X,a₁, ..., 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 - 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, World Scientific, 2001