For review of polynomials, see here. For review of irreducible polynomials, see here. For review of fields, see here. For review of rings, see here. For review of ideals, see here.
Lemma 1:
Let P be a polynomial ∈ F[X] where F is a field.
Let (P) be the set of multiples of P such that (P) = {PQ where Q ∈ F[X]}
Let P be irreducible over F[X]
Then:
F[X]/(P) is a field containing F and a root of P.
Proof:
(1) Since (P) is an ideal of F[X] (see Lemma 1, here), it follows that F[X]/(P) is a ring (see Lemma 3, here)
(2) To prove that F[X]/(P) is a field, we need to show that every nonzero element Q + (P) in F[X]/(P) is invertible and that F[X]/P has unity. [See Definition 3, here for definition of a field]
(3) Since 1 ∈ F[X], it follows that 1 + (P) ∈ F[X]/P and this shows that F[X]/P has unity since (see here for review of the operations of a quotient ring):
[x + (P)]*[1 + (P)] = (x*1) + (P) = x + (P)
(4) Assume that Q + (P) in F[X]/(P) is a nonzero element.
(5) It follows that Q is not in (P) [since if it was Q + (P) = 0 + (P), see Lemma 2, here, but by assumption Q + (P) is nonzero]
(6) So, Q is not divisible by P. [From the definition of (P)]
(7) Since P is irreducible over F[X], P,Q are relatively prime. [see Definition 1, here]
(8) Therefore, there exists P1, Q1 ∈ F[X] such that (See Corollary 3.1, here):
PP1 + QQ1 = 1
(9) Since -PP1 = QQ1 - 1, it follows that:
P divides QQ1 - 1
(10) Since P divides QQ1 - 1, it follows that:
QQ1 - 1 ∈ (P) [See Definition of (P) above]
(11) But then [see here]:
QQ1 + (P) = 1 + (P)
(10) From a previous result (see Lemma 1, here), this shows that:
QQ1 + (P) = 1 + (P)
(11) Since (Q + (P))*(Q1 + (P)) = QQ1 + (P) [see here for review of the operations of a quotient ring], it follows that:
(Q+(P))*(Q1+(P)) = 1 + (P)
(12) This shows that Q1 + (P) is the inverse of Q + (P) in F[X]/(P)
(13) Assume that x ∈ F and x is nonzero
(14) Then, x + (P) is nonzero since no nonzero is divisible by P. [See here for review of polynomials and divisibility]
(15) Thus, F is a subfield of F[X]/(P) since there is a clear mapping from F[X] to F[X]/(P).
(16) Using the operations defined for quotient rings, we get:
P(X + (P)) = P(X) + (P) since:
(a) Let P(X) = aXn + bXn-1 + ... + cX + d
(b) P(X + (P)) = a(X + (P))n + b(X + (P))n-1 + ... + c(X + (P)) + d =
= a[Xn + (P)] + b[Xn-1 + (P)] + ... + cX + (P) + d =
= (aXn + bXn-1 + ... + cX + d) + (P)
(17) Since P(X) ∈ (P), it follows that P(X) + (P) = 0 + (P) [See Lemma 2, here]
(18) Combining step #16 and step #17, gives us:
P(X + (P)) = 0
(19) This shows that X+(P) is a root of P in F[X]/(P).
QED
Theorem: Girard's Theorem
For any nonconstant polynomial P ∈ F[X], there is a field K containing F such that P splits over K into a product of linear factors:
P = a(X - x1)*...*(X - xn) in K[X]
Proof:
(1) Let P be a polynomial such that P ∈ a field F.
(2) By the nature of polynomials (see Theorem 3, here), we can break up P into a product of irreducible factors in F[X] such that:
P = P1*...*Pr
(3) Let s be the number of linear factors in P so that s is between 0 and r.
(4) If (deg P) - s = 0, then each of the factors P1, ..., Pr is linear and K = F.
(5) If (deg P) - s ≥ 1, then at least one of the factors P1, ..., Pr has degreee greater than or equal to 2.
(6) Assume that deg P1 ≥ 2.
(7) Let F1 = F[X]/(P1)
(8) From Lemma 1 above, we know that P1 has a root in F1
(9) So, we can decompose P1 over F1 with at least one linear factor. [See Theorem, here]
(10) The decomposition of P into irreducible factors over F1 is then at least s+1 .
(11) We can repeat this same sequence for all factors of P1 and all factors of P.
(12) In this way, we can our field using Lemma 1 above and find a field K where P can be decomposed into linear factors.
QED
References
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations
, World Scientific, 2001
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