Theorem:
Let f(x) be a rational function in one indeterminate over a field F.
Let V be a root of an irreducible polynomial R(x) ∈ F[X]
If f(V) = 0, then all roots of P are also roots of f.
Proof:
(1) Let f = P/Q where P,Q are polynomials and ∈ F[X] and Q(V) ≠ 0 and P(V) = 0.
(2) Since V is a root an irreducible polynomial R, it follows that R divides P. [see Lemma 2, here]
(3) Let W be any root of R.
(4) It follows that W is also a root of P since:
(a) If W is a root a R, then x - W divides R.
(b) Then it follows that x - W must also divide P since R divides P.
(4) It also follows that if W is a root of R, then W is not a root of Q since:
(a) Assume that W is a root of Q so that Q(W)=0
(b) Then it would follow that R divides Q.
(c) But then Q(V)=0 since R(V)=0
(d) But this is not true since Q(V) ≠ 0.
(e) Therefore, we reject our assumption in step #4a.
(5) Therefore, we have shown that P(W)=0 and Q(W) ≠ 0
(6) Hence f(W)=0 for any root W of the irreducible polynomial R.
QED
References
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations
, World Scientific, 2001
