Corollary 4. If k is a field, then every ideal of k[x] can be written as ⟨f⟩ for some f ∈ k[x]. Furthermore, f is unique up to multiplication by a nonzero constant in k.
Proof. Take an ideal I ⊆ k[x]. If I = {0}, then we are done since I = ⟨0⟩. Otherwise, let f be a nonzero polynomial of minimum degree contained in I. We claim that ⟨f⟩ = I. The inclusion ⟨f⟩ ⊆ I is obvious since I is an ideal. Going the other way, take g ∈ I. By division algorithm, we have g = qf + r, where either r = 0 or deg(r) < deg(f). Since I is an ideal, qf ∈ I and, thus, r = g − qf ∈ I. If r were not zero, then deg(r) < deg(f), which would contradict our choice of f. Thus, r = 0, so that g = qf ∈ ⟨f⟩. This proves that I = ⟨f⟩
Problem/Frage:
Sagt uns dieser Beweis schon irgendwie aus, dass der ggT(,) ein Erzeuger des Ideals sein wird? Falls ja kannst du mir das bitte schonend erklären? Brauche das für einen Vortrag und sehe es leider nicht.
