Lemma 9.6.8 (Classification of simple extensions). If a field extension F/k is generated by one element, then it is k-isomorphic either to the rational function field k(t)/k or to one of the extensions k[t]/(P) for P \in k[t] irreducible.
Proof. Let \alpha \in F be such that F = k(\alpha ); by assumption, such an \alpha exists. There is a morphism of rings
sending the indeterminate t to \alpha . The image is a domain, so the kernel is a prime ideal. Thus, it is either (0) or (P) for P \in k[t] irreducible.
If the kernel is (P) for P \in k[t] irreducible, then the map factors through k[t]/(P), and induces a morphism of fields k[t]/(P) \to F. Since the image contains \alpha , we see easily that the map is surjective, hence an isomorphism. In this case, k[t]/(P) \simeq F.
If the kernel is trivial, then we have an injection k[t] \to F. One may thus define a morphism of the quotient field k(t) into F; given a quotient R(t)/Q(t) with R(t), Q(t) \in k[t], we map this to R(\alpha )/Q(\alpha ). The hypothesis that k[t] \to F is injective implies that Q(\alpha ) \neq 0 unless Q is the zero polynomial. The quotient field of k[t] is the rational function field k(t), so we get a morphism k(t) \to F whose image contains \alpha . It is thus surjective, hence an isomorphism. \square
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