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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