Lemma 10.34.2. Let $R$ be a ring. Let $K$ be a field. If $R \subset K$ and $K$ is of finite type over $R$, then there exists an $f \in R$ such that $R_ f$ is a field, and $R_ f \subset K$ is a finite field extension.

Proof. By Lemma 10.30.2 there exist a nonempty open $U \subset \mathop{\mathrm{Spec}}(R)$ contained in the image $\{ (0)\}$ of $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(R)$. Choose $f \in R$, $f \not= 0$ such that $D(f) \subset U$, i.e., $D(f) = \{ (0)\}$. Then $R_ f$ is a domain whose spectrum has exactly one point and $R_ f$ is a field. Then $K$ is a finitely generated algebra over the field $R_ f$ and hence a finite field extension of $R_ f$ by the Hilbert Nullstellensatz (Theorem 10.34.1). $\square$

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