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 $K/R_ f$ 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). $\square$

Comment #8306 by William Sun on

Other than overkilling with Chevalley's theorem, here is an more elementary argument. $K$ is of finite type over $\operatorname{Frac}(R)$, thus finite and thus integral. Choose generators of $K$ over $R$ and consider their minimal polynomials with coefficients in $\operatorname{Frac}(R)$. By inverting the product $f$ of the (finite number of) denominators, we conclude the generators of $K$ are integral elements over $R_f$. Thus $K$ is integral over $R_f$, and $R_f$ is a field, as desired.

Comment #8930 by on

Dear William, I note that the first step of your proof uses the Hilbert Nullstellensatz (unless I am overlooking something). Now our proof of the Nullstellensatz already uses Chevalley. So this wouldn't avoid using Chevalley. Additionally, there is a step in your proof which uses a lemma that comes after this one, so we can't use that here.

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