Lemma 29.51.7. Let $X$, $Y$ be integral schemes. Let $f : X \to Y$ be locally of finite type. Assume $f$ is dominant. The following are equivalent:

the extension $R(Y) \subset R(X)$ has transcendence degree $0$,

the extension $R(Y) \subset R(X)$ is finite,

there exist nonempty affine opens $U \subset X$ and $V \subset Y$ such that $f(U) \subset V$ and $f|_ U : U \to V$ is finite, and

the generic point of $X$ is the only point of $X$ mapping to the generic point of $Y$.

If $f$ is separated or if $f$ is quasi-compact, then these are also equivalent to

there exists a nonempty affine open $V \subset Y$ such that $f^{-1}(V) \to V$ is finite.

**Proof.**
Choose any affine opens $\mathop{\mathrm{Spec}}(A) = U \subset X$ and $\mathop{\mathrm{Spec}}(R) = V \subset Y$ such that $f(U) \subset V$. Then $R$ and $A$ are domains by definition. The ring map $R \to A$ is of finite type (Lemma 29.15.2). By Lemma 29.8.6 the generic point of $X$ maps to the generic point of $Y$ hence $R \to A$ is injective. Let $K = R(Y)$ be the fraction field of $R$ and $L = R(X)$ the fraction field of $A$. Then $L/K$ is a finitely generated field extension. Hence we see that (1) is equivalent to (2).

Suppose (2) holds. Let $x_1, \ldots , x_ n \in A$ be generators of $A$ over $R$. By assumption there exist nonzero polynomials $P_ i(X) \in R[X]$ such that $P_ i(x_ i) = 0$. Let $f_ i \in R$ be the leading coefficient of $P_ i$. Then we conclude that $R_{f_1 \ldots f_ n} \to A_{f_1 \ldots f_ n}$ is finite, i.e., (3) holds. Note that (3) implies (2). So now we see that (1), (2) and (3) are all equivalent.

Let $\eta $ be the generic point of $X$, and let $\eta ' \in Y$ be the generic point of $Y$. Assume (4). Then $\dim _\eta (X_{\eta '}) = 0$ and we see that $R(X) = \kappa (\eta )$ has transcendence degree $0$ over $R(Y) = \kappa (\eta ')$ by Lemma 29.28.1. In other words (1) holds. Assume the equivalent conditions (1), (2) and (3). Suppose that $x \in X$ is a point mapping to $\eta '$. As $x$ is a specialization of $\eta $, this gives inclusions $R(Y) \subset \mathcal{O}_{X, x} \subset R(X)$, which implies $\mathcal{O}_{X, x}$ is a field, see Algebra, Lemma 10.36.19. Hence $x = \eta $. Thus we see that (1) – (4) are all equivalent.

It is clear that (5) implies (3) with no additional assumptions on $f$. What remains is to prove that if $f$ is either separated or quasi-compact, then the equivalent conditions (1) – (4) imply (5). This follows from Lemma 29.51.5.
$\square$

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