Lemma 29.8.6. Let $f : X \to Y$ be a morphism of integral schemes. The following are equivalent

1. $f$ is dominant,

2. $f$ maps the generic point of $X$ to the generic point of $Y$,

3. for some nonempty affine opens $U \subset X$ and $V \subset Y$ with $f(U) \subset V$ the ring map $\mathcal{O}_ Y(V) \to \mathcal{O}_ X(U)$ is injective,

4. for all nonempty affine opens $U \subset X$ and $V \subset Y$ with $f(U) \subset V$ the ring map $\mathcal{O}_ Y(V) \to \mathcal{O}_ X(U)$ is injective,

5. for some $x \in X$ with image $y = f(x) \in Y$ the local ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is injective, and

6. for all $x \in X$ with image $y = f(x) \in Y$ the local ring map $\mathcal{O}_{Y, y} \to \mathcal{O}_{X, x}$ is injective.

Proof. The equivalence of (1) and (2) follows from Lemma 29.8.5. Let $U \subset X$ and $V \subset Y$ be nonempty affine opens with $f(U) \subset V$. Recall that the rings $A = \mathcal{O}_ X(U)$ and $B = \mathcal{O}_ Y(V)$ are integral domains. The morphism $f|_ U : U \to V$ corresponds to a ring map $\varphi : B \to A$. The generic points of $X$ and $Y$ correspond to the prime ideals $(0) \subset A$ and $(0) \subset B$. Thus (2) is equivalent to the condition $(0) = \varphi ^{-1}((0))$, i.e., to the condition that $\varphi$ is injective. In this way we see that (2), (3), and (4) are equivalent. Similarly, given $x$ and $y$ as in (5) the local rings $\mathcal{O}_{X, x}$ and $\mathcal{O}_{Y, y}$ are domains and the prime ideals $(0) \subset \mathcal{O}_{X, x}$ and $(0) \subset \mathcal{O}_{Y, y}$ correspond to the generic points of $X$ and $Y$ (via the identification of the spectrum of the local ring at $x$ with the set of points specializing to $x$, see Schemes, Lemma 26.13.2). Thus we can argue in the exact same manner as above to see that (2), (5), and (6) are equivalent. $\square$

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