
Lemma 10.29.4. Let $R$ be a domain. Let $\varphi : R \to S$ be a ring map. The following are equivalent:

1. The ring map $R \to S$ is injective.

2. The image $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ contains a dense set of points.

3. There exists a prime ideal $\mathfrak q \subset S$ whose inverse image in $R$ is $(0)$.

Proof. Let $K$ be the field of fractions of the domain $R$. Assume that $R \to S$ is injective. Since localization is exact we see that $K \to S \otimes _ R K$ is injective. Hence there is a prime mapping to $(0)$ by Lemma 10.16.9.

Note that $(0)$ is dense in $\mathop{\mathrm{Spec}}(R)$, so that the last condition implies the second.

Suppose the second condition holds. Let $f \in R$, $f \not= 0$. As $R$ is a domain we see that $V(f)$ is a proper closed subset of $R$. By assumption there exists a prime $\mathfrak q$ of $S$ such that $\varphi (f) \not\in \mathfrak q$. Hence $\varphi (f) \not= 0$. Hence $R \to S$ is injective. $\square$

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