
Lemma 10.29.6. Let $R \to S$ be a ring map. The following are equivalent:

1. The kernel of $R \to S$ consists of nilpotent elements.

2. The minimal primes of $R$ are in the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$.

3. The image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is dense in $\mathop{\mathrm{Spec}}(R)$.

Proof. Let $I = \mathop{\mathrm{Ker}}(R \to S)$. Note that $\sqrt{(0)} = \bigcap _{\mathfrak q \subset S} \mathfrak q$, see Lemma 10.16.2. Hence $\sqrt{I} = \bigcap _{\mathfrak q \subset S} R \cap \mathfrak q$. Thus $V(I) = V(\sqrt{I})$ is the closure of the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$. This shows that (1) is equivalent to (3). It is clear that (2) implies (3). Finally, assume (1). We may replace $R$ by $R/I$ and $S$ by $S/IS$ without affecting the topology of the spectra and the map. Hence the implication (1) $\Rightarrow$ (2) follows from Lemma 10.29.5. $\square$

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