Lemma 10.47.1. Let $R \to S$ be a ring map. Assume

1. $\mathop{\mathrm{Spec}}(R)$ is irreducible,

2. $R \to S$ is flat,

3. $R \to S$ is of finite presentation,

4. the fibre rings $S \otimes _ R \kappa (\mathfrak p)$ have irreducible spectra for a dense collection of primes $\mathfrak p$ of $R$.

Then $\mathop{\mathrm{Spec}}(S)$ is irreducible. This is true more generally with (b) $+$ (c) replaced by “the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is open”.

Proof. The assumptions (b) and (c) imply that the map on spectra is open, see Proposition 10.41.8. Hence the lemma follows from Topology, Lemma 5.8.14. $\square$

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