Lemma 28.3.3. Let $X$ be a scheme. The following are equivalent.

1. The scheme $X$ is irreducible.

2. There exists an affine open covering $X = \bigcup _{i \in I} U_ i$ such that $I$ is not empty, $U_ i$ is irreducible for all $i \in I$, and $U_ i \cap U_ j \not= \emptyset$ for all $i, j \in I$.

3. The scheme $X$ is nonempty and every nonempty affine open $U \subset X$ is irreducible.

Proof. Assume (1). By Schemes, Lemma 26.11.1 we see that $X$ has a unique generic point $\eta$. Then $X = \overline{\{ \eta \} }$. Hence $\eta$ is an element of every nonempty affine open $U \subset X$. This implies that $U = \overline{\{ \eta \} }$ and that any two nonempty affines meet. Thus (1) implies both (2) and (3).

Assume (2). Suppose $X = Z_1 \cup Z_2$ is a union of two closed subsets. For every $i$ we see that either $U_ i \subset Z_1$ or $U_ i \subset Z_2$. Pick some $i \in I$ and assume $U_ i \subset Z_1$ (possibly after renumbering $Z_1$, $Z_2$). For any $j \in I$ the open subset $U_ i \cap U_ j$ is dense in $U_ j$ and contained in the closed subset $Z_1 \cap U_ j$. We conclude that also $U_ j \subset Z_1$. Thus $X = Z_1$ as desired.

Assume (3). Choose an affine open covering $X = \bigcup _{i \in I} U_ i$. We may assume that each $U_ i$ is nonempty. Since $X$ is nonempty we see that $I$ is not empty. By assumption each $U_ i$ is irreducible. Suppose $U_ i \cap U_ j = \emptyset$ for some pair $i, j \in I$. Then the open $U_ i \amalg U_ j = U_ i \cup U_ j$ is affine, see Schemes, Lemma 26.6.8. Hence it is irreducible by assumption which is absurd. We conclude that (3) implies (2). The lemma is proved. $\square$

Comment #7536 by Marco Baracchini on

I think we have to add a bar over $U$ in the second line of the proof:

you proved that $\eta$ is in $U$ for each $U$ open affine in $X$, then $X=\bar{\{\eta\}}=\bar{U}$, then each open affine set is dense in $X$.

Since open affine are basis for the topology, each open non empty set is dense in $X$, then each open non empty set in $U$ is dense in $U$ and we conclude that $U$ is irreducible for each $U$ open (why do we require affine?) set of $X$.

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