## 28.3 Integral, irreducible, and reduced schemes

Definition 28.3.1. Let $X$ be a scheme. We say $X$ is integral if it is nonempty and for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is an integral domain.

Lemma 28.3.2. Let $X$ be a scheme. The following are equivalent.

1. The scheme $X$ is reduced, see Schemes, Definition 26.12.1.

2. There exists an affine open covering $X = \bigcup U_ i$ such that each $\Gamma (U_ i, \mathcal{O}_ X)$ is reduced.

3. For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is reduced.

4. For every open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is reduced.

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. $\square$

Lemma 28.3.4. A scheme $X$ is integral if and only if it is reduced and irreducible.

Proof. If $X$ is irreducible, then every affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ is irreducible. If $X$ is reduced, then $R$ is reduced, by Lemma 28.3.2 above. Hence $R$ is reduced and $(0)$ is a prime ideal, i.e., $R$ is an integral domain.

If $X$ is integral, then for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is reduced and hence $X$ is reduced by Lemma 28.3.2. Moreover, every nonempty affine open is irreducible. Hence $X$ is irreducible, see Lemma 28.3.3. $\square$

In Examples, Section 108.6 we construct a connected affine scheme all of whose local rings are domains, but which is not integral.

Comment #4228 by Lin on

Correct me If am wrong, it seems that (2) to (1) of lemma [01OM], 2nd paragraph, follows directly from the first line. If $U_j \cap U_i \not= \emptyset$, then $U_j$ cannot contain a point in $Z_2$, hence $U_j \subseteq Z_1$, ( otherwise the $Z_1, Z_2$ forms a decomposition. ) ?

Comment #4229 by Lin on

Correct me If am wrong, it seems that (2) to (1) of lemma [01OM], 2nd paragraph, follows directly from the first line. If $U_j \cap U_i \not= \emptyset$, then $U_j$ cannot contain a point in $Z_2$, hence $U_j \subseteq Z_1$, ( otherwise the $Z_1, Z_2$ forms a decomposition. ) ?

Comment #4409 by on

Hmm... I think you are saying the same thing as the proof says.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).