The Stacks project

27.3 Integral, irreducible, and reduced schemes

Definition 27.3.1. 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 27.3.2. Let $X$ be a scheme. The following are equivalent.

  1. The scheme $X$ is reduced, see Schemes, Definition 25.12.1.

  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 27.3.3. 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 25.11.1 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 25.6.8. Hence it is irreducible by assumption which is absurd. We conclude that (3) implies (2). The lemma is proved. $\square$

Lemma 27.3.4. 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 27.3.2 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 27.3.2. Moreover, every nonempty affine open is irreducible. Hence $X$ is irreducible, see Lemma 27.3.3. $\square$

In Examples, Section 104.5 we construct a connected affine scheme all of whose local rings are domains, but which is not integral.


Comments (2)

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 , then cannot contain a point in , hence , ( otherwise the 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 , then cannot contain a point in , hence , ( otherwise the forms a decomposition. ) ?


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01OJ. Beware of the difference between the letter 'O' and the digit '0'.