Definition 28.9.1. Let $X$ be a scheme. We say $X$ is regular, or nonsingular if for every $x \in X$ there exists an affine open neighbourhood $U \subset X$ of $x$ such that the ring $\mathcal{O}_ X(U)$ is Noetherian and regular.
28.9 Regular schemes
Recall, see Algebra, Definition 10.60.10, that a local Noetherian ring $(R, \mathfrak m)$ is said to be regular if $\mathfrak m$ can be generated by $\dim (R)$ elements. Recall that a Noetherian ring $R$ is said to be regular if every local ring $R_{\mathfrak p}$ of $R$ is regular, see Algebra, Definition 10.110.7.
Lemma 28.9.2. Let $X$ be a scheme. The following are equivalent:
$X$ is regular,
$X$ is locally Noetherian and all of its local rings are regular, and
$X$ is locally Noetherian and for any closed point $x \in X$ the local ring $\mathcal{O}_{X, x}$ is regular.
Proof. By the discussion in Algebra preceding Algebra, Definition 10.110.7 we know that the localization of a regular local ring is regular. The lemma follows by combining this with Lemma 28.5.2, with the existence of closed points on locally Noetherian schemes (Lemma 28.5.9), and the definitions. $\square$
Lemma 28.9.3. Let $X$ be a scheme. The following are equivalent:
The scheme $X$ is regular.
For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is Noetherian and regular.
There exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is Noetherian and regular.
There exists an open covering $X = \bigcup X_ j$ such that each open subscheme $X_ j$ is regular.
Moreover, if $X$ is regular then every open subscheme is regular.
Lemma 28.9.4. A regular scheme is normal.
Proof. See Algebra, Lemma 10.157.5. $\square$
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