Definition 28.14.1. Let $X$ be a scheme. We say $X$ is a G-scheme1 if for every $x \in X$ there exists an affine open neighbourhood $x \in U \subset X$ such that the ring $\mathcal{O}_ X(U)$ is a G-ring (see More on Algebra, Definition 15.51.1).
28.14 G-schemes
We remark that a ring $R$ is a G-ring if $R$ is Noetherian and for every prime ideal $\mathfrak p$ of $R$ the ring map $R_\mathfrak p \to (R_\mathfrak p)^\wedge $ is regular.
Lemma 28.14.2. Let $X$ be a scheme. The following are equivalent:
$X$ is a G-scheme,
$X$ is locally Noetherian and for all $x \in X$ the ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{X, x}^\wedge $ is regular, and
$X$ is locally Noetherian and for any closed point $x \in X$ the ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{X, x}^\wedge $ is regular.
Proof. It is clear that (1) implies (2) and (2) implies (3). Assume (3). Let $x \in X$ be any point. By Lemma 28.5.9 there exists a specialization $x \leadsto x'$ with $x'$ closed in $X$. Then $\mathcal{O}_{X, x'}$ is a G-ring by More on Algebra, Lemma 15.51.7. Since $\mathcal{O}_{X, x}$ is the localization of $\mathcal{O}_{X, x'}$ at a prime ideal (Schemes, Lemma 26.13.2), we see that $\mathcal{O}_{X, x} \to \mathcal{O}_{X, x}^\wedge $ is a regular ring map. Since $x \in X$ was arbitrary we conclude that for any affine open $U \subset X$ the Noetherian (Lemma 28.5.2) ring $\mathcal{O}_ X(U)$ is a G-ring. This proves that (1) holds. $\square$
Lemma 28.14.3. Let $X$ be a scheme. The following are equivalent:
The scheme $X$ is a G-scheme.
For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is a G-ring.
There exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is a G-ring.
There exists an open covering $X = \bigcup X_ j$ such that each open subscheme $X_ j$ is a G-scheme.
Moreover, if $X$ is a G-scheme then every open subscheme is a G-scheme.
[1] This may be nonstandard terminology. If $G$ is a finite group or group scheme then sometimes a $G$-scheme denotes a scheme equipped with an action of $G$.
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