# The Stacks Project

## Tag 033H

### 27.7. Normal schemes

Recall that a ring $R$ is said to be normal if all its local rings are normal domains, see Algebra, Definition 10.36.11. A normal domain is a domain which is integrally closed in its field of fractions, see Algebra, Definition 10.36.1. Thus it makes sense to define a normal scheme as follows.

Definition 27.7.1. A scheme $X$ is normal if and only if for all $x \in X$ the local ring $\mathcal{O}_{X, x}$ is a normal domain.

This seems to be the definition used in EGA, see [EGA, 0, 4.1.4]. Suppose $X = \mathop{\rm Spec}(A)$, and $A$ is reduced. Then saying that $X$ is normal is not equivalent to saying that $A$ is integrally closed in its total ring of fractions. However, if $A$ is Noetherian then this is the case (see Algebra, Lemma 10.36.16).

Lemma 27.7.2. Let $X$ be a scheme. The following are equivalent:

1. The scheme $X$ is normal.
2. For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$ is normal.
3. There exists an affine open covering $X = \bigcup U_i$ such that each $\mathcal{O}_X(U_i)$ is normal.
4. There exists an open covering $X = \bigcup X_j$ such that each open subscheme $X_j$ is normal.

Moreover, if $X$ is normal then every open subscheme is normal.

Proof. This is clear from the definitions. $\square$

Lemma 27.7.3. A normal scheme is reduced.

Proof. Immediate from the definitions. $\square$

Lemma 27.7.4. Let $X$ be an integral scheme. Then $X$ is normal if and only if for every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$ is a normal domain.

Proof. This follows from Algebra, Lemma 10.36.10. $\square$

Lemma 27.7.5. Let $X$ be a scheme such that any quasi-compact open has a finite number of irreducible components. The following are equivalent:

1. $X$ is normal, and
2. $X$ is a disjoint union of normal integral schemes.

Proof. It is immediate from the definitions that (2) implies (1). Let $X$ be a normal scheme such that every quasi-compact open has a finite number of irreducible components. If $X$ is affine then $X$ satisfies (2) by Algebra, Lemma 10.36.16. For a general $X$, let $X = \bigcup X_i$ be an affine open covering. Note that also each $X_i$ has but a finite number of irreducible components, and the lemma holds for each $X_i$. Let $T \subset X$ be an irreducible component. By the affine case each intersection $T \cap X_i$ is open in $X_i$ and an integral normal scheme. Hence $T \subset X$ is open, and an integral normal scheme. This proves that $X$ is the disjoint union of its irreducible components, which are integral normal schemes. There are only finitely many by assumption. $\square$

Lemma 27.7.6. Let $X$ be a Noetherian scheme. The following are equivalent:

1. $X$ is normal, and
2. $X$ is a finite disjoint union of normal integral schemes.

Proof. This is a special case of Lemma 27.7.5 because a Noetherian scheme has a Noetherian underlying topological space (Lemma 27.5.5 and Topology, Lemma 5.9.2. $\square$

Lemma 27.7.7. Let $X$ be a locally Noetherian scheme. The following are equivalent:

1. $X$ is normal, and
2. $X$ is a disjoint union of integral normal schemes.

Proof. Omitted. Hint: This is purely topological from Lemma 27.7.6. $\square$

Remark 27.7.8. Let $X$ be a normal scheme. If $X$ is locally Noetherian then we see that $X$ is integral if and only if $X$ is connected, see Lemma 27.7.7. But there exists a connected affine scheme $X$ such that $\mathcal{O}_{X, x}$ is a domain for all $x \in X$, but $X$ is not irreducible, see Examples, Section 100.5. This example is even a normal scheme (proof omitted), so beware!

Lemma 27.7.9. Let $X$ be an integral normal scheme. Then $\Gamma(X, \mathcal{O}_X)$ is a normal domain.

Proof. Set $R = \Gamma(X, \mathcal{O}_X)$. It is clear that $R$ is a domain. Suppose $f = a/b$ is an element of its fraction field which is integral over $R$. Say we have $f^d + \sum_{i = 0, \ldots, d - 1} a_i f^i = 0$ with $a_i \in R$. Let $U \subset X$ be affine open. Since $b \in R$ is not zero and since $X$ is integral we see that also $b|_U \in \mathcal{O}_X(U)$ is not zero. Hence $a/b$ is an element of the fraction field of $\mathcal{O}_X(U)$ which is integral over $\mathcal{O}_X(U)$ (because we can use the same polynomial $f^d + \sum_{i = 0, \ldots, d - 1} a_i|_U f^i = 0$ on $U$). Since $\mathcal{O}_X(U)$ is a normal domain (Lemma 27.7.2), we see that $f_U = (a|_U)/(b|_U) \in \mathcal{O}_X(U)$. It is easy to see that $f_U|_V = f_V$ whenever $V \subset U \subset X$ are affine open. Hence the local sections $f_U$ glue to a global section $f$ as desired. $\square$

The code snippet corresponding to this tag is a part of the file properties.tex and is located in lines 747–922 (see updates for more information).

\section{Normal schemes}
\label{section-normal}

\noindent
Recall that a ring $R$ is said to be normal if all its local rings
are normal domains,
see Algebra, Definition \ref{algebra-definition-ring-normal}.
A normal domain is a domain which is integrally closed in its field
of fractions, see
Algebra, Definition \ref{algebra-definition-domain-normal}.
Thus it makes sense to define a normal scheme as follows.

\begin{definition}
\label{definition-normal}
A scheme $X$ is {\it normal} if and only if for all $x \in X$ the local ring
$\mathcal{O}_{X, x}$ is a normal domain.
\end{definition}

\noindent
This seems to be the definition used in EGA, see \cite[0, 4.1.4]{EGA}.
Suppose $X = \Spec(A)$, and $A$ is reduced. Then saying that $X$ is
normal is not equivalent to saying that $A$ is integrally closed in its
total ring of fractions. However, if $A$ is Noetherian then this is the case
(see Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal}).

\begin{lemma}
\label{lemma-locally-normal}
Let $X$ be a scheme. The following are equivalent:
\begin{enumerate}
\item The scheme $X$ is normal.
\item For every affine open $U \subset X$ the ring $\mathcal{O}_X(U)$
is normal.
\item There exists an affine open covering $X = \bigcup U_i$ such that
each $\mathcal{O}_X(U_i)$ is normal.
\item There exists an open covering $X = \bigcup X_j$
such that each open subscheme $X_j$ is normal.
\end{enumerate}
Moreover, if $X$ is normal then every open subscheme
is normal.
\end{lemma}

\begin{proof}
This is clear from the definitions.
\end{proof}

\begin{lemma}
\label{lemma-normal-reduced}
A normal scheme is reduced.
\end{lemma}

\begin{proof}
Immediate from the definitions.
\end{proof}

\begin{lemma}
\label{lemma-integral-normal}
Let $X$ be an integral scheme.
Then $X$ is normal if and only if for every affine open
$U \subset X$ the ring $\mathcal{O}_X(U)$ is a normal domain.
\end{lemma}

\begin{proof}
This follows from
Algebra, Lemma \ref{algebra-lemma-normality-is-local}.
\end{proof}

\begin{lemma}
\label{lemma-normal-locally-finite-nr-irreducibles}
Let $X$ be a scheme such that any quasi-compact open has a finite number
of irreducible components. The following are equivalent:
\begin{enumerate}
\item $X$ is normal, and
\item $X$ is a disjoint union of normal integral schemes.
\end{enumerate}
\end{lemma}

\begin{proof}
It is immediate from the definitions that (2) implies (1).
Let $X$ be a normal scheme such that every quasi-compact open
has a finite number of irreducible components.
If $X$ is affine then $X$ satisfies (2) by
Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal}.
For a general $X$, let $X = \bigcup X_i$ be
an affine open covering. Note that also each $X_i$ has
but a finite number of irreducible components, and the lemma holds
for each $X_i$. Let $T \subset X$ be an irreducible component.
By the affine case each intersection $T \cap X_i$ is open in $X_i$
and an integral normal scheme.
Hence $T \subset X$ is open, and an integral normal scheme.
This proves that $X$ is the disjoint union of its irreducible components,
which are integral normal schemes. There are only finitely many
by assumption.
\end{proof}

\begin{lemma}
\label{lemma-normal-Noetherian}
Let $X$ be a Noetherian scheme.
The following are equivalent:
\begin{enumerate}
\item $X$ is normal, and
\item $X$ is a finite disjoint union of normal integral schemes.
\end{enumerate}
\end{lemma}

\begin{proof}
This is a special case of
Lemma \ref{lemma-normal-locally-finite-nr-irreducibles} because a Noetherian
scheme has a Noetherian underlying topological space
(Lemma \ref{lemma-Noetherian-topology}
and
Topology, Lemma \ref{topology-lemma-Noetherian}.
\end{proof}

\begin{lemma}
\label{lemma-normal-locally-Noetherian}
Let $X$ be a locally Noetherian scheme.
The following are equivalent:
\begin{enumerate}
\item $X$ is normal, and
\item $X$ is a disjoint union of integral normal schemes.
\end{enumerate}
\end{lemma}

\begin{proof}
Omitted. Hint: This is purely topological from
Lemma \ref{lemma-normal-Noetherian}.
\end{proof}

\begin{remark}
\label{remark-normal-connected-irreducible}
Let $X$ be a normal scheme. If $X$ is locally Noetherian then we see that
$X$ is integral if and only if $X$ is connected, see
Lemma \ref{lemma-normal-locally-Noetherian}.
But there exists a connected affine scheme $X$ such that
$\mathcal{O}_{X, x}$ is a domain for all $x \in X$, but $X$ is not
irreducible, see Examples, Section
\ref{examples-section-connected-locally-integral-not-integral}.
This example is even a normal scheme (proof omitted), so beware!
\end{remark}

\begin{lemma}
\label{lemma-normal-integral-sections}
Let $X$ be an integral normal scheme.
Then $\Gamma(X, \mathcal{O}_X)$ is a normal domain.
\end{lemma}

\begin{proof}
Set $R = \Gamma(X, \mathcal{O}_X)$.
It is clear that $R$ is a domain.
Suppose $f = a/b$ is an element of its fraction field
which is integral over $R$. Say we have
$f^d + \sum_{i = 0, \ldots, d - 1} a_i f^i = 0$ with
$a_i \in R$. Let $U \subset X$ be affine open.
Since $b \in R$ is not zero and since $X$ is integral we see
that also $b|_U \in \mathcal{O}_X(U)$ is not zero.
Hence $a/b$ is an element of the fraction field of
$\mathcal{O}_X(U)$ which is integral over $\mathcal{O}_X(U)$
(because we can use the same polynomial
$f^d + \sum_{i = 0, \ldots, d - 1} a_i|_U f^i = 0$ on $U$).
Since $\mathcal{O}_X(U)$ is a normal domain
(Lemma \ref{lemma-locally-normal}), we see that
$f_U = (a|_U)/(b|_U) \in \mathcal{O}_X(U)$. It is easy to
see that $f_U|_V = f_V$ whenever $V \subset U \subset X$ are
affine open. Hence the local sections $f_U$ glue to a global
section $f$ as desired.
\end{proof}

Comment #2543 by Grayson Jorgenson on May 15, 2017 a 10:59 pm UTC

There appears to be a minor typo in the proof of Lemma 27.7.9: the summations are indexed $i = 1,...,d$, but should be indexed $i = 0,...,d-1$ or the exponents changed from $f^i$ to $f^{d - i}$.

Comment #2576 by Johan (site) on May 25, 2017 a 7:49 pm UTC

Thanks Grayson. Fixed here.

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