# The Stacks Project

## Tag 02QT

Lemma 41.10.1. Let $(S, \delta)$ be as in Situation 41.8.1. Let $X$ be locally of finite type over $S$. Let $Z \subset X$ be a closed subscheme.

1. Let $Z' \subset Z$ be an irreducible component and let $\xi \in Z'$ be its generic point. Then $$\text{length}_{\mathcal{O}_{X, \xi}} \mathcal{O}_{Z, \xi} < \infty$$
2. If $\dim_\delta(Z) \leq k$ and $\xi \in Z$ with $\delta(\xi) = k$, then $\xi$ is a generic point of an irreducible component of $Z$.

Proof. Let $Z' \subset Z$, $\xi \in Z'$ be as in (1). Then $\dim(\mathcal{O}_{Z, \xi}) = 0$ (for example by Properties, Lemma 27.10.3). Hence $\mathcal{O}_{Z, \xi}$ is Noetherian local ring of dimension zero, and hence has finite length over itself (see Algebra, Proposition 10.59.6). Hence, it also has finite length over $\mathcal{O}_{X, \xi}$, see Algebra, Lemma 10.51.12.

Assume $\xi \in Z$ and $\delta(\xi) = k$. Consider the closure $Z' = \overline{\{\xi\}}$. It is an irreducible closed subscheme with $\dim_\delta(Z') = k$ by definition. Since $\dim_\delta(Z) = k$ it must be an irreducible component of $Z$. Hence we see (2) holds. $\square$

The code snippet corresponding to this tag is a part of the file chow.tex and is located in lines 3212–3228 (see updates for more information).

\begin{lemma}
\label{lemma-multiplicity-finite}
Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
Let $X$ be locally of finite type over $S$.
Let $Z \subset X$ be a closed subscheme.
\begin{enumerate}
\item Let $Z' \subset Z$ be an irreducible component and
let $\xi \in Z'$ be its generic point.
Then
$$\text{length}_{\mathcal{O}_{X, \xi}} \mathcal{O}_{Z, \xi} < \infty$$
\item If $\dim_\delta(Z) \leq k$ and $\xi \in Z$ with
$\delta(\xi) = k$, then $\xi$ is a generic point of an
irreducible component of $Z$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $Z' \subset Z$, $\xi \in Z'$ be as in (1).
Then $\dim(\mathcal{O}_{Z, \xi}) = 0$ (for example by
Properties, Lemma \ref{properties-lemma-codimension-local-ring}).
Hence $\mathcal{O}_{Z, \xi}$ is Noetherian
local ring of dimension zero, and hence has finite length over
itself (see
Algebra, Proposition \ref{algebra-proposition-dimension-zero-ring}).
Hence, it also has finite length over $\mathcal{O}_{X, \xi}$, see
Algebra, Lemma \ref{algebra-lemma-pushdown-module}.

\medskip\noindent
Assume $\xi \in Z$ and $\delta(\xi) = k$.
Consider the closure $Z' = \overline{\{\xi\}}$. It is an irreducible
closed subscheme with $\dim_\delta(Z') = k$ by definition.
Since $\dim_\delta(Z) = k$ it must be an irreducible component
of $Z$. Hence we see (2) holds.
\end{proof}

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