Lemma 42.9.1. Let $(S, \delta )$ be as in Situation 42.7.1. 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 28.10.3). Hence $\mathcal{O}_{Z, \xi }$ is Noetherian local ring of dimension zero, and hence has finite length over itself (see Algebra, Proposition 10.60.7). Hence, it also has finite length over $\mathcal{O}_{X, \xi }$, see Algebra, Lemma 10.52.12.

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. $\square$

Comment #7442 by old friend on

"Hence, it also has finite length over OX,ξ, see Algebra, Lemma 10.52.12." In this case, the two lengths be would even be equal right? We have OX,ξ ---> OZ,ξ is surjective. If this is the case, citing tag 00IX would be helpful too.

Comment #7443 by David Holmes on

Dear old friend, What you propose seems to me to work, and 00IX seems simpler to apply here than 02M0.

Comment #7446 by old friend on

Thanks David Holmes for confirming. I was confused by the definition given in e.g. Fulton or 3264. I think stacks insists on using O_X,ξ rather than O_Z,ξ because they later define cycle associated to a coherent sheaf F on X. There the length of F_ξ only makes sense over O_X,ξ but when F is O_Z, we could take length over O_Z,ξ.

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