The Stacks project

32.18 Limits and dimensions of fibres

The following lemma is most often used in the situation of Lemma 32.10.1 to assure that if the fibres of the limit have dimension $\leq d$, then the fibres at some finite stage have dimension $\leq d$.

Lemma 32.18.1. Let $I$ be a directed set. Let $(f_ i : X_ i \to S_ i)$ be an inverse system of morphisms of schemes over $I$. Assume

  1. all the morphisms $S_{i'} \to S_ i$ are affine,

  2. all the schemes $S_ i$ are quasi-compact and quasi-separated,

  3. the morphisms $f_ i$ are of finite type, and

  4. the morphisms $X_{i'} \to X_ i \times _{S_ i} S_{i'}$ are closed immersions.

Let $f : X = \mathop{\mathrm{lim}}\nolimits _ i X_ i \to S = \mathop{\mathrm{lim}}\nolimits _ i S_ i$ be the limit. Let $d \geq 0$. If every fibre of $f$ has dimension $\leq d$, then for some $i$ every fibre of $f_ i$ has dimension $\leq d$.

Proof. For each $i$ let $U_ i = \{ x \in X_ i \mid \dim _ x((X_ i)_{f_ i(x)}) \leq d\} $. This is an open subset of $X_ i$, see Morphisms, Lemma 29.28.4. Set $Z_ i = X_ i \setminus U_ i$ (with reduced induced scheme structure). We have to show that $Z_ i = \emptyset $ for some $i$. If not, then $Z = \mathop{\mathrm{lim}}\nolimits Z_ i \not= \emptyset $, see Lemma 32.4.3. Say $z \in Z$ is a point. Note that $Z \subset X$ is a closed subscheme. Set $s = f(z)$. For each $i$ let $s_ i \in S_ i$ be the image of $s$. We remark that $Z_ s$ is the limit of the schemes $(Z_ i)_{s_ i}$ and $Z_ s$ is also the limit of the schemes $(Z_ i)_{s_ i}$ base changed to $\kappa (s)$. Moreover, all the morphisms

\[ Z_ s \longrightarrow (Z_{i'})_{s_{i'}} \times _{\mathop{\mathrm{Spec}}(\kappa (s_{i'}))} \mathop{\mathrm{Spec}}(\kappa (s)) \longrightarrow (Z_ i)_{s_ i} \times _{\mathop{\mathrm{Spec}}(\kappa (s_ i))} \mathop{\mathrm{Spec}}(\kappa (s)) \longrightarrow X_ s \]

are closed immersions by assumption (4). Hence $Z_ s$ is the scheme theoretic intersection of the closed subschemes $(Z_ i)_{s_ i} \times _{\mathop{\mathrm{Spec}}(\kappa (s_ i))} \mathop{\mathrm{Spec}}(\kappa (s))$ in $X_ s$. Since all the irreducible components of the schemes $(Z_ i)_{s_ i} \times _{\mathop{\mathrm{Spec}}(\kappa (s_ i))} \mathop{\mathrm{Spec}}(\kappa (s))$ have dimension $> d$ and contain $z$ we conclude that $Z_ s$ contains an irreducible component of dimension $> d$ passing through $z$ which contradicts the fact that $Z_ s \subset X_ s$ and $\dim (X_ s) \leq d$. $\square$

Lemma 32.18.2. Notation and assumptions as in Situation 32.8.1. If

  1. $f$ is a quasi-finite morphism, and

  2. $f_0$ is locally of finite type,

then there exists an $i \geq 0$ such that $f_ i$ is quasi-finite.

Proof. Follows immediately from Lemma 32.18.1. $\square$

Lemma 32.18.3. Assumptions and notation as in Situation 32.8.1. Let $d \geq 0$. If

  1. $f$ has relative dimension $\leq d$ (Morphisms, Definition 29.29.1), and

  2. $f_0$ is locally of finite type,

then there exists an $i$ such that $f_ i$ has relative dimension $\leq d$.

Proof. Follows immediately from Lemma 32.18.1. $\square$

Lemma 32.18.4. Notation and assumptions as in Situation 32.8.1. If

  1. $f$ has relative dimension $d$, and

  2. $f_0$ is locally of finite presentation,

then there exists an $i \geq 0$ such that $f_ i$ has relative dimension $d$.

Proof. By Lemma 32.18.1 we may assume all fibres of $f_0$ have dimension $\leq d$. By Morphisms, Lemma 29.28.6 the set $U_0 \subset X_0$ of points $x \in X_0$ such that the dimension of the fibre of $X_0 \to Y_0$ at $x$ is $\leq d - 1$ is open and retrocompact in $X_0$. Hence the complement $E = X_0 \setminus U_0$ is constructible. Moreover the image of $X \to X_0$ is contained in $E$ by Morphisms, Lemma 29.28.3. Thus for $i \gg 0$ we have that the image of $X_ i \to X_0$ is contained in $E$ (Lemma 32.4.10). Then all fibres of $X_ i \to Y_ i$ have dimension $d$ by the aforementioned Morphisms, Lemma 29.28.3. $\square$

Lemma 32.18.5. Let $S$ be a quasi-compact and quasi-separated scheme. Let $f : X \to S$ be a morphism of finite presentation. Let $d \geq 0$ be an integer. If $Z \subset X$ be a closed subscheme such that $\dim (Z_ s) \leq d$ for all $s \in S$, then there exists a closed subscheme $Z' \subset X$ such that

  1. $Z \subset Z'$,

  2. $Z' \to X$ is of finite presentation, and

  3. $\dim (Z'_ s) \leq d$ for all $s \in S$.

Proof. By Proposition 32.5.4 we can write $S = \mathop{\mathrm{lim}}\nolimits S_ i$ as the limit of a directed inverse system of Noetherian schemes with affine transition maps. By Lemma 32.10.1 we may assume that there exist a system of morphisms $f_ i : X_ i \to S_ i$ of finite presentation such that $X_{i'} = X_ i \times _{S_ i} S_{i'}$ for all $i' \geq i$ and such that $X = X_ i \times _{S_ i} S$. Let $Z_ i \subset X_ i$ be the scheme theoretic image of $Z \to X \to X_ i$. Then for $i' \geq i$ the morphism $X_{i'} \to X_ i$ maps $Z_{i'}$ into $Z_ i$ and the induced morphism $Z_{i'} \to Z_ i \times _{S_ i} S_{i'}$ is a closed immersion. By Lemma 32.18.1 we see that the dimension of the fibres of $Z_ i \to S_ i$ all have dimension $\leq d$ for a suitable $i \in I$. Fix such an $i$ and set $Z' = Z_ i \times _{S_ i} S \subset X$. Since $S_ i$ is Noetherian, we see that $X_ i$ is Noetherian, and hence the morphism $Z_ i \to X_ i$ is of finite presentation. Therefore also the base change $Z' \to X$ is of finite presentation. Moreover, the fibres of $Z' \to S$ are base changes of the fibres of $Z_ i \to S_ i$ and hence have dimension $\leq d$. $\square$


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