The Stacks project

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

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 $\dim (X_ s) \leq d$. $\square$

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

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

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. $\square$

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 $\leq d$. $\square$