The Stacks project

Lemma 3.9.9. Let $\alpha $ be an ordinal as in Lemma 3.9.2 above. The category $\mathit{Sch}_\alpha $ satisfies the following properties:

  1. If $X, Y, S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$, then for any morphisms $f : X \to S$, $g : Y \to S$ the fibre product $X \times _ S Y$ in $\mathit{Sch}_\alpha $ exists and is a fibre product in the category of schemes.

  2. Given any at most countable collection $S_1, S_2, \ldots $ of elements of $\mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$, the coproduct $\coprod _ i S_ i$ exists in $\mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ and is a coproduct in the category of schemes.

  3. For any $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ and any open immersion $U \to S$, there exists a $V \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ with $V \cong U$.

  4. For any $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ and any closed immersion $T \to S$, there exists an $S' \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ with $S' \cong T$.

  5. For any $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ and any finite type morphism $T \to S$, there exists an $S' \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ with $S' \cong T$.

  6. Suppose $S$ is a scheme which has an open covering $S = \bigcup _{i \in I} S_ i$ such that there exists a $T \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ with (a) $\text{size}(S_ i) \leq \text{size}(T)^{\aleph _0}$ for all $i \in I$, and (b) $|I| \leq \text{size}(T)^{\aleph _0}$. Then $S$ is isomorphic to an object of $\mathit{Sch}_\alpha $.

  7. For any $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ and any morphism $f : T \to S$ locally of finite type such that $T$ can be covered by at most $\text{size}(S)^{\aleph _0}$ open affines, there exists an $S' \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ with $S' \cong T$. For example this holds if $T$ can be covered by at most $|\mathbf{R}| = 2^{\aleph _0} = \aleph _0^{\aleph _0}$ open affines.

  8. For any $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ and any monomorphism $T \to S$ which is either locally of finite presentation or quasi-compact, there exists an $S' \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ with $S' \cong T$.

  9. Suppose that $T \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ is affine. Write $R = \Gamma (T, \mathcal{O}_ T)$. Then any of the following schemes is isomorphic to a scheme in $\mathit{Sch}_\alpha $:

    1. For any ideal $I \subset R$ with completion $R^* = \mathop{\mathrm{lim}}\nolimits _ n R/I^ n$, the scheme $\mathop{\mathrm{Spec}}(R^*)$.

    2. For any finite type $R$-algebra $R'$, the scheme $\mathop{\mathrm{Spec}}(R')$.

    3. For any localization $S^{-1}R$, the scheme $\mathop{\mathrm{Spec}}(S^{-1}R)$.

    4. For any prime $\mathfrak p \subset R$, the scheme $\mathop{\mathrm{Spec}}(\overline{\kappa (\mathfrak p)})$.

    5. For any subring $R' \subset R$, the scheme $\mathop{\mathrm{Spec}}(R')$.

    6. Any scheme of finite type over a ring of cardinality at most $|R|^{\aleph _0}$.

    7. And so on.

Proof. Statements (1) and (2) follow directly from the definitions. Statement (3) follows as the size of an open subscheme $U$ of $S$ is clearly smaller than or equal to the size of $S$. Statement (4) follows from (5). Statement (5) follows from (7). Statement (6) follows as the size of $S$ is $\leq \max \{ |I|, \sup _ i \text{size}(S_ i)\} \leq \text{size}(T)^{\aleph _0}$ by Lemma 3.9.5. Statement (7) follows from (6). Namely, for any affine open $V \subset T$ we have $\text{size}(V) \leq \text{size}(S)$ by Lemma 3.9.7. Thus, we see that (6) applies in the situation of (7). Part (8) follows from Lemma 3.9.8.

Statement (9) is translated, via Lemma 3.9.4, into an upper bound on the cardinality of the rings $R^*$, $S^{-1}R$, $\overline{\kappa (\mathfrak p)}$, $R'$, etc. Perhaps the most interesting one is the ring $R^*$. As a set, it is the image of a surjective map $R^{\mathbf{N}} \to R^*$. Since $|R^{\mathbf{N}}| = |R|^{\aleph _0}$, we see that it works by our choice of $Bound(\kappa )$ being at least $\kappa ^{\aleph _0}$. Phew! (The cardinality of the algebraic closure of a field is the same as the cardinality of the field, or it is $\aleph _0$.) $\square$

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