# The Stacks Project

## Tag 000H

### 3.9. Constructing categories of schemes

We will discuss how to apply this to produce, given an initial set of schemes, a ''small'' category of schemes closed under a list of natural operations. Before we do so, we introduce the size of a scheme. Given a scheme $S$ we define $$\text{size}(S) = \max(\aleph_0, \kappa_1, \kappa_2),$$ where we define the cardinal numbers $\kappa_1$ and $\kappa_2$ as follows:

1. We let $\kappa_1$ be the cardinality of the set of affine opens of $S$.
2. We let $\kappa_2$ be the supremum of all the cardinalities of all $\Gamma(U, \mathcal{O}_S)$ for all $U \subset S$ affine open.

Lemma 3.9.1. For every cardinal $\kappa$, there exists a set $A$ such that every element of $A$ is a scheme and such that for every scheme $S$ with $\text{size}(S) \leq \kappa$, there is an element $X \in A$ such that $X \cong S$ (isomorphism of schemes).

Proof. Omitted. Hint: think about how any scheme is isomorphic to a scheme obtained by glueing affines. $\square$

We denote $Bound$ the function which to each cardinal $\kappa$ associates $$\tag{3.9.1.1} Bound(\kappa) = \max\{\kappa^{\aleph_0}, \kappa^+\}.$$ We could make this function grow much more rapidly, e.g., we could set $Bound(\kappa) = \kappa^\kappa$, and the result below would still hold. For any ordinal $\alpha$, we denote $\textit{Sch}_\alpha$ the full subcategory of category of schemes whose objects are elements of $V_\alpha$. Here is the result we are going to prove.

Lemma 3.9.2. With notations $\text{size}$, $Bound$ and $\textit{Sch}_\alpha$ as above. Let $S_0$ be a set of schemes. There exists a limit ordinal $\alpha$ with the following properties:

1. We have $S_0 \subset V_\alpha$; in other words, $S_0 \subset \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$.
2. For any $S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ and any scheme $T$ with $\text{size}(T) \leq Bound(\text{size}(S))$, there exists a scheme $S' \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ such that $T \cong S'$.
3. For any countable1 diagram category $\mathcal{I}$ and any functor $F : \mathcal{I} \to \textit{Sch}_\alpha$, the limit $\mathop{\rm lim}\nolimits_\mathcal{I} F$ exists in $\textit{Sch}_\alpha$ if and only if it exists in $\textit{Sch}$ and moreover, in this case, the natural morphism between them is an isomorphism.
4. For any countable diagram category $\mathcal{I}$ and any functor $F : \mathcal{I} \to \textit{Sch}_\alpha$, the colimit $\mathop{\rm colim}\nolimits_\mathcal{I} F$ exists in $\textit{Sch}_\alpha$ if and only if it exists in $\textit{Sch}$ and moreover, in this case, the natural morphism between them is an isomorphism.

Proof. We define, by transfinite induction, a function $f$ which associates to every ordinal an ordinal as follows. Let $f(0) = 0$. Given $f(\alpha)$, we define $f(\alpha + 1)$ to be the least ordinal $\beta$ such that the following hold:

1. We have $\alpha + 1 \leq \beta$ and $f(\alpha) \leq \beta$.
2. For any $S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_{f(\alpha)})$ and any scheme $T$ with $\text{size}(T) \leq Bound(\text{size}(S))$, there exists a scheme $S' \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\beta)$ such that $T \cong S'$.
3. For any countable diagram category $\mathcal{I}$ and any functor $F : \mathcal{I} \to \textit{Sch}_{f(\alpha)}$, if the limit $\mathop{\rm lim}\nolimits_\mathcal{I} F$ or the colimit $\mathop{\rm colim}\nolimits_\mathcal{I} F$ exists in $\textit{Sch}$, then it is isomorphic to a scheme in $\textit{Sch}_\beta$.

To see $\beta$ exists, we argue as follows. Since $\mathop{\rm Ob}\nolimits(\textit{Sch}_{f(\alpha)})$ is a set, we see that $\kappa = \sup_{S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_{f(\alpha)})} Bound(\text{size}(S))$ exists and is a cardinal. Let $A$ be a set of schemes obtained starting with $\kappa$ as in Lemma 3.9.1. There is a set $CountCat$ of countable categories such that any countable category is isomorphic to an element of $CountCat$. Hence in (3) above we may assume that $\mathcal{I}$ is an element in $CountCat$. This means that the pairs $(\mathcal{I}, F)$ in (3) range over a set. Thus, there exists a set $B$ whose elements are schemes such that for every $(\mathcal{I}, F)$ as in (3), if the limit or colimit exists, then it is isomorphic to an element in $B$. Hence, if we pick any $\beta$ such that $A \cup B \subset V_\beta$ and $\beta > \max\{\alpha + 1, f(\alpha)\}$, then (1)--(3) hold. Since every nonempty collection of ordinals has a least element, we see that $f(\alpha + 1)$ is well defined. Finally, if $\alpha$ is a limit ordinal, then we set $f(\alpha) = \sup_{\alpha' < \alpha} f(\alpha')$.

Pick $\beta_0$ such that $S_0 \subset V_{\beta_0}$. By construction $f(\beta) \geq \beta$ and we see that also $S_0 \subset V_{f(\beta_0)}$. Moreover, as $f$ is nondecreasing, we see $S_0 \subset V_{f(\beta)}$ is true for any $\beta \geq \beta_0$. Next, choose any ordinal $\beta_1 > \beta_0$ with cofinality $\text{cf}(\beta_1) > \omega = \aleph_0$. This is possible since the cofinality of ordinals gets arbitrarily large, see Proposition 3.7.2. We claim that $\alpha = f(\beta_1)$ is a solution to the problem posed in the lemma.

The first property of the lemma holds by our choice of $\beta_1 > \beta_0$ above.

Since $\beta_1$ is a limit ordinal (as its cofinality is infinite), we get $f(\beta_1) = \sup_{\beta < \beta_1} f(\beta)$. Hence $\{f(\beta) \mid \beta < \beta_1\} \subset f(\beta_1)$ is a cofinal subset. Hence we see that $$V_\alpha = V_{f(\beta_1)} = \bigcup\nolimits_{\beta < \beta_1} V_{f(\beta)}.$$ Now, let $S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$. We define $\beta(S)$ to be the least ordinal $\beta$ such that $S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_{f(\beta)})$. By the above we see that always $\beta(S) < \beta_1$. Since $\mathop{\rm Ob}\nolimits(\textit{Sch}_{f(\beta + 1)}) \subset \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$, we see by construction of $f$ above that the second property of the lemma is satisfied.

Suppose that $\{S_1, S_2, \ldots\} \subset \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ is a countable collection. Consider the function $\omega \to \beta_1$, $n \mapsto \beta(S_n)$. Since the cofinality of $\beta_1$ is $> \omega$, the image of this function cannot be a cofinal subset. Hence there exists a $\beta < \beta_1$ such that $\{S_1, S_2, \ldots\} \subset \mathop{\rm Ob}\nolimits(\textit{Sch}_{f(\beta)})$. It follows that any functor $F : \mathcal{I} \to \textit{Sch}_\alpha$ factors through one of the subcategories $\textit{Sch}_{f(\beta)}$. Thus, if there exists a scheme $X$ that is the colimit or limit of the diagram $F$, then, by construction of $f$, we see $X$ is isomorphic to an object of $\textit{Sch}_{f(\beta + 1)}$ which is a subcategory of $\textit{Sch}_\alpha$. This proves the last two assertions of the lemma. $\square$

Remark 3.9.3. The lemma above can also be proved using the reflection principle. However, one has to be careful. Namely, suppose the sentence $\phi_{scheme}(X)$ expresses the property ''$X$ is a scheme'', then what does the formula $\phi_{scheme}^{V_\alpha}(X)$ mean? It is true that the reflection principle says we can find $\alpha$ such that for all $X \in V_\alpha$ we have $\phi_{scheme}(X) \leftrightarrow \phi_{scheme}^{V_\alpha}(X)$ but this is entirely useless. It is only by combining two such statements that something interesting happens. For example suppose $\phi_{red}(X, Y)$ expresses the property ''$X$, $Y$ are schemes, and $Y$ is the reduction of $X$'' (see Schemes, Definition 25.12.5). Suppose we apply the reflection principle to the pair of formulas $\phi_1(X, Y) = \phi_{red}(X, Y)$, $\phi_2(X) = \exists Y, \phi_1(X, Y)$. Then it is easy to see that any $\alpha$ produced by the reflection principle has the property that given $X \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ the reduction of $X$ is also an object of $\textit{Sch}_\alpha$ (left as an exercise).

Lemma 3.9.4. Let $S$ be an affine scheme. Let $R = \Gamma(S, \mathcal{O}_S)$. Then the size of $S$ is equal to $\max\{ \aleph_0, |R|\}$.

Proof. There are at most $\max\{|R|, \aleph_0\}$ affine opens of $\mathop{\rm Spec}(R)$. This is clear since any affine open $U \subset \mathop{\rm Spec}(R)$ is a finite union of principal opens $D(f_1) \cup \ldots \cup D(f_n)$ and hence the number of affine opens is at most $\sup_n |R|^n = \max\{|R|, \aleph_0\}$, see [Kunen, Ch. I, 10.13]. On the other hand, we see that $\Gamma(U, \mathcal{O}) \subset R_{f_1} \times \ldots \times R_{f_n}$ and hence $|\Gamma(U, \mathcal{O})| \leq \max\{\aleph_0, |R_{f_1}|, \ldots, |R_{f_n}|\}$. Thus it suffices to prove that $|R_f| \leq \max\{\aleph_0, |R|\}$ which is omitted. $\square$

Lemma 3.9.5. Let $S$ be a scheme. Let $S = \bigcup_{i \in I} S_i$ be an open covering. Then $\text{size}(S) \leq \max\{|I|, \sup_i\{\text{size}(S_i)\}\}$.

Proof. Let $U \subset S$ be any affine open. Since $U$ is quasi-compact there exist finitely many elements $i_1, \ldots, i_n \in I$ and affine opens $U_i \subset U \cap S_i$ such that $U = U_1 \cup U_2 \cup \ldots \cup U_n$. Thus $$|\Gamma(U, \mathcal{O}_U)| \leq |\Gamma(U_1, \mathcal{O})| \otimes \ldots \otimes |\Gamma(U_n, \mathcal{O})| \leq \sup\nolimits_i\{\text{size}(S_i)\}$$ Moreover, it shows that the set of affine opens of $S$ has cardinality less than or equal to the cardinality of the set $$\coprod_{n \in \omega} \coprod_{i_1, \ldots, i_n \in I} \{\text{affine opens of }S_{i_1}\} \times \ldots \times \{\text{affine opens of }S_{i_n}\}.$$ Each of the sets inside the disjoint union has cardinality at most $\sup_i\{\text{size}(S_i)\}$. The index set has cardinality at most $\max\{|I|, \aleph_0\}$, see [Kunen, Ch. I, 10.13]. Hence by [Jech, Lemma 5.8] the cardinality of the coproduct is at most $\max\{\aleph_0, |I|\} \otimes \sup_i\{\text{size}(S_i)\}$. The lemma follows. $\square$

Lemma 3.9.6. Let $f : X \to S$, $g : Y \to S$ be morphisms of schemes. Then we have $\text{size}(X \times_S Y) \leq \max\{\text{size}(X), \text{size}(Y)\}$.

Proof. Let $S = \bigcup_{k \in K} S_k$ be an affine open covering. Let $X = \bigcup_{i \in I} U_i$, $Y = \bigcup_{j \in J} V_j$ be affine open coverings with $I$, $J$ of cardinality $\leq \text{size}(X), \text{size}(Y)$. For each $i \in I$ there exists a finite set $K_i$ of $k \in K$ such that $f(U_i) \subset \bigcup_{k \in K_i} S_k$. For each $j \in J$ there exists a finite set $K_j$ of $k \in K$ such that $g(V_j) \subset \bigcup_{k \in K_j} S_k$. Hence $f(X), g(Y)$ are contained in $S' = \bigcup_{k \in K'} S_k$ with $K' = \bigcup_{i \in I} K_i \cup \bigcup_{j \in J} K_j$. Note that the cardinality of $K'$ is at most $\max\{\aleph_0, |I|, |J|\}$. Applying Lemma 3.9.5 we see that it suffices to prove that $\text{size}(f^{-1}(S_k) \times_{S_k} g^{-1}(S_k)) \leq \max\{\text{size}(X), \text{size}(Y))\}$ for $k \in K'$. In other words, we may assume that $S$ is affine.

Assume $S$ affine. Let $X = \bigcup_{i \in I} U_i$, $Y = \bigcup_{j \in J} V_j$ be affine open coverings with $I$, $J$ of cardinality $\leq \text{size}(X), \text{size}(Y)$. Again by Lemma 3.9.5 it suffices to prove the lemma for the products $U_i \times_S V_j$. By Lemma 3.9.4 we see that it suffices to show that $$|A \otimes_C B| \leq \max\{\aleph_0, |A|, |B|\}.$$ We omit the proof of this inequality. $\square$

Lemma 3.9.7. Let $S$ be a scheme. Let $f : X \to S$ be locally of finite type with $X$ quasi-compact. Then $\text{size}(X) \leq \text{size}(S)$.

Proof. We can find a finite affine open covering $X = \bigcup_{i = 1, \ldots n} U_i$ such that each $U_i$ maps into an affine open $S_i$ of $S$. Thus by Lemma 3.9.5 we reduce to the case where both $S$ and $X$ are affine. In this case by Lemma 3.9.4 we see that it suffices to show $$|A[x_1, \ldots, x_n]| \leq \max\{\aleph_0, |A|\}.$$ We omit the proof of this inequality. $\square$

In Algebra, Lemma 10.104.13 we will show that if $A \to B$ is an epimorphism of rings, then $|B| \leq \max(|A|, \aleph_0)$. The analogue for schemes is the following lemma.

Lemma 3.9.8. Let $f : X \to Y$ be a monomorphism of schemes. If at least one of the following properties holds, then $\text{size}(X) \leq \text{size}(Y)$:

1. $f$ is quasi-compact,
2. $f$ is locally of finite presentation,
3. add more here as needed.

But the bound does not hold for monomorphisms which are locally of finite type.

Proof. Let $Y = \bigcup_{j \in J} V_j$ be an affine open covering of $Y$ with $|J| \leq \text{size}(Y)$. By Lemma 3.9.5 it suffices to bound the size of the inverse image of $V_j$ in $X$. Hence we reduce to the case that $Y$ is affine, say $Y = \mathop{\rm Spec}(B)$. For any affine open $\mathop{\rm Spec}(A) \subset X$ we have $|A| \leq \max(|B|, \aleph_0) = \text{size}(Y)$, see remark above and Lemma 3.9.4. Thus it suffices to show that $X$ has at most $\text{size}(Y)$ affine opens. This is clear if $X$ is quasi-compact, whence case (1) holds. In case (2) the number of isomorphism classes of $B$-algebras $A$ that can occur is bounded by $\text{size}(B)$, because each $A$ is of finite type over $B$, hence isomorphic to an algebra $B[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$ for some $n, m$, and $f_j \in B[x_1, \ldots, x_n]$. However, as $X \to Y$ is a monomorphism, there is a unique morphism $\mathop{\rm Spec}(A) \to X$ over $Y = \mathop{\rm Spec}(B)$ if there is one, hence the number of affine opens of $X$ is bounded by the number of these isomorphism classes.

To prove the final statement of the lemma consider the ring $B = \prod_{n \in \mathbf{N}} \mathbf{F}_2$ and set $Y = \mathop{\rm Spec}(B)$. For every ultrafilter $\mathcal{U}$ on $\mathbf{N}$ we obtain a maximal ideal $\mathfrak m_\mathcal{U}$ with residue field $\mathbf{F}_2$; the map $B \to \mathbf{F}_2$ sends the element $(x_n)$ to $\mathop{\rm lim}\nolimits_\mathcal{U} x_n$. Details omitted. The morphism of schemes $X = \coprod_\mathcal{U} \mathop{\rm Spec}(\mathbf{F}_2) \to Y$ is a monomorphism as all the points are distinct. However the cardinality of the set of affine open subschemes of $X$ is equal to the cardinality of the set of ultrafilters on $\mathbf{N}$ which is $2^{2^{\aleph_0}}$. We conclude as $|B| = 2^{\aleph_0} < 2^{2^{\aleph_0}}$. $\square$

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

1. If $X, Y, S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$, then for any morphisms $f : X \to S$, $g : Y \to S$ the fibre product $X \times_S Y$ in $\textit{Sch}_\alpha$ 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 $\mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$, the coproduct $\coprod_i S_i$ exists in $\mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ and is a coproduct in the category of schemes.
3. For any $S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ and any open immersion $U \to S$, there exists a $V \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ with $V \cong U$.
4. For any $S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ and any closed immersion $T \to S$, there exists a $S' \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ with $S' \cong T$.
5. For any $S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ and any finite type morphism $T \to S$, there exists a $S' \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ 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 $T \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ 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 $\textit{Sch}_\alpha$.
7. For any $S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ 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 a $S' \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ 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 $S \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ and any monomorphism $T \to S$ which is either locally of finite presentation or quasi-compact, there exists a $S' \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ with $S' \cong T$.
9. Suppose that $T \in \mathop{\rm Ob}\nolimits(\textit{Sch}_\alpha)$ is affine. Write $R = \Gamma(T, \mathcal{O}_T)$. Then any of the following schemes is isomorphic to a scheme in $\textit{Sch}_\alpha$:
1. For any ideal $I \subset R$ with completion $R^* = \mathop{\rm lim}\nolimits_n R/I^n$, the scheme $\mathop{\rm Spec}(R^*)$.
2. For any finite type $R$-algebra $R'$, the scheme $\mathop{\rm Spec}(R')$.
3. For any localization $S^{-1}R$, the scheme $\mathop{\rm Spec}(S^{-1}R)$.
4. For any prime $\mathfrak p \subset R$, the scheme $\mathop{\rm Spec}(\overline{\kappa(\mathfrak p)})$.
5. For any subring $R' \subset R$, the scheme $\mathop{\rm Spec}(R')$.
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 $\aleph_0$.) $\square$

Remark 3.9.10. Let $R$ be a ring. Suppose we consider the ring $\prod_{\mathfrak p \in \mathop{\rm Spec}(R)} \kappa(\mathfrak p)$. The cardinality of this ring is bounded by $|R|^{2^{|R|}}$, but is not bounded by $|R|^{\aleph_0}$ in general. For example if $R = \mathbf{C}[x]$ it is not bounded by $|R|^{\aleph_0}$ and if $R = \prod_{n \in \mathbf{N}} \mathbf{F}_2$ it is not bounded by $|R|^{|R|}$. Thus the ''And so on'' of Lemma 3.9.9 above should be taken with a grain of salt. Of course, if it ever becomes necessary to consider these rings in arguments pertaining to fppf/étale cohomology, then we can change the function $Bound$ above into the function $\kappa \mapsto \kappa^{2^\kappa}$.

In the following lemma we use the notion of an fpqc covering which is introduced in Topologies, Section 33.8.

Lemma 3.9.11. Let $f : X \to Y$ be a morphism of schemes. Assume there exists an fpqc covering $\{g_j : Y_j \to Y\}_{j \in J}$ such that $g_j$ factors through $f$. Then $\text{size}(Y) \leq \text{size}(X)$.

Proof. Let $V \subset Y$ be an affine open. By definition there exist $n \geq 0$ and $a : \{1, \ldots, n\} \to J$ and affine opens $V_i \subset Y_{a(i)}$ such that $V = g_{a(1)}(V_1) \cup \ldots \cup g_{a(n)}(V_n)$. Denote $h_j : Y_j \to X$ a morphism such that $f \circ h_j = g_j$. Then $h_{a(1)}(V_1) \cup \ldots \cup h_{a(n)}(V_n)$ is a quasi-compact subset of $f^{-1}(V)$. Hence we can find a quasi-compact open $W \subset f^{-1}(V)$ which contains $h_{a(i)}(V_i)$ for $i = 1, \ldots, n$. In particular $V = f(W)$.

On the one hand this shows that the cardinality of the set of affine opens of $Y$ is at most the cardinality of the set $S$ of quasi-compact opens of $X$. Since every quasi-compact open of $X$ is a finite union of affines, we see that the cardinality of this set is at most $\sup |S|^n = \max(\aleph_0, |S|)$. On the other hand, we have $\mathcal{O}_Y(V) \subset \prod_{i = 1, \ldots, n} \mathcal{O}_{Y_{a(i)}}(V_i)$ because $\{V_i \to V\}$ is an fpqc covering. Hence $\mathcal{O}_Y(V) \subset \mathcal{O}_X(W)$ because $V_i \to V$ factors through $W$. Again since $W$ has a finite covering by affine opens of $X$ we conclude that $|\mathcal{O}_Y(V)|$ is bounded by the size of $X$. The lemma now follows from the definition of the size of a scheme. $\square$

In the following lemma we use the notion of an fppf covering which is introduced in Topologies, Section 33.7.

Lemma 3.9.12. Let $\{f_i : X_i \to X\}_{i \in I}$ be an fppf covering of a scheme. There exists an fppf covering $\{W_j \to X\}_{j \in J}$ which is a refinement of $\{X_i \to X\}_{i \in I}$ such that $\text{size}(\coprod W_j) \leq \text{size}(X)$.

Proof. Choose an affine open covering $X = \bigcup_{a \in A} U_a$ with $|A| \leq \text{size}(X)$. For each $a$ we can choose a finite subset $I_a \subset I$ and for $i \in I_a$ a quasi-compact open $W_{a, i} \subset X_i$ such that $U_a = \bigcup_{i \in I_a} f_i(W_{a, i})$. Then $\text{size}(W_{a, i}) \leq \text{size}(X)$ by Lemma 3.9.7. We conclude that $\text{size}(\coprod_a \coprod_{i \in I_a} W_{i, a}) \leq \text{size}(X)$ by Lemma 3.9.5. $\square$

1. Both the set of objects and the morphism sets are countable. In fact you can prove the lemma with $\aleph_0$ replaced by any cardinal whatsoever in (3) and (4).

The code snippet corresponding to this tag is a part of the file sets.tex and is located in lines 310–855 (see updates for more information).

\section{Constructing categories of schemes}
\label{section-categories-schemes}

\noindent
We will discuss how to apply this to produce, given an initial
set of schemes, a small'' category of schemes closed under
a list of natural operations. Before we do so, we introduce the
size of a scheme. Given a scheme $S$ we define
$$\text{size}(S) = \max(\aleph_0, \kappa_1, \kappa_2),$$
where we define the cardinal numbers $\kappa_1$ and $\kappa_2$ as follows:
\begin{enumerate}
\item We let $\kappa_1$ be the cardinality of the set of affine opens of $S$.
\item We let $\kappa_2$ be the supremum of all the cardinalities of
all $\Gamma(U, \mathcal{O}_S)$ for all $U \subset S$ affine open.
\end{enumerate}

\begin{lemma}
\label{lemma-bounded-size}
For every cardinal $\kappa$, there exists a set $A$ such
that every element of $A$ is a scheme and such that for every
scheme $S$ with $\text{size}(S) \leq \kappa$, there is
an element $X \in A$ such that $X \cong S$ (isomorphism
of schemes).
\end{lemma}

\begin{proof}
Omitted. Hint: think about how any scheme is isomorphic to a scheme
obtained by glueing affines.
\end{proof}

\noindent
We denote $Bound$ the function which to each
cardinal $\kappa$ associates

\label{equation-bound}
Bound(\kappa) = \max\{\kappa^{\aleph_0}, \kappa^+\}.

We could make this function grow much more rapidly, e.g., we could
set $Bound(\kappa) = \kappa^\kappa$, and the result below would still hold.
For any ordinal $\alpha$, we denote $\Sch_\alpha$ the full
subcategory of category of schemes whose objects are elements of
$V_\alpha$. Here is the result we are going to prove.

\begin{lemma}
\label{lemma-construct-category}
With notations $\text{size}$, $Bound$ and $\Sch_\alpha$ as above.
Let $S_0$ be a set of schemes. There exists a limit ordinal
$\alpha$ with the following properties:
\begin{enumerate}
\item
\label{item-inclusion}
We have $S_0 \subset V_\alpha$; in other words,
$S_0 \subset \Ob(\Sch_\alpha)$.
\item
\label{item-bounded}
For any $S \in \Ob(\Sch_\alpha)$ and any
scheme $T$ with $\text{size}(T) \leq Bound(\text{size}(S))$,
there exists a scheme $S' \in \Ob(\Sch_\alpha)$
such that $T \cong S'$.
\item
\label{item-limit}
For any countable\footnote{Both the set of objects and
the morphism sets are countable. In fact you can prove the lemma with
$\aleph_0$ replaced by any cardinal whatsoever in (3) and (4).} diagram
category $\mathcal{I}$ and
any functor $F : \mathcal{I} \to \Sch_\alpha$, the limit
$\lim_\mathcal{I} F$ exists in $\Sch_\alpha$ if and
only if it exists in $\Sch$ and moreover, in this case,
the natural morphism between them is an isomorphism.
\item
\label{item-colimit}
For any countable diagram category $\mathcal{I}$ and
any functor $F : \mathcal{I} \to \Sch_\alpha$, the colimit
$\colim_\mathcal{I} F$ exists in $\Sch_\alpha$ if and
only if it exists in $\Sch$ and moreover, in this case,
the natural morphism between them is an isomorphism.
\end{enumerate}
\end{lemma}

\begin{proof}
We define, by transfinite induction, a function $f$ which associates
to every ordinal an ordinal as follows. Let $f(0) = 0$.
Given $f(\alpha)$, we define $f(\alpha + 1)$ to be the least
ordinal $\beta$ such that the following hold:
\begin{enumerate}
\item We have $\alpha + 1 \leq \beta$ and $f(\alpha) \leq \beta$.
\item For any $S \in \Ob(\Sch_{f(\alpha)})$ and any
scheme $T$ with $\text{size}(T) \leq Bound(\text{size}(S))$,
there exists a scheme $S' \in \Ob(\Sch_\beta)$
such that $T \cong S'$.
\item For any countable diagram category $\mathcal{I}$ and
any functor $F : \mathcal{I} \to \Sch_{f(\alpha)}$, if
the limit $\lim_\mathcal{I} F$ or the colimit
$\colim_\mathcal{I} F$ exists in $\Sch$,
then it is isomorphic to a scheme in $\Sch_\beta$.
\end{enumerate}
To see $\beta$ exists, we argue as follows. Since
$\Ob(\Sch_{f(\alpha)})$ is a set, we see that
$\kappa = \sup_{S \in \Ob(\Sch_{f(\alpha)})} Bound(\text{size}(S))$
exists and is a cardinal.
Let $A$ be a set of schemes obtained starting with $\kappa$
as in Lemma \ref{lemma-bounded-size}.
There is a set $CountCat$ of countable
categories such that any countable category is isomorphic to
an element of $CountCat$. Hence in (3) above we may assume
that $\mathcal{I}$ is an element in $CountCat$. This means that
the pairs $(\mathcal{I}, F)$ in (3) range over a set.
Thus, there exists a set $B$ whose elements are schemes
such that for every $(\mathcal{I}, F)$ as in (3), if the
limit or colimit exists, then it is isomorphic to an element in $B$.
Hence, if we pick any $\beta$ such that $A \cup B \subset V_\beta$
and $\beta > \max\{\alpha + 1, f(\alpha)\}$, then (1)--(3) hold.
Since every nonempty collection of ordinals has a least element,
we see that $f(\alpha + 1)$ is well defined. Finally, if $\alpha$
is a limit ordinal, then we set
$f(\alpha) = \sup_{\alpha' < \alpha} f(\alpha')$.

\medskip\noindent
Pick $\beta_0$ such that $S_0 \subset V_{\beta_0}$.
By construction $f(\beta) \geq \beta$ and we see that
also $S_0 \subset V_{f(\beta_0)}$. Moreover, as $f$ is
nondecreasing, we see $S_0 \subset V_{f(\beta)}$ is true for any
$\beta \geq \beta_0$.
Next, choose any ordinal $\beta_1 > \beta_0$ with cofinality
$\text{cf}(\beta_1) > \omega = \aleph_0$. This is possible
since the cofinality of ordinals gets arbitrarily large, see
Proposition \ref{proposition-exist-ordinals-large-cofinality}.
We claim that
$\alpha = f(\beta_1)$ is a solution to the problem posed in the lemma.

\medskip\noindent
The first property of the lemma holds by our choice
of $\beta_1 > \beta_0$ above.

\medskip\noindent
Since $\beta_1$ is a limit ordinal (as its cofinality is infinite),
we get $f(\beta_1) = \sup_{\beta < \beta_1} f(\beta)$.
Hence $\{f(\beta) \mid \beta < \beta_1\} \subset f(\beta_1)$ is a
cofinal subset. Hence we see that
$$V_\alpha = V_{f(\beta_1)} = \bigcup\nolimits_{\beta < \beta_1} V_{f(\beta)}.$$
Now, let $S \in \Ob(\Sch_\alpha)$. We define
$\beta(S)$ to be the least ordinal $\beta$ such that
$S \in \Ob(\Sch_{f(\beta)})$. By the above we see
that always $\beta(S) < \beta_1$. Since
$\Ob(\Sch_{f(\beta + 1)}) \subset \Ob(\Sch_\alpha)$, we
see by construction of $f$ above that the second property of the lemma
is satisfied.

\medskip\noindent
Suppose that $\{S_1, S_2, \ldots\} \subset \Ob(\Sch_\alpha)$
is a countable collection. Consider the function
$\omega \to \beta_1$, $n \mapsto \beta(S_n)$. Since the cofinality
of $\beta_1$ is $> \omega$, the image of this function cannot be a
cofinal subset. Hence there exists a $\beta < \beta_1$ such
that $\{S_1, S_2, \ldots\} \subset \Ob(\Sch_{f(\beta)})$.
It follows that any functor $F : \mathcal{I} \to \Sch_\alpha$
factors through one of the subcategories $\Sch_{f(\beta)}$.
Thus, if there exists a scheme $X$ that is the colimit or limit
of the diagram $F$, then, by construction of $f$, we see
$X$ is isomorphic to an object
of $\Sch_{f(\beta + 1)}$ which is a subcategory of
$\Sch_\alpha$. This proves the last two assertions of
the lemma.
\end{proof}

\begin{remark}
\label{remark-how-to-use-reflection}
The lemma above can also be proved using the reflection principle.
However, one has to be careful. Namely, suppose the sentence
$\phi_{scheme}(X)$ expresses the property $X$ is a scheme'', then
what does the formula $\phi_{scheme}^{V_\alpha}(X)$ mean?
It is true that the reflection principle says we can find $\alpha$ such that
for all $X \in V_\alpha$ we have
$\phi_{scheme}(X) \leftrightarrow \phi_{scheme}^{V_\alpha}(X)$
but this is entirely useless. It is only by combining two such
statements that something interesting happens. For example suppose
$\phi_{red}(X, Y)$ expresses the property $X$, $Y$ are schemes,
and $Y$ is the reduction of $X$'' (see
Schemes, Definition \ref{schemes-definition-reduced-induced-scheme}).
Suppose we apply the reflection principle to the pair of
formulas $\phi_1(X, Y) = \phi_{red}(X, Y)$,
$\phi_2(X) = \exists Y, \phi_1(X, Y)$. Then it is easy to see that
any $\alpha$ produced by the reflection principle has the property that
given $X \in \Ob(\Sch_\alpha)$ the reduction of
$X$ is also an object of $\Sch_\alpha$ (left as an exercise).
\end{remark}

\begin{lemma}
\label{lemma-bound-affine}
Let $S$ be an affine scheme.
Let $R = \Gamma(S, \mathcal{O}_S)$.
Then the size of $S$ is equal to $\max\{ \aleph_0, |R|\}$.
\end{lemma}

\begin{proof}
There are at most $\max\{|R|, \aleph_0\}$ affine opens of
$\Spec(R)$. This is clear since any affine open
$U \subset \Spec(R)$ is a finite union of principal
opens $D(f_1) \cup \ldots \cup D(f_n)$ and hence the number
of affine opens is at most $\sup_n |R|^n = \max\{|R|, \aleph_0\}$,
see \cite[Ch. I, 10.13]{Kunen}. On the other hand, we see that
$\Gamma(U, \mathcal{O}) \subset R_{f_1} \times \ldots \times R_{f_n}$
and hence $|\Gamma(U, \mathcal{O})| \leq \max\{\aleph_0, |R_{f_1}|, \ldots, |R_{f_n}|\}$. Thus
it suffices to prove that $|R_f| \leq \max\{\aleph_0, |R|\}$
which is omitted.
\end{proof}

\begin{lemma}
\label{lemma-bound-size}
Let $S$ be a scheme. Let $S = \bigcup_{i \in I} S_i$ be
an open covering. Then
$\text{size}(S) \leq \max\{|I|, \sup_i\{\text{size}(S_i)\}\}$.
\end{lemma}

\begin{proof}
Let $U \subset S$ be any affine open. Since $U$ is quasi-compact
there exist finitely many elements $i_1, \ldots, i_n \in I$
and affine opens $U_i \subset U \cap S_i$ such that
$U = U_1 \cup U_2 \cup \ldots \cup U_n$. Thus
$$|\Gamma(U, \mathcal{O}_U)| \leq |\Gamma(U_1, \mathcal{O})| \otimes \ldots \otimes |\Gamma(U_n, \mathcal{O})| \leq \sup\nolimits_i\{\text{size}(S_i)\}$$
Moreover, it shows that the set of affine opens of $S$ has
cardinality less than or equal to the cardinality of the set
$$\coprod_{n \in \omega} \coprod_{i_1, \ldots, i_n \in I} \{\text{affine opens of }S_{i_1}\} \times \ldots \times \{\text{affine opens of }S_{i_n}\}.$$
Each of the sets inside the disjoint union has cardinality at most
$\sup_i\{\text{size}(S_i)\}$. The index set has cardinality at most
$\max\{|I|, \aleph_0\}$, see \cite[Ch. I, 10.13]{Kunen}.
Hence by \cite[Lemma 5.8]{Jech} the cardinality
of the coproduct is at most $\max\{\aleph_0, |I|\} \otimes \sup_i\{\text{size}(S_i)\}$. The lemma follows.
\end{proof}

\begin{lemma}
\label{lemma-bound-size-fibre-product}
Let $f : X \to S$, $g : Y \to S$ be morphisms of schemes.
Then we have
$\text{size}(X \times_S Y) \leq \max\{\text{size}(X), \text{size}(Y)\}$.
\end{lemma}

\begin{proof}
Let $S = \bigcup_{k \in K} S_k$ be an affine open covering.
Let $X = \bigcup_{i \in I} U_i$, $Y = \bigcup_{j \in J} V_j$
be affine open coverings with $I$, $J$ of cardinality
$\leq \text{size}(X), \text{size}(Y)$.
For each $i \in I$ there exists a finite set $K_i$ of $k \in K$
such that $f(U_i) \subset \bigcup_{k \in K_i} S_k$.
For each $j \in J$ there exists a finite set $K_j$ of $k \in K$
such that $g(V_j) \subset \bigcup_{k \in K_j} S_k$.
Hence $f(X), g(Y)$ are contained in
$S' = \bigcup_{k \in K'} S_k$ with
$K' = \bigcup_{i \in I} K_i \cup \bigcup_{j \in J} K_j$.
Note that the cardinality of $K'$
is at most $\max\{\aleph_0, |I|, |J|\}$. Applying
Lemma \ref{lemma-bound-size}
we see that it suffices to prove that
$\text{size}(f^{-1}(S_k) \times_{S_k} g^{-1}(S_k)) \leq \max\{\text{size}(X), \text{size}(Y))\}$ for $k \in K'$.
In other words, we may assume that $S$ is affine.

\medskip\noindent
Assume $S$ affine.
Let $X = \bigcup_{i \in I} U_i$, $Y = \bigcup_{j \in J} V_j$
be affine open coverings with $I$, $J$ of cardinality
$\leq \text{size}(X), \text{size}(Y)$.
Again by
Lemma \ref{lemma-bound-size}
it suffices to prove the lemma for the products
$U_i \times_S V_j$. By
Lemma \ref{lemma-bound-affine}
we see that it suffices to show that
$$|A \otimes_C B| \leq \max\{\aleph_0, |A|, |B|\}.$$
We omit the proof of this inequality.
\end{proof}

\begin{lemma}
\label{lemma-bound-finite-type}
Let $S$ be a scheme.
Let $f : X \to S$ be locally of finite type with $X$ quasi-compact.
Then $\text{size}(X) \leq \text{size}(S)$.
\end{lemma}

\begin{proof}
We can find a finite affine open covering $X = \bigcup_{i = 1, \ldots n} U_i$
such that each $U_i$ maps into an affine open $S_i$ of $S$. Thus by
Lemma \ref{lemma-bound-size}
we reduce to the case where both $S$ and $X$ are affine. In this case by
Lemma \ref{lemma-bound-affine}
we see that it suffices to show
$$|A[x_1, \ldots, x_n]| \leq \max\{\aleph_0, |A|\}.$$
We omit the proof of this inequality.
\end{proof}

\noindent
In
Algebra, Lemma \ref{algebra-lemma-epimorphism-cardinality}
we will show that if $A \to B$ is an epimorphism of rings, then
$|B| \leq \max(|A|, \aleph_0)$.
The analogue for schemes is the following lemma.

\begin{lemma}
\label{lemma-bound-monomorphism}
Let $f : X \to Y$ be a monomorphism of schemes.
If at least one of the following properties
holds, then $\text{size}(X) \leq \text{size}(Y)$:
\begin{enumerate}
\item $f$ is quasi-compact,
\item $f$ is locally of finite presentation,
\item add more here as needed.
\end{enumerate}
But the bound does not hold for monomorphisms
which are locally of finite type.
\end{lemma}

\begin{proof}
Let $Y = \bigcup_{j \in J} V_j$ be an affine open covering of $Y$
with $|J| \leq \text{size}(Y)$. By Lemma \ref{lemma-bound-size}
it suffices to bound the size of the inverse image of $V_j$ in $X$.
Hence we reduce to the case that $Y$ is affine, say $Y = \Spec(B)$.
For any affine open $\Spec(A) \subset X$ we have
$|A| \leq \max(|B|, \aleph_0) = \text{size}(Y)$, see remark above
and Lemma \ref{lemma-bound-affine}. Thus it suffices to show
that $X$ has at most $\text{size}(Y)$ affine opens. This is clear
if $X$ is quasi-compact, whence case (1) holds.
In case (2) the number of isomorphism classes of $B$-algebras $A$
that can occur is bounded by $\text{size}(B)$, because each
$A$ is of finite type over $B$, hence isomorphic to an algebra
$B[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$
for some $n, m$, and $f_j \in B[x_1, \ldots, x_n]$. However, as
$X \to Y$ is a monomorphism, there is a unique morphism
$\Spec(A) \to X$ over $Y = \Spec(B)$ if there is one,
hence the number of affine
opens of $X$ is bounded by the number of these isomorphism classes.

\medskip\noindent
To prove the final statement of the lemma consider the ring
$B = \prod_{n \in \mathbf{N}} \mathbf{F}_2$ and set $Y = \Spec(B)$.
For every ultrafilter $\mathcal{U}$ on $\mathbf{N}$ we obtain a maximal
ideal $\mathfrak m_\mathcal{U}$ with residue field $\mathbf{F}_2$;
the map $B \to \mathbf{F}_2$ sends the element $(x_n)$ to
$\lim_\mathcal{U} x_n$. Details omitted.
The morphism of schemes $X = \coprod_\mathcal{U} \Spec(\mathbf{F}_2) \to Y$
is a monomorphism as all the points are distinct. However the cardinality
of the set of affine open subschemes of $X$ is equal to the cardinality
of the set of ultrafilters on $\mathbf{N}$ which is
$2^{2^{\aleph_0}}$. We conclude as $|B| = 2^{\aleph_0} < 2^{2^{\aleph_0}}$.
\end{proof}

\begin{lemma}
\label{lemma-what-is-in-it}
Let $\alpha$ be an ordinal as in Lemma \ref{lemma-construct-category} above.
The category $\Sch_\alpha$ satisfies the following
properties:
\begin{enumerate}
\item If $X, Y, S \in \Ob(\Sch_\alpha)$, then
for any morphisms $f : X \to S$, $g : Y \to S$ the fibre
product $X \times_S Y$ in $\Sch_\alpha$ exists
and is a fibre product in the category of schemes.
\item Given any at most countable collection $S_1, S_2, \ldots$
of elements of $\Ob(\Sch_\alpha)$, the coproduct
$\coprod_i S_i$ exists in $\Ob(\Sch_\alpha)$ and
is a coproduct in the category of schemes.
\item For any $S \in \Ob(\Sch_\alpha)$ and
any open immersion $U \to S$, there exists a
$V \in \Ob(\Sch_\alpha)$ with $V \cong U$.
\item For any $S \in \Ob(\Sch_\alpha)$ and
any closed immersion $T \to S$, there exists a
$S' \in \Ob(\Sch_\alpha)$ with $S' \cong T$.
\item For any $S \in \Ob(\Sch_\alpha)$ and
any finite type morphism $T \to S$, there exists a
$S' \in \Ob(\Sch_\alpha)$ with $S' \cong T$.
\item Suppose $S$ is a scheme which has an open covering
$S = \bigcup_{i \in I} S_i$ such that there exists
a $T \in \Ob(\Sch_\alpha)$ 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 $\Sch_\alpha$.
\item For any $S \in \Ob(\Sch_\alpha)$ 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 a
$S' \in \Ob(\Sch_\alpha)$ 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.
\item For any $S \in \Ob(\Sch_\alpha)$ and
any monomorphism $T \to S$ which is either locally of finite presentation
or quasi-compact, there exists a
$S' \in \Ob(\Sch_\alpha)$ with $S' \cong T$.
\item Suppose that $T \in \Ob(\Sch_\alpha)$ is
affine. Write $R = \Gamma(T, \mathcal{O}_T)$.
Then any of the following schemes is isomorphic to a scheme
in $\Sch_\alpha$:
\begin{enumerate}
\item For any ideal $I \subset R$ with completion
$R^* = \lim_n R/I^n$, the scheme $\Spec(R^*)$.
\item For any finite type $R$-algebra $R'$, the
scheme $\Spec(R')$.
\item For any localization $S^{-1}R$, the scheme $\Spec(S^{-1}R)$.
\item For any prime $\mathfrak p \subset R$, the scheme
$\Spec(\overline{\kappa(\mathfrak p)})$.
\item For any subring $R' \subset R$, the scheme
$\Spec(R')$.
\item Any scheme of finite type over a ring of cardinality at most
$|R|^{\aleph_0}$.
\item And so on.
\end{enumerate}
\end{enumerate}
\end{lemma}

\begin{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 \ref{lemma-bound-size}. Statement (7) follows from (6).
Namely, for any affine open $V \subset T$ we have
$\text{size}(V) \leq \text{size}(S)$ by
Lemma \ref{lemma-bound-finite-type}.
Thus, we see that (6) applies in the situation of (7).
Part (8) follows from
Lemma \ref{lemma-bound-monomorphism}.

\medskip\noindent
Statement (9) is translated, via Lemma \ref{lemma-bound-affine},
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 $\aleph_0$.)
\end{proof}

\begin{remark}
\label{remark-what-is-not-in-it}
Let $R$ be a ring. Suppose we consider the ring
$\prod_{\mathfrak p \in \Spec(R)} \kappa(\mathfrak p)$.
The cardinality of this ring is bounded by $|R|^{2^{|R|}}$,
but is not bounded by $|R|^{\aleph_0}$ in general.
For example if $R = \mathbf{C}[x]$ it is not bounded by
$|R|^{\aleph_0}$ and if $R = \prod_{n \in \mathbf{N}} \mathbf{F}_2$
it is not bounded by $|R|^{|R|}$.
Thus the And so on'' of Lemma \ref{lemma-what-is-in-it} above
should be taken with a grain of salt. Of course, if it ever becomes
necessary to consider these rings in arguments pertaining to
fppf/\'etale cohomology, then we can change the function
$Bound$ above into the function $\kappa \mapsto \kappa^{2^\kappa}$.
\end{remark}

\noindent
In the following lemma we use the notion of an fpqc covering which
is introduced in Topologies, Section \ref{topologies-section-fpqc}.

\begin{lemma}
\label{lemma-bound-by-covering}
Let $f : X \to Y$ be a morphism of schemes. Assume there exists an
fpqc covering $\{g_j : Y_j \to Y\}_{j \in J}$ such that $g_j$ factors
through $f$. Then $\text{size}(Y) \leq \text{size}(X)$.
\end{lemma}

\begin{proof}
Let $V \subset Y$ be an affine open. By definition there exist
$n \geq 0$ and $a : \{1, \ldots, n\} \to J$ and affine opens
$V_i \subset Y_{a(i)}$ such that
$V = g_{a(1)}(V_1) \cup \ldots \cup g_{a(n)}(V_n)$.
Denote $h_j : Y_j \to X$ a morphism such that $f \circ h_j = g_j$.
Then $h_{a(1)}(V_1) \cup \ldots \cup h_{a(n)}(V_n)$ is
a quasi-compact subset of $f^{-1}(V)$. Hence we can find a
quasi-compact open $W \subset f^{-1}(V)$ which contains
$h_{a(i)}(V_i)$ for $i = 1, \ldots, n$.
In particular $V = f(W)$.

\medskip\noindent
On the one hand this shows that the cardinality of the set of
affine opens of $Y$ is at most the cardinality of the set $S$ of
quasi-compact opens of $X$. Since every quasi-compact open of
$X$ is a finite union of affines, we see that the cardinality
of this set is at most $\sup |S|^n = \max(\aleph_0, |S|)$.
On the other hand, we have
$\mathcal{O}_Y(V) \subset \prod_{i = 1, \ldots, n} \mathcal{O}_{Y_{a(i)}}(V_i)$
because $\{V_i \to V\}$ is an fpqc covering. Hence
$\mathcal{O}_Y(V) \subset \mathcal{O}_X(W)$ because $V_i \to V$
factors through $W$. Again since $W$ has a finite covering by
affine opens of $X$ we conclude that $|\mathcal{O}_Y(V)|$
is bounded by the size of $X$. The lemma now follows
from the definition of the size of a scheme.
\end{proof}

\noindent
In the following lemma we use the notion of an fppf covering which
is introduced in Topologies, Section \ref{topologies-section-fppf}.

\begin{lemma}
\label{lemma-bound-fppf-covering}
Let $\{f_i : X_i \to X\}_{i \in I}$ be an fppf covering of a scheme.
There exists an fppf covering $\{W_j \to X\}_{j \in J}$
which is a refinement of $\{X_i \to X\}_{i \in I}$ such that
$\text{size}(\coprod W_j) \leq \text{size}(X)$.
\end{lemma}

\begin{proof}
Choose an affine open covering $X = \bigcup_{a \in A} U_a$ with
$|A| \leq \text{size}(X)$. For each $a$ we can choose
a finite subset $I_a \subset I$ and for $i \in I_a$ a quasi-compact open
$W_{a, i} \subset X_i$ such that $U_a = \bigcup_{i \in I_a} f_i(W_{a, i})$.
Then $\text{size}(W_{a, i}) \leq \text{size}(X)$ by
Lemma \ref{lemma-bound-finite-type}.
We conclude that
$\text{size}(\coprod_a \coprod_{i \in I_a} W_{i, a}) \leq \text{size}(X)$
by Lemma \ref{lemma-bound-size}.
\end{proof}

Comment #316 by w.xu on September 26, 2013 a 7:02 pm UTC

In the proof of lemma 3.9.8, the underlying set of X is written as a union of affine opens whose index set is just X. And then it says it has cardinality at most size(Y). I can only see that the cardinality of X is less than the cardinality of Y. The definition of size() cannot imply the statement.

Comment #318 by Johan (site) on October 22, 2013 a 3:09 pm UTC

@#316: Since the cardinality of X is less than the cardinality of Y (for just points) we see that the index set of the union (namely the set of points of X) has cardinality at most the cardinality of the set of points of Y. What is used is that the cardinality of the set of points of a scheme is at most the size of the scheme... I guess what you are saying is that that is wrong... Hmm... I think you are right. Oops! Will fix this soon.

Comment #319 by Johan (site) on October 22, 2013 a 7:54 pm UTC

OK, there was no fix for the lemma. What I did was to impose a further condition on the morphism, requiring it to be either quasi-compact or locally of finite type. You can find the commit here. Luckily we only used the lemma as stated in one place.

You win a Stacks project mug. Thanks!

## Add a comment on tag 000H

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).