## 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

3.9.1.1
$$\label{sets-equation-bound} Bound(\kappa ) = \max \{ \kappa ^{\aleph _0}, \kappa ^+\} .$$

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 $\mathit{Sch}_\alpha$ 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 $\mathit{Sch}_\alpha$ 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, $S_0 \subset \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$.

2. For any $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ and any scheme $T$ with $\text{size}(T) \leq Bound(\text{size}(S))$, there exists a scheme $S' \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ such that $T \cong S'$.

3. For any countable1 diagram category $\mathcal{I}$ and any functor $F : \mathcal{I} \to \mathit{Sch}_\alpha$, the limit $\mathop{\mathrm{lim}}\nolimits _\mathcal {I} F$ exists in $\mathit{Sch}_\alpha$ if and only if it exists in $\mathit{Sch}$ and moreover, in this case, the natural morphism between them is an isomorphism.

4. For any countable index category $\mathcal{I}$ and any functor $F : \mathcal{I} \to \mathit{Sch}_\alpha$, the colimit $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} F$ exists in $\mathit{Sch}_\alpha$ if and only if it exists in $\mathit{Sch}$ 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 $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\alpha )})$ and any scheme $T$ with $\text{size}(T) \leq Bound(\text{size}(S))$, there exists a scheme $S' \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\beta )$ such that $T \cong S'$.

3. For any countable index category $\mathcal{I}$ and any functor $F : \mathcal{I} \to \mathit{Sch}_{f(\alpha )}$, if the limit $\mathop{\mathrm{lim}}\nolimits _\mathcal {I} F$ or the colimit $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} F$ exists in $\mathit{Sch}$, then it is isomorphic to a scheme in $\mathit{Sch}_\beta$.

To see $\beta$ exists, we argue as follows. Since $\mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\alpha )})$ is a set, we see that $\kappa = \sup _{S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\alpha )})} Bound(\text{size}(S))$ 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 $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$. We define $\beta (S)$ to be the least ordinal $\beta$ such that $S \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\beta )})$. By the above we see that always $\beta (S) < \beta _1$. Since $\mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\beta + 1)}) \subset \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$, we see by construction of $f$ above that the second property of the lemma is satisfied.

Suppose that $\{ S_1, S_2, \ldots \} \subset \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ 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 $\{ S_1, S_2, \ldots \} \subset \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_{f(\beta )})$. It follows that any functor $F : \mathcal{I} \to \mathit{Sch}_\alpha$ factors through one of the subcategories $\mathit{Sch}_{f(\beta )}$. 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 $\mathit{Sch}_{f(\beta + 1)}$ which is a subcategory of $\mathit{Sch}_\alpha$. 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 26.12.5). 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 $X \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}_\alpha )$ the reduction of $X$ is also an object of $\mathit{Sch}_\alpha$ (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 $\mathop{\mathrm{Spec}}(R)$. This is clear since any affine open $U \subset \mathop{\mathrm{Spec}}(R)$ 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 [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. $\square$

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 [Ch. I, 10.13, Kunen]. Hence by [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. $\square$

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 10.107.13 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 $Y = \mathop{\mathrm{Spec}}(B)$. For any affine open $\mathop{\mathrm{Spec}}(A) \subset X$ 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 $\mathop{\mathrm{Spec}}(A) \to X$ over $Y = \mathop{\mathrm{Spec}}(B)$ 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 $Y = \mathop{\mathrm{Spec}}(B)$. 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 $\mathop{\mathrm{lim}}\nolimits _\mathcal {U} x_ n$. Details omitted. The morphism of schemes $X = \coprod _\mathcal {U} \mathop{\mathrm{Spec}}(\mathbf{F}_2) \to Y$ 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 $\mathit{Sch}_\alpha$ 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 $\aleph _0$.) $\square$

Remark 3.9.10. Let $R$ be a ring. Suppose we consider the ring $\prod _{\mathfrak p \in \mathop{\mathrm{Spec}}(R)} \kappa (\mathfrak p)$. 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 34.9.

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 34.7.

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).

Comment #316 by w.xu on

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 on

@#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 on

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!

Comment #7906 by chris on

This page would benefit from a forward reference to what a scheme is.

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