The Stacks project

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

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

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. $\square$

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. $\square$

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

Moreover, it shows that the set of affine opens of $S$ has cardinality less than or equal to the cardinality of the set

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. $\square$

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

We omit the proof of this inequality. $\square$

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

We omit the proof of this inequality. $\square$

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 $|B| = 2^{\aleph _0} < 2^{2^{\aleph _0}}$. $\square$

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 $\aleph _0$.) $\square$

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. $\square$

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. $\square$