The Stacks project

An heavy proof for this is :

Let $\kappa$ be a cardinal. In the following set theoretic reasoning, all functions have to be regarded as their graphs, and a scheme $S$ is represented as the 4-tuple $S=(X,\mathcal{T},\mathcal{F},\rho)$ where $(X,\mathcal{T})$ is the underlying topological space of $S$ and $\mathcal{O}_S=(\mathcal{F},\rho)$ is the sheaf of rings of S. To be clear on this representation, there exists a set of rings $\mathcal{C}$ and a set of morphisms $\mathcal{D}$ such that $\mathcal{F}:\mathcal{T}\rightarrow\mathcal{C}$ is the "rule" that associates to every open set its ring of regular functions and $\rho:\{(U,V)\in\mathcal{T}^2\;:\;V\subset U\}\rightarrow\mathcal{D}$ is the "rule" that associates to each inclusion of open sets its restriction morphism. We then set $\epsilon_1=\mathfrak{P}(\kappa^\kappa)$, $\epsilon_2=\mathfrak{P}(\mathfrak{P}(\mathfrak{P}(\kappa^\kappa)))$, $\epsilon_3=\mathfrak{P}(\mathfrak{P}(\mathfrak{P}(\kappa^\kappa))\times\mathfrak{P}(\kappa^\kappa)\times\mathfrak{P}((\kappa^\kappa)^3)\times\mathfrak{P}((\kappa^\kappa)^3))$ and $\epsilon_4=\mathfrak{P}(\mathfrak{P}(\kappa^\kappa)^2\times\mathfrak{P}((\kappa^\kappa)^2))$. We set $A=\{(A,B,C,D)\in\epsilon_1\times\epsilon_2\times\epsilon_3\times\epsilon_4\;:\;(A,B,C,D)\;\text{is a scheme}\}$.\~\ Now, we let $S=(X,\mathcal{T},\mathcal{F},\rho)$ be a scheme with $\text{size}(S)\leq\kappa$.\~\First of, if $U\in\mathcal{T}$, we write $\mathcal{F}(U)=(E_U,+_U,\times_U)$ its ring of regular functions. In this setting, we have $\forall U\in\mathcal{T},\;\lvert E_U\rvert\leq\kappa_2^{\kappa_1}\leq\kappa^\kappa$. Indeed, since $S$ is a scheme, the open affines of $S$ form a basis of $\mathcal{T}$, so if $U\in\mathcal{T}$, then $E_U$ injects (via restrictions to sub affines of U) in $\prod_{V\;\text{affine},\;V\subset U} E_V$ as $\mathcal{O}_S$ is a sheaf. Furthermore, $\lvert X\rvert\leq \kappa_1 2^{\kappa_2}\leq\kappa2^\kappa\leq2^\kappa\leq\kappa^\kappa$ (the third inequality holds as $\kappa$ is infinite) since $X$ is covered by its affines as it is a scheme and since for an affine $U\cong Spec(A)$ (for some $A$), we have $\Gamma(U,\mathcal{O}_S)\cong A$ (and so $\lvert U \rvert=\lvert Spec(A)\rvert\leq2^{\lvert A\rvert}$). Using the axiom of choice, we can now choose a family of injections $(i_U:g\in E_U\longmapsto\widehat{g}\in\kappa^\kappa)_{U\in\mathcal{T}}$ and an injection $i_X:x\in X\longmapsto\widehat{x}\in\kappa^\kappa$. We set $\widehat{X}=i_X(X)$, $\widehat{\mathcal{T}}=\{i_X(U)\;:\;U\in\mathcal{T}\}$ and for all $U\in\mathcal{T}$, we define $\widehat{E_U}=i_U(E_U)$, $\widehat{+_U}:(\widehat{g},\widehat{h})\in\widehat{E_U}^2\longmapsto\widehat{g+_U h}\in\widehat{E_U}$ and $\widehat{\times_U}:(\widehat{g},\widehat{h})\in\widehat{E_U}^2\longmapsto\widehat{g\times_U h}\in\widehat{E_U}$ (both are well defined as the injections are bijective on their images). Now we set $\widehat{\mathcal{F}}:\widehat{U}\in\widehat{\mathcal{T}}\longmapsto(\widehat{E_U},\widehat{+_U},\widehat{\times_U})\in\mathfrak{P}(\kappa^\kappa)\times\mathfrak{P}((\kappa^\kappa)^3)\times\mathfrak{P}((\kappa^\kappa)^3)$ and $\widehat{\rho}:(\widehat{U},\widehat{V})\in\{(\widehat{A},\widehat{B})\;:\;(A,B)\in\mathcal{T}^2,\;B\subset A\}\longmapsto(\widehat{\rho_V^U}:\widehat{g}\in\widehat{E_U}\longmapsto\widehat{\rho_V^U(g)}\in\widehat{E_V})\in\mathfrak{P}((\kappa^\kappa)^2)$.\~\ By construction, $\widehat{X}\in\epsilon_1$, $\widehat{\mathcal{T}}\in\epsilon_2$, $\widehat{\mathcal{F}}\in\epsilon_3$ as $\widehat{\mathcal{T}}\subset\mathfrak{P}(\mathfrak{P}(\kappa^\kappa))$ and $\widehat{\rho}\in\epsilon_4$ as $\{(\widehat{A},\widehat{B})\;:\;(A,B)\in\mathcal{T}^2,\;B\subset A\}\subset\widehat{\mathcal{T}}\subset\mathfrak{P}(\mathfrak{P}(\kappa^\kappa))$. What we just did is build a replica of $S$, $\widehat{S}:=(\widehat{X},\widehat{\mathcal{T}},\widehat{\mathcal{F}},\widehat{\rho})$ belonging to A as it naturally is a scheme, trivially isomorphic as schemes to $S$.

It is weird, curly brackets do not appear. Many sets defined by comprehension are not displayed correctly. I do not know how to fix it.

Going to leave as is.