The Stacks project

Decided to remove the ${}^*$'s. Thanks. See this.

Thanks and fixed here.

Yes, that is clear from the proof. Going to leave as is.

Thanks and fixed here.

Thanks and fixed here.

Going to leave as is. (Note that all ideals in a Cohen ring $\Lambda$ are given by powers of $p$ so the thing about kernels of maps from a Cohen ring is clear.)

It seems that invoking Lemma 00F4 is not necessary, for $S_\lambda \rightarrow S$ is surjective, which immediately makes $R \rightarrow S$ finitely presented.

Thanks and fixed here.

Thanks. I changed the statement of the lemma. Unfortunately, now it seems a bit harder to parse the statement. Changes are here.

Going to leave as is.

Thanks and fixed here.

@#8747. Note that $\mathop{\mathrm{Hom}}\nolimits _ R (S^{-1}M, N) = \mathop{\mathrm{Hom}}\nolimits _ {S^{-1}R} (S^{-1}M, N)$ by Lemma 10.9.5. So it's fine as stated and the proof gives the stated result. Going to leave as is.

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.

This is a good exercise. Prove it using the following steps: (a) first prove it for $k$-points, (b) then prove it for $K$-points for any field extension $K/k$, (c) show that (b) is enough to conclude. OK?

OK, I got a nice exposition of the heart of this proof isolated in a separate lemma which I have accepted and edited here. Hopefully this will be online soon.

No. I changed Lemma 10.120.17 in stead.

Thanks. Fixed it by only allowing nonzero $\mathfrak m$ in this commit.

Yes, but we don't even need it for the argument to work.

OK, I think it should be $(g, -f)$. Fixed here.

Thanks and fixed here.