The Stacks project

I think my comment was wrong. There is a filtration after base change, and you can use it to prove the result. However, that is not the shortest proof, and I do not know if the filtration comes from a sequence of sections. I need to double-check what is in "N'eron models" -- they have effectively evicted us from our offices while they repair the HVAC system.

Thanks Jason!

Since we've defined the normalization only for a quasi-compact and quasi-separated morphism, shouldn't one remark at some point during the proof that this is indeed the case for $g$? (It follows from 26.21.13 and 26.21.14.)

Also, instead of "with a unique morphism $h:X'\to Z$. We omit the verification that the diagram commutes," one could write "with a unique morphism $h:X'\to Z$ over $X$. On the other hand, by https://stacks.math.columbia.edu/tag/01LQ#comment-8448 , we have $h\circ f'=g$."

I propose to add the following remark to the webpage: Let $S$ be a scheme and let $\varphi:\mathcal{A}\to\mathcal{B}$ be a morphism of quasi-coherent $\mathcal{O}_S$-algebras. As it was explained in #8438 (ii), we obtain a morphism $\underline{\text{Spec}}_S\mathcal{B}\to \underline{\text{Spec}}_S\mathcal{A}$ over $S$. This gives a transformation of functors $\text{Hom}_S(-,\underline{\text{Spec}}_S\mathcal{B})\to\text{Hom}_S(-,\underline{\text{Spec}}_S\mathcal{A})$ that induce a transformation of functors $G'\to G$, where $G$ is our old friend and $G'$ is the functor taking a scheme $f:T\to S$ over $S$ to $\text{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{B},f_*\mathcal{O}_T)$. We claim that at $f:T\to S$ the transformation $G'\to G$ is given by precomposition by $\varphi$. We begin with a morphism $\mathcal{B}\to f_*\mathcal{O}_X$ of $\mathcal{O}_S$-algebras that gives a morphism $g:T\to\underline{\text{Spec}}_S\mathcal{B}$ over $S$. In turn, we get a composite $T\to\underline{\text{Spec}}_S\mathcal{B}\to\underline{\text{Spec}}_S\mathcal{A}$. By #8438 (i), this gives a morphism of $\mathcal{O}_S$-algebras $\mathcal{A}\to f_*\mathcal{O}_T$ that on sections over an open affine $U\subset S$ reads as $\mathcal{A}(U)\xrightarrow{\varphi}\mathcal{B}(U)\to\mathcal{O}_X(f^{-1}U)$, i.e., it equals the composite $\mathcal{A}\to\mathcal{B}\to f_*\mathcal{O}_X$, which is what we wanted to show.

By the discussion in "N'eron models" around finite flat morphisms, after a faithfully flat base change there is a filtration of f_*O_X whose associated graded pieces are f-pushforwards of pushforwards of invertible O_S-modules along sections of f (i.e., multisections are "scheme-theoretic unions" of sections after faithfully flat base change). The composition of those sections with sigma give a corresponding filtration of the structure sheaf of the relative Proj (by ideal sheaves) compatible with the filtration on the pushforward of O_X. You can use this to prove that the morphism from the structure sheaf of the relative Proj to the pushforward of O_X is surjective.

I think instead of "we have $a_i \in R_f$ and $x^d + \sum_{i = 1, \ldots, d} \varphi(a_i) x^{d - i} = 0$ in $S_f$" it would be more informative to write "there are $b_i\in R_f$ such that $x^d + \sum_{i = 1, \ldots, d} \varphi(b_i) x^{d - i} = 0$ in $S_f$." At least in the calculation I derived I end up with different $a_i$'s (this doesn't affect the rest of the proof).

I agree with MAO Zhouhang that it's useful to mention here that $D_{I^\infty\text{-}\mathrm{torsion}}(A)$ is the category of $A$ complexes with $I^\infty$-torsion cohomology.

Argh, the reference to Lemma 26.21.11 is still wrong because the Proj is over $S$ and not over $X$. The best thing would be for people to work out for themselves why $\sigma$ is a closed immersion (by doing a local computation). But we can also see it using theory. Write $X$ as $\text{Proj}_S(\mathcal{A})$ where $\mathcal{A} = (f_*\mathcal{O}_X)[T]$. (Namely, we have $\text{Proj}(S[T]) = \text{Spec}(S)$ and there is a relative version of this.) Then with $\mathcal{B} = \text{Sym}^*_{\mathcal{O}_S}(f_*\mathcal{O}_X)$ there is a map $\psi : \mathcal{B} \to \mathcal{A}$ of graded $\mathcal{O}_S$-algebras. (Namely, given a ring map $R \to S$ there is a graded $R$-algebra map $\text{Sym}_R^*(S) \to S[T]$, $s \mapsto sT$ and there is a relative version of this.) The map $\psi$ is surjective in all sufficiently high degrees. (Namely, in degrees $\geq 1$.) Whence the corresponding morphism $r_\psi : X \to \mathbf{P}(f_*\mathcal{O}_X)$ is everywhere defined and a closed immersion, see Lemma 27.18.3.

Okay, sorry, the conditions $\rho_U^U=\text{id}_{X_U}$ and $\theta^U_U=\text{id}_{\mathcal{F}_U}$ already follow from the current hypotheses (maybe one could mention this?) .

In 27.2.2.1, I think the middle and last arrows should be labelled as $(i_V^{-1})^*(\theta_U|_{f^{-1}(V)})$ and $(i_V^{-1})^*\theta_V^{-1}$, respectively.

Instead of saying "by covering the intersection $U_1\cap U_2=\bigcup V_j$ by elements $V_j$ of $\mathcal{B}$ and taking" I think it would be better to write "by considering any $V\in\mathcal{B}$ such that $V\subset U_1\cap U_2$ and defining", since actually using some covering is not enough to later construct $\varphi_{12}$.

To give just the minimum amount of hints to avoid the first use of "we omit," one could:

1. After "condition (d) says exactly that this is compatible in case we have a triple of elements $W\subset V\subset U$ of $\mathcal{B}$," add "i.e., $(\theta')^U_W=(\theta')^V_W\circ(\theta')^U_V|_{f^{-1}(W)}$."
2. If we used the notation $\varphi^{V_j}_{12}$ for what the proof currently denotes $\varphi_{12}|_{V_j}$, then instead of "we omit the verification that these maps do indeed glue to a $\varphi_{12}$" one could write "since $\varphi^V_{12}|_W=\varphi^W_{12}$ for $V,W\in\mathcal{B}$ with $W\subset V\subset U_1\cap U_2$, by Sheaves, 6.33.1, we get a $\varphi_{12}$."

On the other hand, I think that for 27.2.2.1 to follow, in the lemma hypotheses we would need to add "suppose $\rho_U^U=\text{id}_{X_U}$ and $\theta^U_U=\text{id}_{\mathcal{F}_U}$ for $U\in\mathcal{B}$." The proof of 27.2.2.1 may be added after "we get our $\mathcal{F}$ on $X$" by adding "along with isomorphisms $\theta_U:\mathcal{F}'_U\to\mathcal{F}|_{f^{-1}(U)}$, $U\in\mathcal{B}$, such that $\theta_U|_{f^{-1}(U\cap V)}=\theta_V|_{f^{-1}(U\cap V)}\circ \varphi_{U,V}$ for $U,V\in\mathcal{B}$. In particular, for $U,V\in\mathcal{B}$ with $V\subset U$, we have $\theta_V^{-1}\circ\theta_U|_{f^{-1}(V)}=\varphi_{U,V}=(\theta')^U_V=i^*_V\theta^U_V$, which is 27.2.2.1 (in the middle equality we used the definition of $\varphi_{U,V}$, plus $(\theta')^V_V=\text{id}_{\mathcal{F}'_V}$)."

Potential typo in the statement of the lemma: it is written "[l]et $U\subset X$ be an open", but condition (3) is that $U\cap p(i(Z))=\emptyset$, which only makes sense if $U\subset S$.

Minor style typo: in the statement perhaps one would want to enclose the name of the categories between curly braces to fit the style in 29.11.6.

2nd proof (maybe worth of mention?): The functor $\underline{\text{Spec}}_S$ is fully faithful by remark (ii) of https://stacks.math.columbia.edu/tag/01LQ#comment-8438 , and it is essentially surjective by 29.11.3.

Final remarks on why I am writing all these comments in this section instead of just introducing a pull request on GitHub: In my opinion, it would be better to change this whole section to the study of the representability of $G$ instead of $F$, and maybe mentioning the existence of $F$ just at the end plus the categorical remark in #8435. I do think so because (a) $G$ is easier to remember and more motivated than $F$, see #8435 (I also think it's what #3430 had on his mind), and (b) proofs with $G$ are less wordy than those with $F$. However, I was unsure whether such a big edit would be accepted or if only a proper subset of all the edits would pass. In consequence, I deemed it better to point it all out here.

I think it would be nice to add the following two remarks somewhere in this section. I will be adopting the POV explained in #8435 of representability of $G$ instead of $F$.

(i) So far, we have an explicit description of the map $$\tag{1}\label{map} \text{Hom}_S(T,\underline{\text{Spec}}_S\mathcal{A}) \to\operatorname{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T),$$ Namely, it is the one induced by the universal element $\xi=(\mathcal{A}\to\pi_*\mathcal{O}_{\underline{\text{Spec}}_S\mathcal{A}})\in G(\pi)$. However, we don't have yet an explicit description of the inverse of \eqref{map}. Here's one: Given an $\mathcal{O}_S$-linear map $\mathcal{A}\to f_*\mathcal{O}_T$, the induced map $T\to\underline{\text{Spec}}_S\mathcal{A}$ is the unique morphism over $S$ such that for all open affine $U\subset S$, the restricted map over $U$ equals $f^{-1}(U)\to\operatorname{Spec}(\mathcal{A}(U)) \xrightarrow{i_U^{-1}}\pi^{-1}(U)$, where $i_U$ is the map in 27.3.4 and the first arrow comes from Schemes, 26.6.4. Proof: First we note that $U\subset S\mapsto\text{Hom}_U(f^{-1}(U),\pi^{-1}(U))$ is a sheaf on $S$ (basically because $U\subset S\mapsto\text{Hom}_U(f^{-1}(U),\underline{\text{Spec}}_S\mathcal{A})$ is a sheaf). Then we note that for any open $U\subset S$, we have an induced map $$\text{Hom}_U(f^{-1}(U),\pi^{-1}(U))\to\text{Hom}_{\mathcal{O}_U\text{-Alg}}(\mathcal{A}|_U,\underbrace{f_*\mathcal{O}_T|_U}_{(f|_{f^{-1}(U)})_*\mathcal{O}_{f^{-1}(U)}}).$$ Namely, it is the one obtained by the universal element $\iota^*\xi=(\mathcal{A}|_U\to(\pi_*\mathcal{O}_{\underline{\text{Spec}}_S\mathcal{A}})|_U)$ of $G_\iota$, where $\iota:\pi^{-1}(U)\to\underline{\text{Spec}}_S\mathcal{A}$ is the inclusion and we are using the notation given in https://stacks.math.columbia.edu/tag/001L#comment-8427 . From here, it is not difficult to verify that the diagram $$\require{AMScd} \begin{CD} \text{Hom}_S(T,\underline{\text{Spec}}_S\mathcal{A})@>>> \operatorname{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T)\\ @VVV@VVV\\ \text{Hom}_U(f^{-1}(U),\pi^{-1}(U))@>>> \text{Hom}_{\mathcal{O}_U\text{-Alg}}(\mathcal{A}|_U,f_*\mathcal{O}_T|_U) \end{CD}$$ commutes. In other words, we have a morphism of sheaves $\text{Hom}_{(-)}(f^{-1}(-),\pi^{-1}(-))\to\mathcal{H}om_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T)$ on $S$. Thus, we can characterize the inverse of \eqref{map} locally on $S$ and we finish by the proof in https://stacks.math.columbia.edu/tag/01LT#comment-8436

(ii) An explanation of why $\underline{\text{Spec}}_S$ is a functor: If we have a morphism of $\mathcal{O}_S$-algebras $\mathcal{A}\to\mathcal{B}\cong\pi_{\mathcal{B},*}\mathcal{O}_{\underline{\text{Spec}}_S\mathcal{B}}$ (we've used 27.4.6, (3)) then we obtain a canonical map $\underline{\text{Spec}}_S\mathcal{B}\to\underline{\text{Spec}}_S\mathcal{A}$ over $S$. By remark (i), it is the unique morphism of schemes over $S$ such that for every open affine $U\subset S$, we have that $\pi_\mathcal{B}^{-1}(U)\to\pi_\mathcal{A}^{-1}(U)$ identifies with $\text{Spec}(\mathcal{B}(U))\to\text{Spec}(\mathcal{A}(U))$ via the isomorphisms of Lemma 27.3.4. It follows that $\underline{\text{Spec}}_S$ is a contravariant functor from the category of quasi-coherent $\mathcal{O}_S$-algebras to the category of schemes over $S$. On the other hand, equation \eqref{map} specializes to $$\text{Hom}_S(\underline{\text{Spec}}_S\mathcal{B}, \underline{\text{Spec}}_S\mathcal{A}) \xrightarrow{\cong}\operatorname{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},\mathcal{B}),$$ so the functor is fully faithful.

'The second main result is that for a morphism $f$ if ringed sites....' right above the displayed equality of derived completions has a typo. The 'if' should be 'of'.

I would propose writing the whole proof in this way: I'm going to adopt the POV of representability of $G$ instead of $F$, see https://stacks.math.columbia.edu/tag/01LQ#comment-8435 . We have $\mathcal{A}=\widetilde{A_R}$ (where $A_R$ just means $A$ regarded as an $R$-module, notation introduced in the paragraph before 10.14.3). To show that $\pi:\operatorname{Spec} A\to\operatorname{Spec} R$ represents $$\begin{align*} G:\text{Sch}^\text{opp}/S&\to\text{Sets}\\ (f:T\to S)&\mapsto \operatorname{Hom}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T), \end{align*}$$ we check that the identity map $\varphi_\text{univ}:\widetilde{A_R}\to\pi_*\widetilde{A}=\widetilde{A_R}$ is a universal element for $G$. Fix $f:T\to S$. We have a map $\text{Hom}_S(T,\operatorname{Spec}A)\to G(f)$, induced by $\varphi_\text{univ}$. To construct the inverse, for an $\mathcal{O}_S$-linear map $\varphi:\widetilde{A_R}\to f_*\mathcal{O}_T$ (an element of $G(f)$), we have a map on global sections $A\to\Gamma(T,\mathcal{O}_T)$, which is $R$-linear. Therefore, it induces a morphism $T\to\operatorname{Spec}A$ over $\operatorname{Spec} R$, by Schemes, 26.6.4. On the one hand, it is easy to see that the composite $\text{Hom}_S(T,\operatorname{Spec}A)\to G(f)\to\text{Hom}_S(T,\operatorname{Spec}A)$ is the identity (use uniqueness in Schemes, 26.6.4). On the other hand, to see that $G(f)\to \text{Hom}_S(T,\operatorname{Spec}A)\to G(f)$ is the identity, we note that all the involved maps are compatible with restrictions to distinguished open subsets of $S$ (i.e., given $U\subset S$ distinguished open, one restricts to $f^{-1}(U)\to U$ and to $\pi^{-1}(U)\to U$). Hence, it suffices to see that $$\begin{matrix} G(f)&\longrightarrow&\text{Hom}_S(T,\operatorname{Spec}A)&\longrightarrow& G(f)\\ &\searrow&&\swarrow\\ &&\text{Hom}_{R\text{-Alg}}(A,\Gamma(T,\mathcal{O}_T)) \end{matrix}$$ commutes. But this is easy.

I think it's also interesing to think about the presheaf on $\text{Sch}/S$ that the object $\pi:\underline{\text{Spec}}_S\mathcal{A}\to S$ represents. Let $\mathcal{C}$ be a category, and let $c$ be an object of $\mathcal{C}$. Let $G:\mathcal{C}^\text{opp}/c\to\text{Sets}$ be a presheaf on $\mathcal{C}/c$. Consider the induced functor $G':\mathcal{C}^\text{opp}\to\text{Sets}$ that sends $d\in\text{Ob}(\mathcal{C})$ to $\coprod_{f \in \mathcal{C}(d,c)} G(f)$. Then $G$ is representable if and only if $G'$ is representable (not difficult to show). In our situation, $\mathcal{C}=\text{Sch}$, $c=S$, the functor $G$ sends $f:T\to S$ to $\text{Mor}_{\mathcal{O}_S\text{-Alg}}(\mathcal{A},f_*\mathcal{O}_T)\cong\text{Mor}_{\mathcal{O}_T\text{-Alg}}(f^*\mathcal{A},\mathcal{O}_T)$, and $G'=F$ is 27.4.0.1.

That is, in Görtz & Wedhorn words (see (11.2) of the 2nd ed.), this construction can be regarded as a globalized version of the natural isomorphim $$\text{Hom}_{\operatorname{Spec}R}(T,\operatorname{Spec} A) \cong \text{Hom}_{R\text{-Alg}}(A,\Gamma(T,\mathcal{O}_T)),$$ where $A$ is an $R$-algebra.

I'm a little confused about the statement of the lemma and its proof. In the statement, we see $U$ is an algebraic space étale over $X$. But in the proof, after citing lemma 17.5.2, we see that $U$, having an underlying topological space, is a scheme.

I guess that previous lemma 65.20.1 is used to prove the space case. Since the $\mathcal{F}\vert_U \in U_{\mathrm{space,étale}}$, we may assume $U=X$. Then consider the sheaf $\mathcal{G}$ assigning $V\rightarrow X$ to a singleton if $\sigma\vert_V=0$ and assigning $V$ to $\emptyset$ otherwise. Then using 65.20.1 we get the open subspace $W$.

There is also a typo for the conclusion $(3)$: where $\bar{s}=(U\rightarrow X)\circ \bar{u}$.

I think the expression $(\underline{\text{Spec}}_S(\mathcal{A}), \varphi) \in F(\underline{\text{Spec}}_S(\mathcal{A}))$ should be $(\pi, \varphi)\in F(\underline{\text{Spec}}_S(\mathcal{A}))$ instead. Here's an alternative way to phrase the proof that doesn't require invoking $X$: "Let $U\subset S$ be open affine. Since the functor $F_U$ is represented by $\pi^{-1}(U)$, let $(\pi^U,\varphi^U)\in F_U(\pi^{-1}(U))$ be the universal element. By the proof of 26.15.4 (we verified the hypotheses in the proof of 27.4.3), it suffices to see that $(\pi,\varphi)|_U=(\pi^U,\varphi^U)$. This follows from the construction of $\pi$ and $\varphi$."

In line 4, it should read "such that (X_{K'})_{red} is geometrically reduced"