The Stacks project

In the fourth line of the proof, the $(1,f_n,f_{n-1}\circ f_n,\dots,f_2\circ\cdots\circ f_n$ is missing a parenthesis.

Seems like a typo: After Defn. 04FS, is the base scheme $S$ just $\mathbf A^1_{\mathbf C}$ instead of its $\operatorname{Spec}$?

I think a slightly more general variant of Remark 03B8 is the following: given (1) in Lemma 015Z, the converse to its conclusion is true at least if $\mathcal{A}$ has enough injectives. Indeed, let $f:B\to B'$ be a monomorphism in $\mathcal{B}$, and let $A=\mathrm{Ker}\ v(f)$. Choose a monomorphism $g:A\to I$ with $I$ injective (it exists by assumption). Then $g$ extends to $g':v(B)\to I$, whence by adjunction a morphism $B\to u(I)$. If $u(I)$ is injective, this morphism extends to $h:B'\to u(I)$, hence by adjunction a morphism $k:v(B')\to I$ extending $g'$. But then $k$ "extends'' $g$, which forces $A=0$ since $g$ was a monomorphism.

Maybe one can weaken this hypothesis on $\mathcal{A}$?

Details on the diagram chasing parts: (1) $\Rightarrow$ (2) comes from a four lemma (2 injective 1 surjective version). It shows the map is injective, and surjectivity follows from finite generation. (4) $\Rightarrow$ (1) comes from the snake lemma: we have the SES $\ker f_1 \to \textrm{coker} f_3 \to \textrm{coker} f_2$, and since $f_2$, $f_1$ are isomorphisms, the kernel or cokernel is $0$, so the cokernel of $f_3$ is 0, which means it's surjective.

Another item in Lema 10.18.3 (or 07BJ) that can be useful is: (2 bis) $\varphi(\mathfrak{m}_R^k)\subseteq \mathfrak{m}_S$ for some $k>0$. This item gives the equivalence between being local and being continuos (in the adic topology). The proof of (2bis) => (2) is straightforward: Let $r\in\mathfrak{m}_R$ be such that $\varphi(r)$ is invertible, then $\varphi(r^k)$ is invertible, but $\varphi(r^k)\in \mathfrak{m}_S$

Also, maybe it is worth adding to the statement that “taking germ,” $\Gamma(X,\Omega_{\mathcal{O}_2/\mathcal{O}_1})\to\Omega_{\mathcal{O}_{2,x}/\mathcal{O}_{1,x}}$, is compatible with universal derivations? (Symbolically, $(dt)_x=d(t_x)$). This follows from the compatibility with universal derivations either coming from Lemma 17.28.6 ($f^{-1}d_{\mathcal{O}_2/\mathcal{O}_1}\cong d_{f^{-1}\mathcal{O}_2/f^{-1}\mathcal{O}_1})$ applied to the inclusion $\{x\}\to X$ or possibly from Lemma 10.131.5 (where I guess it holds $d_{R/S}=\operatorname{colim}_i d_{R_i/S_i}$).

By a "scheme-theoretically dominant" morphism, do you mean a morphism whose scheme-theoretic image is the full target?

There is a tricky typo in Example 0662. The indices $p,q$ for the second spectral seqeunce are interchanged.

Maybe one could link at the beginning of this section the definition of regular local ring? This is 10.60.10. (It was not immediate for me to trace back the definition by its name since its contained on the section "dimension".)

Okay, so it is indeed not clear because "fully faithful" has not been shown explicitly. Instead, the proof shows $H^0 \circ Can = id$ and $Can \circ H^0 = id$ according to Definition 4.2.17. The latter equality results from the proof of essential surjectivity.

Regarding #8691: Ignore last paragraph. Serre does not use that map in his book. Also, the map $\mathcal{K}_X(\Omega_{X/k})\to\nu_*\mathcal{K}_{X'}(\Omega_{X'/k})$ from the case I explained is just the identity, as $\mathcal{K}_X(\Omega_{X/k})$ and $\mathcal{K}_{X'}(\Omega_{X'/k})$ are the sheaves constantly $\Omega_{X/k,\eta}\cong\Omega_{X'/k,\eta'}$, where $\eta\in X$, $\eta'\in X'$ is the generic point.

The standard proof of Lemma 01CD seems to use the tensor-hom adjunction (Lemma 01CN), and I find it difficult to take a different approach.

In 38.32.1, can we expect even more that if there are two given compactifications, then the third one dominating both of them can be embedded (as open subschemes) into the two given ones?

In the remark after Definition 03MQ, $U_j$ should be $Z_j$.

Should $\mathcal{C}$ be $\mathcal{K}[2-a]$ instead? Second-to-last line of the proof, the (a-1) in the upper-index should be (a+1) instead?

In the proof of Lemma 07HT, the expression $\text{d} : B \to K/K^2 +(K \cap J(1))^{[2]}$ seems to be missing parentheses, and the divided power structure on $J(1)$ has already been denoted by $\delta(1)$ in the statement of the lemma. In Lemma 07HV, "two maps $B\to C$" should be "two maps $B\to D$".

In the proof of Lemma 07KI, I believe the $\text{Cris}(C/A)$ should be $\text{Cris}^{\wedge}(C/A)$.

In the general case, if $X$ is any scheme locally with finitely many irreducible components, then $X^\nu=\coprod Z_i^\nu$, where $Z_i\subset X$ are the irreducible components of $X$, Tag 29.54.6, so that $\nu_*\mathcal{O}_{X^\nu}=\prod\nu_{i,*}\mathcal{O}_{Z_i^\nu}$, where $\nu_i:Z_i^\nu\to X^\nu\to X$. Thus, $(\nu_*\mathcal{O}_{X^\nu})_x = \prod(\nu_{i,*}\mathcal{O}_{Z_i^\nu})_x,$ where we have used that “taking stalks” is exact, and the fact that the product $\prod\nu_{i,*}\mathcal{O}_{Z_i^\nu}$ is locally finite. Using the facts (i) $\nu_i$ equals the composite $Z_i^\nu\to Z_i\to X$ (second proof idea of Tag 29.54.6), (ii) for a closed immersion of topological spaces $f:Z\to T$ and a sheaf $\mathcal{F}$ over $Z$, the natural map $(f_*\mathcal{F})_{z}\to\mathcal{F}_z$ is an isomorphism (Tag 6.32.1) and (iii) the last equation in #8693, we get $$(\nu_*\mathcal{O}_{X^\nu})_x=\prod_i\bigcap_{y\in\nu_i^{-1}(x)}\mathcal{O}_{Z_i^\nu,y},$$ where the $i$-th intersection is taken inside the function field of $X_i^\nu$. (Slogan: the stalk $(\nu_*\mathcal{O}_{X^\nu})_x$ is rational functions on $X^\nu$ defined at all points of $\nu^{-1}(x)$.)

There is a fourth way to write $(\nu_*\mathcal{O}_{X^\nu})_x$, which is a consequence of the integrality of $\mathcal{O}_X(U)\to\nu_*\mathcal{O}_{X^\nu}(U)$ for $U\subset X$ open affine. Namely, if $U=\operatorname{Spec} A$ is an open affine neighborhood of $x$, $\mathfrak{q}$ is the prime in $A$ associated with $x$ and $B$ is the integral closure of $A$ in the total ring of fractions of $A_{red}$, then

(4) $S^{-1}B$, where $\displaystyle S=\bigcap_{\substack{\mathfrak{p}\subset B\\\text{lying over }\mathfrak{q}}} B-\mathfrak{p}$.

This is Bourbaki, Commutative Algebra, Ch. V, § 2, no. 1, Proposition 2.

As far as I can tell, (4) is independent of (1)-(3) (the latter do not prove (4) nor are proven by it).

In particular, using the fact:

for $B$ some domain and $S_i\subset B$, $i\in I$, a multiplicatively closed subset, $(\bigcap S_i)^{-1}B=\bigcap S_i^{-1}B$,

one gets: if $X$ is irreducible (hence $X_{red}$ and $X^\nu$ are integral), then $$(\nu_*\mathcal{O}_{X^\nu})_x =\bigcap_{y\in\nu^{-1}(x)}\mathcal{O}_{X^\nu,y},$$ where $\mathcal{O}_{X^\nu,y}$ are subrings of the function field of $X^\nu$ and thus we can intersect them. (Intuitively, $(\nu_*\mathcal{O}_{X^\nu})_x$ identifies with the rational functions on $X^\nu$ defined on all points of the fibre $\nu^{-1}(x)$.)

Also, in my reformulation of part (1), instead of "canonical" one can say that:

The map $\mathcal{K}_Y(\mathcal{F})\to f_*\mathcal{K}_X(\mathcal{G})$ is the unique morphism of $\mathcal{K}_Y$-modules that makes the following diagram commute: $$\require{AMScd} \begin{CD} \mathcal{F}@>>>f_*\mathcal{G}\\ @VVV@VVV\\ \mathcal{K}_Y(\mathcal{F})@>>> f_*\mathcal{K}_X(\mathcal{G}) \end{CD}$$

The fact that the constructed morphism indeed makes the square commute can be seen leveraging the factorization $\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{K}_Y \to f_*\mathcal{G}\otimes_{f_*\mathcal{O}_X}f_*\mathcal{K}_X \to f_*(\mathcal{G}\otimes_{\mathcal{O}_X}\mathcal{K}_X)$.

Uniqueness follows from the fact that by precomposing the $\mathcal{K}_Y$-linear map $can:\mathcal{K}_Y(\mathcal{F})\to f_*\mathcal{K}_X(\mathcal{G})$ with the unit $\mathcal{F}\to\mathcal{K}_Y(\mathcal{F})$ we are obtaining the adjunct $\mathcal{O}_Y$-linear morphism $\mathcal{F}\to f_*\mathcal{K}_X(\mathcal{G})$ of $can$; thus, it is uniquely determined.