The Stacks project

[It seems to me that the following remark is correct, but i should write it out with even more care to be 100%. Sorry if there is any mistake.] In remark 6.7.2 the argument that -with the definition of sheaf given here- the image of the empty set must be a final object of the value category does not seem complete to me -and perhaps the conclusion even fails. We do have that « the “collection of sections $s_i$” from the definition above actually form an element of the empty product which is the final object of the category the sheaf has values in ». But then we must justify that this yields a morphism from $\mathcal F(\emptyset)$ to this product (a final object of the value category) and that this morphism is an isomorphism -for instance because it is an equalizer of identity morphisms. Actually this requires some reinterpretation of the definition in terms of elements given here, because if $I=\emptyset$ there is no $s_i$ and the ensuing uniqueness condition is vacuous. For the argument to work we can reformulate the definition in terms of equalizer diagram. Alternatively we can -as done in Hartshorne, p61- formulate separately the locality axiom, this shows that the short "combination" of the locality axiom with the gluing axiom into a single axiom like here is not equivalent to the conjunction of the 2 standard axioms. The formulations look equivalent but the one given here is actually weaker because the empty family argument does not work: so a sheaf as defined here can have a nonfinal object as value on the empty set.

@8593 Thanks! I guessed that we may use the affiness of disjoint union of affine schemes but I didn't know how can we emdody it. Good Idea ~ Thank you.

@#8592 You almost get there. Pick another $x'\in V$, then there exists an $s''\in \Gamma(V,\mathcal{L}^{m'})$ such that $x'\in V_{s''}$ is affine. Now set $s\in \Gamma(X,\mathcal{L}^{mm'})$ to be $s| _U=s'^{\otimes m'}$ and $s| _{V}=s''^{\otimes m}$. You get that $X_s=U_ {s'}\coprod V_ {s''}$ is affine.

O.K. Can you give more concrete hint? To show that $\mathcal{L}$ is ample, it suffices to show that for every $x\in X$, there exists an $n \ge 1$ and $s \in \Gamma(X, \mathcal{L}^{n}) = \Gamma(U, \mathcal{L}^n) \oplus \Gamma(V, \mathcal{L}^n)$ (Sheaf axiom) such that $x\in X_s$ and $X_s$ is affine. Fix $x\in X = U \cup V$. We may assume that $x\in U$. Since $\mathcal{L}|_U$ is ample, by the definition, there exists an $m \ge 1$ and $s' \in \Gamma(U, \mathcal{L}^m)$ such that $x\in U_{s'}$ and $U_{s'}$ is affine in $U$. From this, how can we find such $n$ and $s \in \Gamma(X, \mathcal{L}^{n})$ (lifting) as above so that $x\in X_s$ and $X_s$ is affine.

What should I catch? What whould be key point to breakthrough this difficulty? :)

@#8587 If you insist that $U$ and $V$ are disjoint, then your claim is true. Just unravel the definition 28.26.1. Otherwise consider $(X,L)=(\mathbb{P}^1,\mathcal{O}(-1))$. On each standard affine $D_+(x_0)$ and $D_+(x_1)$ the line bundle $L$ is trivial, hence it is ample on each affine. However $L$ has no global section hence it cannot be ample.

Unnecessary period at the end of (3).

In tag 03FS (3), the fiber product should be $V \times_Y X \to V$.

How do we get "Observe that $g'*O{X'}=f^∗g_∗O_{S'}"? Maybe it is an application of some base change theorem?

Can I ask about ampleness of invertible sheaf?
Let $X= U \cup V$ be a scheme which is disjoint union of two open subschemes. Let $\mathcal{L}$ be an invertible $\mathcal{O}_X$-module. If each $\mathcal{L}|_{U}$ and $\mathcal{L}|_V$ is ample, then $\mathcal{L}$ is ample?

The phrase "acting on X by translations" should say "acting on Y by translations".

I don't know if this might be only me, but I got confused by the current phrasing of the proof to see how one obtains a globally splitting s.e.s. from a global retraction of $i$ over $S$. After posting my confusion here, R. van Dobben de Bruyn explained in a comment what I was missing.

Also, $\mathbf{P}(\mathcal{O}^{\oplus n}_S) = \mathbf{P}^{n-1}_S$.

Correction: "In general, let $\mathcal{L}$ be an ample invertible sheaf on $S$."

After "if $g : X \to Z$ is a left inverse of $i$, then $i^*c_g$ is a right inverse of the map $i^*\Omega_{X/S} \to \Omega_{Z/S}$," one could say "by Modules, Lemma 17.28.13."

Okay, I forgot to justify the last claim from the statement in the alternative proof: one sees that the resulting pulled-back map $i^*\Omega_{X/S}\to\Omega_{Z/S}$ equals $c_i$ (Lemma 29.32.8) by applying Modules, Lemma 17.28.13, taking $S=S'=S''$ and $X''\to X'\to X$ to be the morphisms of ringed spaces $(Z,\mathcal{O}_Z)\to (X,\mathcal{O}_X/\mathcal{I})\to(X,\mathcal{O}_X)$.

I'm not sure if any of this is any better the already existing proof. I just felt it was nice to connect Lemma 29.32 to the similar-looking Modules, Lemma 17.28.9.

Last thing: maybe one could make explicit in the statement that the maps are $\mathcal{O}_Z$-linear.

Suggested alternative proof:

We can assume that $i$ is a closed immersion by shrinking $X$ if necessary. Let $\mathcal{I}\subset\mathcal{O}_X$ be the ideal sheaf associated to $Z$. Consider the sequence Modules, Lemma 17.28.9, by setting $\mathcal{O}_2\to\mathcal{O}_2'$ equal to $\mathcal{O}_X\to\mathcal{O}_X/\mathcal{I}$ and $\mathcal{O}_1=f^{-1}\mathcal{O}_S$, where $f:X\to S$ is the structure morphism. Since $i^*$ is right exact, we can pull back along $i$ and get an exact sequence $$\mathcal{C}_{Z/X} \to i^*\Omega_{X/S}\otimes_{\mathcal{O}_Z}\mathcal{O}_Z \to i^*\Omega_{(\mathcal{O}_X/\mathcal{I})/f^{-1}\mathcal{O}_S} \to 0,$$ where in the second term we have used Modules, Lemma 17.16.4. Hence, the second term is $i^*\Omega_{X/S}$. On the other hand, by Modules, Lemma 17.28.6, the third term equals $$i^{-1}\Omega_{(\mathcal{O}_X/\mathcal{I})/f^{-1}\mathcal{O}_S}\otimes_{i^{-1}\mathcal{O}_X}\mathcal{O}_Z \cong\Omega_{Z/S}\otimes_{i^{-1}\mathcal{O}_X}\mathcal{O}_Z\cong\Omega_{Z/S},$$ where the last step is done using formula $\mathcal{F}\otimes_{\mathcal{O}_Y}\mathcal{O}_Y/\mathcal{J}\cong\mathcal{F}/\mathcal{J}\mathcal{F}$, for $(Y,\mathcal{O}_Y)$ a ringed space, $\mathcal{J}\subset\mathcal{O}_Y$ an ideal sheaf and $\mathcal{F}$ an $\mathcal{O}_Y$-module.

Just curious: is there some specific reason why this isn't stated in Modules, Section 17.28? Here's the proof for arbitrary ringed spaces $X,Y,S$: The sequence is the same in the ringed spaces case, this time the maps come from Modules, Lemma 17.28.12. Call $g:X\to S$ and $h:Y\to S$ to the structure morphisms. By taking induced maps in stalks at $x\in X$ and using Modules, Lemma 17.28.7, we obtain a sequence $$\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{Y,f(x)}} \Omega_{\mathcal{O}_{Y,f(x)}/\mathcal{O}_{S,g(x)}}\to\Omega_{\mathcal{O}_{X,x}/\mathcal{O}_{S,g(x)}}\to\Omega_{\mathcal{O}_{X,x}/\mathcal{O}_{Y,f(x)}} \to 0.$$ It suffices to see that the maps of the sequence are the same as the ones in Algebra, Lemma 10.131.7. This is because (i) the “characterizing property” at the end of Modules, Lemma 17.28.12 and (ii) by means of the isomorphism from Sheaves, Lemma 6.26.4, we can identify $(f^*s)_x=s_x\otimes 1$, for a local section $s$ of a sheaf of $\mathcal{O}_Y$-modules.

Isn't it a special case of Lemma 08EU?

Writing the ideal $q'$ as $q' = (q,T-a_0)$ is not precise, because $a_0$ is an element of $\kappa$, while $q'$ is an ideal of $S[T]/(Q)$.

SS:

1. The zeroth Čech cohomology of a presheaf of sets on a site is separated.
2. If a presheaf of sets on a site is separated, then its zeroth Čech cohomology is a sheaf and the canonical map of presheaves from the presheaf to its zeroth Čech cohomology is injective.
3. The canonical map of presheaves from a sheaf to its zeroth Čech cohomology is an isomorphism.
4. The zeroth Čech cohomology of the the zeroth Čech cohomology of a presheaf of sets on a site is always a sheaf, i.e., double application of zeroth Čech cohomology over a presheaf of sets on a site produces always a sheaf.