# Tag Info

2

No, this is not true, not even for closed points. Consider $\Bbb P^1_k$. Then $A=k$ and $\mathcal{O}_{\Bbb P^1_k,pt}\cong k[t]_{(t)}$, and $j^{-1}$ of the maximal ideal is the zero ideal, so the requested localization of $A$ is just $A$, which is very far from $k[t]_{(t)}$. Certainly this will be an isomorphism if $X$ is quasiaffine - I don't see other good/...

2

Let $f : \mathcal F \to \mathcal O_X$ non-zero and $s \in F(X)$ so that the image of $s$ is not zero. Since $\mathcal F$ is torsion, there is $t \in \mathcal O_X$ so that $ts = 0$. It implies that $s$ is zero outside the locus $Z(t)$. In particular, $f(s)$ is zero outside $Z(t)$ and by density $f(s)$ is zero everywhere, a contradiction.

2

For a prime ideal $p$, $A_p := (A\setminus p)^{-1} A$ : you invert elements not in $p$ to make "everything be about $p$". (Recall that $A_p$ is local with maximal ideal $pA_p$, so if $p$ was inverted, that wouldn't make sense)

0

As Schemer said in the comments, if your sheaf $F$ is supported on a point then all intermediate cohomology vanishes, but it is not a vector bundle. However, this is practically the only problem that can occur. Specifically, Hartshorne, in Lemma 6.3 Chapter 3 of Ample Subvarieties of Algebraic Varieties, gives the following result: Let $E$ be a coherent ...

4

Yes, even with no assumptions on $f$. Suppose we have an exact sequence $$0\to M_n\to\cdots\to M_1\to M_0\to0$$ of flat $A$-modules. If $N$ is any $A$-module, then the sequence $$0\to M_n\otimes_AN\to\cdots\to M_1\otimes_AN\to M_0\otimes_AN\to0$$ is exact. In the case of $f:\mathop{\mathrm{Spec}}B\to\mathop{\mathrm{Spec}}A$, pullback is given by $-\otimes_AB$...

2

The answer is no for general affine morphisms: take any affine scheme over a field with nontrivial Picard group for $X$ and the spectrum of said field for $Y$. For instance, let $X$ be an affine open in an elliptic curve over some algebraically closed field of characteristic zero. As written in the question body, the answer is yes for the finite case as ...

1

Let me concentrate on your first question; this should clarify the authors' claim. We'll see if that's enough for you to figure out the rest. Even though later we will be interested in the rather specific $k$-vector space $\hom(V,W)$ of linear maps, for now, it is conceptually easier to consider any finite-dimensional $k$-vector space $V$. I like to think ...

1

No, this construction will never give a Calabi--Yau variety. For simplicity assume $X$ is smooth. By construction, the variety $V=\mathbb P_X(F)$ is a projective bundle, so it is uniruled (except possibly in the trivial case when $F$ has rank 1). It is a basic property of smooth projective uniruled varieties $V$ that $$H^0(V,K_V^m)=0 \quad \text{ for all ... 1 Let X be a topological space where any two non-empty open set has non-empty intersection. Fix a point p\in X and define the following for any open set U$$\mathcal{F}(U)=\left\{\begin{array}{ll} 0 & \text{if $U=\emptyset$ or $p\in U$}\\ A &\text{ otherwise}\end{array}\right. $$where A is any non-trivial abelian group. Restriction maps are ... 4 There are two things to keep in mind. First, we always need to fix an ample line bundle to speak about stability. Secondly, Intersection Theory is made a way that you can work (at least if X is smooth) on the Grothendieck group K_0(X), so one might consider locally free resolution. Fix an ample line bundle H. To define the degree of any coherent sheaf ... 2 The hypothesis that the divisor be effective is used in giving the interpretation that the sections have poles of order \le a_i on V_i. If you modify the "description" to say that when a_i<0, we require a zero of order \ge |a_i| along V_i, then it works fine. In the case of your final short exact sequence, you need D effective here because ... 2 The key facts here are: Every abelian group is a \mathbb Z-module. An \mathcal R-module \mathcal F is a sheaf of abelian groups equipped with a morphism \mathcal R \to \operatorname{End}_{\text{Sh}(X)}(\mathcal F). Let \mathbb Z_X^- be the presheaf U \mapsto \mathbb Z and let \mathbb Z_X be its sheafification. Let \mathcal F be a sheaf of ... 4 There's an exact sequence$$0\to \pi^*\Omega^1_X\to \Omega^1_{\widetilde{X}}\to i_*\Omega^1_{E/Y}\to 0$$where i:E\to \widetilde{X} is the inclusion of the exceptional divisor. This corresponds to the fact that pulling back differential forms gives you differential forms which are constant along the fibers. Checking that this is exact may be done from ... 0 Oops, this is rather silly. If \Delta (X)\subset U,V\subset Y so that X\to U\subset Y,X\to V\subset Y are both the same locally closed immersion \Delta, and also noting that if we ever have a morphism f:X\to Y with U\subset X maps into V\supset f(U), then f^* commutes with restriction, we get \Delta^*(\mathcal{I}/{\mathcal{I}^2})|_X=\Delta_X^*... 0 The crucial question is: What are the stalks of \mathcal{F}? For any open U \subset \mathbb{R}, one has$$\mathcal{F}(U) = \bigoplus_{x \in U \cap [0,1)} x \cdot \mathbb{Z}, and for any inclusion $V \subset U$, the restriction maps $\mathcal{F}(U) \to \mathcal{F}(V)$ factors out all summands that do not belong to $V$. So let $\overline f \in \mathcal{F}... 3 Let me add one way to create some examples: If$\mathcal{F}$is a sheaf (of for example abelian groups, rings, etc.), then$\mathcal{F}(\emptyset)$will be the terminal object in the category where the sheaf takes its values. That means you can find examples of presheaves which are not sheaves by preventing$\mathcal{F}(\emptyset)$to be the terminal ... 0 Yes. For instance, the numerable topology on Top controls bundles that are classified by maps to a classifying space. Open covers in the numerable topology are those open covers that admit a subordinate partition of unity. If G is a topological group, it has a classifying space BG, as well as a classifying stack G-Bun of principal G-bundles, defined as ... 2 The answer lies in the definition of "generated by finitely many sections". As I understand,$\mathcal{F}|_U$is generated by finitely many sections if, there are sections$s_1,\dots, s_n$over$U$such that$\mathcal{F}|_U$is the smallest sheaf of$\mathcal{O}_U$-modules that contains all the$s_i$. In other words, if$\mathcal{G}$is a subsheaf of$\...

2

That’s not the right definition. In any category with a terminal object $1$, a constant morphism is a morphism that factors through the terminal object. In particular there is exactly one such morphism $X \to Y$ for every global point $1 \to Y$; note that these are very different from points of the Zariski spectrum, for schemes, because the terminal object ...

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