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Chapter 2: Problem 1

Let \(\left(X, C_{X}\right)\) be a locally ringed space, and let \(f: \mathscr{L} \rightarrow . ll\) be a surjective map of invertible sheaves on \(X\). Show that \(f\) is an isomorphism. [Hint: Reduce to a question of modules over a local ring by looking at the stalks.]

Short Answer

Expert verified
The given surjective map \(f\) is an isomorphism because its kernel is the zero sheaf, making \(f\) injective and hence an isomorphism, as it was a surjection to begin with.

Step by step solution

01

Understanding locally ringed spaces

A locally ringed space \((X, C_X)\) consists of a topological space \(X\) endowed with a sheaf \(C_X\) of rings on \(X\) such that all stalks \(C_{X,x}\) are local rings. In our case, we have a surjection \(f:\mathscr{L}\rightarrow\mathscr{M}\) between sheaves of \(C_X\)-modules \(\mathscr{L}\) and \(\mathscr{M}\).
02

Reduction to Modules over a Local Ring

In order to show that the map \(f\) is an isomorphism, let's reduce the situation to algebra by looking at the stalks. For \(x \in X\), we denote by \(f_x:\mathscr{L}_x\rightarrow\mathscr{M}_x\) the map on the stalks. Since \(f\) is a morphism of \(C_X\)-modules, \(f_x\) is a surjective \(C_{X,x}\)-module homomorphism.
03

Considering the Kernel

We consider the stalk at \(x\) of the kernel sheaf \(K = ker (f)\), denoted \(K_x\). The canonical injection from \(K\) to \(\mathscr{L}\) induces an injection \(K_x\rightarrow\mathscr{L}_x\). We then have an exact sequence (0->K_x->\mathscr{L}_x->\mathscr{M}_x->0), where -> denotes morphisms.
04

Conclusion: Isomorphism of Sheaves

Since stalks \(K_x\) are zero for all \(x\) in \(X\), it follows that \(K\) is the zero sheaf and hence \(f\) is injective. Considering that we started with a surjection, we conclude that \(f\) is an isomorphism.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Invertible Sheaves
Invertible sheaves are an essential tool in algebraic geometry and sheaf theory. To understand them, think of them as generalizations of line bundles.
Specifically, an invertible sheaf over a topological space is a sheaf that can be both multiplied by or inverted through tensor products, similar to how you might consider multiplying numbers.
This allows them to act like modules over the structure sheaf of a ringed space. An interesting feature of invertible sheaves is their "invertibility." For a given sheaf \(\mathscr{L}\), there exists another sheaf \(\mathscr{L}^{-1}\) such that the tensor product \(\mathscr{L} \otimes \mathscr{L}^{-1}\) is isomorphic to the structure sheaf. This provides powerful ways to manipulate and understand the underlying space.
In the context of the locally ringed space \(\left(X, C_{X}\right)\), invertible sheaves play a critical role in simplifying complex problems by focusing on local details through tensor operations.
Stalks of Sheaves
To visualize stalks of sheaves, consider them as localized data on a topological space. Each stalk at a point \(x\) collects all the information from sections defined around \(x\).
This is the core concept of sheaf theory: gluing local data to understand global structures.Stalks of sheaves take abstract, global information and make it "local" by focusing on individual points. This has substantial implications, especially in algebraic geometry, where understanding the nuances around each point is crucial.
For a morphism \(f: \mathscr{L} \to \mathscr{M}\) of sheaves, studying sheaves via stalks \(f_x: \mathscr{L}_x \to \mathscr{M}_x\) can often simplify the demonstration of properties, as shown in the original exercise. This process can reduce complex maps between sheaves to problems involving modules over rings, which are typically simpler to manage.
Local Rings
Local rings are a fundamental building block in algebraic geometry and commutative algebra. Imagine them as rings with a unique maximal ideal, which simplifies many algebraic structures.
In practical terms, a local ring can help analyze and focus on the properties at a specific point in a space.Every stalk \(C_{X,x}\) in a locally ringed space is a local ring. This means that, at each point \(x\), the stalk has very controlled, predictable behavior. This is incredibly helpful when you're reducing sheaf theory problems to algebraic ones, as every problem becomes much more manageable.
With local rings as the backdrop, homomorphisms of modules can be understood more clearly. In the context of the solution above, examining surjective homomorphisms of modules over these local rings allows one to conclude certain properties about the original morphism between sheaves.

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Most popular questions from this chapter

If \(V, W\) are two varieties over an algebraically closed field \(k,\) and if \(V \times W\) is their product, as defined in (I, Ex. 3.15,3.16 ), and if \(t\) is the functor of (2.6) then \(t(V \times W)=t(V) \times_{\text {spec } k} t(W)\).

Let \(X\) be an integral scheme of finite type over a field \(k\), having function field \(K\) We say that a valuation of \(K / k\) (see \(I, \$ 6\) ) has center \(x\) on \(X\) if its valuation ring \(R\) dominates the local ring \(C_{x . X}\) (a) If \(X\) is separated over \(k\), then the center of any valuation of \(K / k\) on \(X\) (if it exists) is unique. (b) If \(X\) is proper over \(k\), then every valuation of \(K / k\) has a unique center on \(X\) \(*(\mathrm{c})\) Prove the converses of \((\mathrm{a})\) and \((\mathrm{b}) .[\text { Hint }: \text { While parts }(\mathrm{a}) \text { and }(\mathrm{b})\) follow quite easily from (4.3) and \((4.7),\) their converses will require some comparison of valuations in different fields. (d) If \(X\) is proper over \(k\), and if \(k\) is algebraically closed, show that \(\Gamma\left(X, C_{X}\right)=k\) This result generalizes (I, 3.4a). [Hint: Let \(a \in \Gamma\left(X, \mathscr{C}_{X}\right),\) with \(a \notin k .\) Show that there is a valuation ring \(R\) of \(K / k\) with \(a^{-1} \in \mathrm{m}_{R} .\) Then use (b) to get a contradiction. Note. If \(X\) is a variety over \(k,\) the criterion of (b) is sometimes taken as the definition of a complete variety.

Let \(X\) be a scheme of finite type over a field \(k\) (not necessarily algebraically closed). (a) Show that the following three conditions are equivalent (in which case we say that \(X\) is geometrically irreducible). (i) \(X \times_{k} \bar{k}\) is irreducible, where \(\bar{k}\) denotes the algebraic closure of \(k .\) abuse of notation, we write \(X \times_{k} \bar{k}\) to denote \(X \times_{\text {spec } k}\) Spec \(\bar{k} .\) (ii) \(X \times_{k} k_{s}\) is irreducible, where \(k_{s}\) denotes the separable closure of \(k\) (iii) \(X \times_{k} K\) is irreducible for every extension field \(K\) of \(k\) (b) Show that the following three conditions are equivalent (in which case we say \(X\) is geometrically reduced) (i) \(X \times_{k} \bar{k}\) is reduced. (ii) \(X \times_{k} k_{p}\) is reduced, where \(k_{p}\) denotes the perfect closure of \(k\) (iii) \(X \times_{k} K\) is reduced for all extension fields \(K\) of \(k\) (c) We say that \(X\) is geometrically integral if \(X \times_{k} \bar{k}\) is integral. Give examples of integral schemes which are neither geometrically irreducible nor geometrically reduced.

Let \(\left\\{y_{i}\right\\}\) be a direct system of sheaves on a noetherian topological space \(X\). In this case show that the presheaf \(U \mapsto \varliminf_{1}, y_{1}(U)\) is already a sheaf. In particular, \(\Gamma\left(X, \lim _{\longrightarrow} \overline{\mathscr{F}}_{1}\right)=\lim _{\longrightarrow} \Gamma(X, \mathscr{F},)\)

Let \(A\) be a ring, let \(S=A\left[x_{0}, \ldots, x_{r}\right]\) and let \(X=\) Proj \(S\). We have seen that a homogeneous ideal \(I\) in \(S\) defines a closed subscheme of \(X\) (Ex. 3.12 ), and that conversely every closed subscheme of \(X\) arises in this way (5.16) (a) For any homogeneous ideal \(I \subseteq S\), we define the saturation \(I\) of \(I\) to be \(\left\\{s \in S | \text { for each } i=0, \ldots, r, \text { there is an } n \text { such that } x_{i}^{n} s \in I\right\\} .\) We say that \(I\) is saturated if \(I=I .\) Show that \(T\) is a homogeneous ideal of \(S\). (b) Two homogeneous ideals \(I_{1}\) and \(I_{2}\) of \(S\) define the same closed subscheme of \(X\) if and only if they have the same saturation. (c) If \(Y\) is any closed subscheme of \(X\), then the ideal \(\Gamma_{*}\left(\mathscr{I}_{Y}\right)\) is saturated. Hence it is the largest homogeneous ideal defining the subscheme \(Y\) (d) There is a \(1-1\) correspondence between saturated ideals of \(S\) and closed subschemes of \(X\).
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