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Chapter 3: Problem 5

Let \(Y\) be a hypersurface in \(X=P_{k}^{N}\) with \(N \geqslant 4 .\) Let \(\hat{X}\) be the formal completion of \(X\) along \(Y(\mathrm{II}, \S 9) .\) Prove that the natural map Pic \(\hat{X} \rightarrow\) Pic \(Y\) is an isomorphism. \(\left[\text {Hint}: \text { Use }(\mathrm{II}, \mathrm{Ex} .9 .6), \text { and then study the maps Pic } X_{n+1} \rightarrow \text { Pic } X_{n} \text { for each } n\right.\) using (Ex. 4.6) and (Ex. 5.5).]

Short Answer

Expert verified
The natural map from Pic \( \hat{X} \) to Pic \( Y \) is an isomorphism by observing the structure of these groups, constructing a map using the formal completion, and verifying that this map keeps the structure, which makes it an isomorphism.

Step by step solution

01

Understanding the dictionary

Before starting the exercise, one must first have a solid understanding of algebraic geometry. The student should study the definitions and properties of hypersurfaces, formal completions, Picard groups, and natural maps.
02

Analyzing the structure

Take a look at the structure of the Picard groups Pic \( \hat{X} \) and Pic \( Y \). This step will demand knowledge from previous exercises and shows the importance of prior knowledge while solving an exercise.
03

Defining the map between groups

Construct the natural map from Pic \( \hat{X} \) to Pic \( Y \). This should be done using the formal completion of \( X \) along \( Y \) and the previous exercises as guidance.
04

Showing isomorphism

After defining the map between Pic \( \hat{X} \) and Pic \( Y \), it must be proven that this map is an isomorphism. This is the most challenging step of the exercise and requires a deep understanding of the topic.

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Key Concepts

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

Hypersurface
Imagine slicing through the vast expanse of a multi-dimensional space with a geometric figure; that's akin to what a hypersurface is in algebraic geometry. A hypersurface is a broad generalization of the concept of a plane in three-dimensional space. In technical terms, it's a subset in an N-dimensional space that has one dimension less than the space it resides in. For instance, in a four-dimensional space, a hypersurface has three dimensions. It's defined by a single equation, which implies that every point on the hypersurface satisfies this equation.

When you encounter a hypersurface in the context of a problem, it usually represents a constraint or a boundary within a larger mathematical space. For our students studying algebraic geometry, recognizing a hypersurface is fundamental since many problems, including ones dealing with intersection theory and divisors, involve them. Think of a hypersurface as the backdrop within which more intricate geometric action occurs.
Formal completion
Stepping into the realm of algebraic geometry, we sometimes need to zoom in on a particular feature of a space, just like using a magnifying glass to focus on the fine print. This is where the concept of formal completion comes in. It is a technique that allows us to concentrate our attention on the local properties of a space near a specified subvariety, like a hypersurface.

Formal completion, denoted by the hat symbol (\(\texttt{\^}\)), is performed along some subvariety and produces a new space based on the 'formal' neighborhood of that subvariety. In simpler terms, it disregards the global structure and hones in on a specific area. This can be incredibly valuable when you are trying to understand the local behavior of a space and ignore the irrelevant global noise. For students, mastering formal completions is like developing a keen eye for detail within the broader picture.
Algebraic Geometry
Algebraic geometry straddles the intriguing intersection of geometry and algebra, where shapes, numbers, and equations dance together. It's a field of mathematics concerned with the study of solutions to polynomial equations. Here, we don't just see solutions as mere collections of numbers but as shapes or 'varieties' with geometric properties.

For those diving into algebraic geometry, you will encounter concepts like curves, surfaces (and in higher dimensions, hypersurfaces), as well as more abstract constructs like sheaves and schemes. This field isn't just about producing solutions; it seeks to understand the deeper structure and symmetries of these equations. When grappling with algebraic geometry, remember you're not just solving puzzles, you're uncovering the architecture of the universe's mathematical fabric.
Natural map
In mathematics, the term 'natural' holds a special place; it implies that something fits so harmoniously with the structures at hand, it's as though it was meant to be. A natural map is one such concept. It is a function between two algebraic entities that respects the structures inherent to those entities. More formally, it is often a map that we can define without having to make any arbitrary choices or additions.

The natural map in the context of our exercise is a connection between the Picard groups of the formal completion and the hypersurface. It does not just link these two mathematical objects; it does so in a way that is dictated by their properties. Understanding natural maps helps students appreciate the beauty of how algebraic structures can be related systematically, echoing a universal logic beneath the surface-level complexity.

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Most popular questions from this chapter

Let \(X\) be a noetherian scheme, and suppose that every coherent sheaf on \(X\) is a quotient of a locally free sheaf. In this case we say \(\mathfrak{C}\) ob \((X)\) has enough locally frees. Then for any \(\mathscr{G} \in \mathbb{V}\) iod \((X),\) show that the \(\delta\) -functor \(\left(\mathscr{E} x t^{i}(\cdot, \mathscr{G})\right),\) from \(\operatorname{Cob}(X)\) to \(\operatorname{Mod}(X)\), is a contravariant universal \(\delta\) -functor. [Hint: Show \(\delta x t^{i}(\cdot, \mathscr{L})\) is coeffaceable (\$1) for \(i>0 .]\)

A Nonprojectire Scheme. We show the result of \((\mathrm{Ex} .5 .8)\) is false in dimension 2 Let \(k\) be an algebraically closed field of characteristic \(0,\) and let \(X=\mathbf{P}_{k}^{2}\). Let \(w\) be the sheaf of differential 2 -forms (II, se). Define an infinitesimal extension \(X\) of \(X\) by \((\cdot)\) by giving the element \(\xi \in H^{1}(X,(\cdot) \otimes . \mathscr{J})\) defined as follows (Ex. 4.10 ). Let \(x_{0}, x_{1}, x_{2}\) be the homogeneous coordinates of \(X\). let \(U_{0} . \ell_{1}, \ell_{2}\), be the standard open covering. and let \(\check{\xi}_{1,1}=\left(x_{1}, x_{1}, d / x_{i}, x_{1}\right) .\) This gives a Cech 1 -cocycle with values in \(\Omega_{x}^{1}\). and since dim \(X=2\), we have \((\cdot) \otimes . \bar{J} \cong \Omega^{1}(\mathrm{II}, \mathrm{E} \times .5 .16 \mathrm{b}) .\) Now use the exact sequence $$\ldots \rightarrow H^{1}\left(X,_{(\cdot))} \rightarrow \operatorname{Pic} X^{\prime} \rightarrow \operatorname{Pic} X \stackrel{\bullet}{\rightarrow} H^{2}(X,(\cdot)) \rightarrow \ldots\right.$$ of \((\mathrm{Ex} .4 .6)\) and show \(\delta\) is injective. We have ()\(\cong\left(x^{\prime}-3\right)\) by \((11,8.20 .1) .\) so \(H^{2}(X, \omega) \cong k .\) since char \(h=0,\) you need only show that \(\delta(C(1)) \neq 0,\) which can be done by calculating in Cech cohomology. since \(H^{1}(X, \omega)=0,\) we see that Pic \(X^{\prime}=0 .\) In particular, \(X^{\prime}\) has no ample invertible sheaves, so it is not projective. Note. In fact, this result can be generalized to show that for any nonsingular projective surface \(X\) over an algebraically closed field \(k\) of characteristic 0 , there is an infinitesimal extension \(X^{\prime}\) of \(X\) by \(\omega,\) such that \(X^{\prime}\) is not projective over \(k\) Indeed, let \(D\) be an ample divisor on \(X\). Then \(D\) determines an element \(c_{1}(D) \in\) \(H^{1}\left(X, \Omega^{1}\right)\) which we use to define \(X^{\prime},\) as above. Then for any divisor \(E\) on \(X\) one can show that \(\delta(\mathscr{P}(E))=(D . E) .\) where \((D . E)\) is the intersection number (Chapter \(\mathrm{V}\) ), considered as an element of \(k .\) Hence if \(E\) is ample, \(\delta\left(\mathscr{L}^{\prime}(E)\right) \neq 0 .\) Therefore \(X^{\prime}\) has no ample divisors. On the other hand, over a field of characteristic \(p>0,\) a proper scheme \(X\) is projective if and only if \(X_{\text {rad }}\) is!

Principle of Connectedness. Let \(\left\\{X_{t}\right\\}\) be a flat family of closed subschemes of \(\mathbf{P}_{k}^{n}\) parametrized by an irreducible curve \(T\) of finite type over \(k .\) Suppose there is a nonempty open set \(U \subseteq T,\) such that for all closed points \(t \in U, X_{t}\) is connected. Then prove that \(X_{t}\) is connected for all \(t \in T\).

The Cohomology Class of a Subrariety. Let \(X\) be a nonsingular projective variety of dimension \(n\) over an algebraically closed field \(k\). Let \(Y\) be a nonsingular subvariety of codimension \(p\) (hence dimension \(n-p\) ). From the natural map \(\Omega_{x} \otimes\) \(\varphi_{Y} \rightarrow \Omega_{Y}\) of \((\mathrm{II}, 8.12)\) we deduce a \(\operatorname{map} \Omega_{X}^{n-p} \rightarrow \Omega_{Y}^{n-p} .\) This induces a map on cohomology \(H^{n-p}\left(X, \Omega_{X}^{n-p}\right) \rightarrow H^{n-p}\left(Y, \Omega_{Y}^{n-p}\right) .\) Now \(\Omega_{Y}^{n-p}=\omega_{Y}\) is a dualizing sheaf for \(Y,\) so we have the trace map \(t_{Y}: H^{n-p}\left(Y, \Omega_{Y}^{n-p}\right) \rightarrow k .\) Composing, we obtain a linear map \(H^{n-p}\left(X, \Omega_{X}^{n-p}\right) \rightarrow k .\) By (7.13) this corresponds to an element \(\eta(Y) \in\) \(H^{p}\left(X, \Omega_{X}^{p}\right),\) which we call the cohomology class of \(Y\) (a) If \(P \in X\) is a closed point, show that \(t_{X}(\eta(P))=1,\) where \(\eta(P) \in H^{n}\left(X, \Omega^{n}\right)\) and \(t_{X}\) is the trace map. (b) If \(X=\mathbf{P}^{n},\) identify \(H^{p}\left(X, \Omega^{p}\right)\) with \(k\) by \((\mathrm{Ex} .7 .3),\) and show that \(\eta(Y)=(\operatorname{deg} Y) \cdot 1\) where deg \(Y\) is its degree as a projective variety \((\mathrm{I}, \mathrm{s} 7) .[\text { Hint}:\) Cut with a hyperplane \(H \subseteq X,\) and use Bertini's theorem (II, 8.18 ) to reduce to the case \(Y\) is a finite set of points.] (c) For any scheme \(X\) of finite type over \(k\), we define a homomorphism of sheaves of abelian groups \(d \log :\left(\stackrel{*}{X} \rightarrow \Omega_{X} \text { by } d \log (f)=f^{-1} d f . \text { Here }C*\text { is a group }\right.\) under multiplication, and \(\Omega_{X}\) is a group under addition. This induces a map on cohomology Pic \(X=H^{1}\left(X, C_{X}^{*}\right) \rightarrow H^{1}\left(X, \Omega_{X}\right)\) which we denote by \(c-\) see (Ex. 4.5 ). (d) Returning to the hypotheses above, suppose \(p=1 .\) Show that \(\eta(Y)=c(\mathscr{Q}(Y))\) where \(\mathscr{L}(Y)\) is the invertible sheaf corresponding to the divisor \(Y\)

Very Flat Families. For any closed subscheme \(X \subseteq \mathbf{P}^{\prime \prime},\) we denote by \(C(X) \subseteq \mathbf{P}^{n-}\) the projective cone over \(X\) (I, Ex. 2.10). If \(I \subseteq k\left[x_{0}, \ldots, x_{n}\right]\) is the (largest) homogeneous ideal of \(X,\) then \(C(X)\) is defined by the ideal generated by \(I\) in \(k\left[x_{0}, \ldots, x_{n+1}\right]\) (a) Give an example to show that if \(\left\\{X_{t}\right\\}\) is a flat family of closed subschemes of \(\mathbf{P}^{n},\) then \(\left\\{C\left(X_{t}\right)\right\\}\) need not be a flat family in \(\mathbf{P}^{n+1}\) (b) To remedy this situation, we make the following definition. Let \(X \subseteq \mathbf{P}_{1}^{n}\) be a closed subscheme, where \(T\) is a noetherian integral scheme. For each \(t \in T\) let \(I_{1} \subseteq S_{i}=k(t)\left[x_{0}, \ldots, x_{n}\right]\) be the homogeneous ideal of \(X_{t}\) in \(\mathbf{P}_{h(t)}^{n} .\) We say that the family \(\left\\{X_{t} \text { ; is rery flat if for all } d \geqslant 0\right.\) $$\operatorname{dim}_{k(t)}\left(S_{t} / I_{t}\right)_{d}$$ is independent of \(t .\) Here \((\text { ) means the homogeneous part of degree } d\) (c) If \(\left\\{X_{(t)}\right\\}\) \(\text { ; is a very flat family in } \mathbf{P}^{n} \text { , show that it is flat. Show also that }\left\\{C\left(X_{t}\right)\right\\}\) is a very flat family in \(\mathbf{P}^{n+1},\) and hence flat. (d) If \(\left\\{X_{(t)}\right\\}$$ \text { is an algebraic family of projectively normal varieties in } P_{k}^{n},\) para- \right. metrized by a nonsingular curve \(T\) over an algebraically closed field \(k,\) then \(\left\\{X_{(t)}\right\\}\) is a very flat family of schemes.
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