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Chapter 2: Problem 2

Let \(S\) be a scheme, let \(X\) be a reduced scheme over \(S\), and let \(Y\) be a separated scheme over \(S\). Let \(f\) and \(g\) be two \(S\) -morphisms of \(X\) to \(Y\) which agree on an open dense subset of \(X .\) Show that \(f=g .\) Give examples to show that this result fails if either (a) \(X\) is nonreduced, or (b) \(Y\) is nonseparated. [Hint: Consider the \(\left.\operatorname{map} h: X \rightarrow Y \times_{S} Y \text { obtained from } f \text { and } g .\right].\)

Short Answer

Expert verified
Yes, if two \(S\)-morphisms \(f\) and \(g\) from a reduced scheme \(X\) to a separated scheme \(Y\) agree on an open dense subset of \(X\), they are indeed identical. However, this result fails if either (a) \(X\) is nonreduced, or (b) \(Y\) is nonseparated. Examples of failure for cases (a) and (b) have been shown in step 4.

Step by step solution

01

Constructing the Map \(h\)

First construct the map \(h\) by making use of morphisms \(f\) and \(g\). This map \(h: X \rightarrow Y \times_{S} Y \) is obtained from \(f\) and \(g\). The map \(h\) sends a point \(x\) in \(X\) to \( (f(x), g(x))\) in \(Y \times_{S} Y\).
02

Set the Map Equal to Diagonal Map

The next step is to evoke the diagonal map \( \Delta: Y \rightarrow Y \times_{S} Y \), where \(\Delta(y) = (y, y)\). Given that \(f\) and \(g\) agree on an open, dense subset of \(X\), it means that on this subset, \(h\) and \( \Delta \circ f \) (or \( \Delta \circ g \)) are the same. This is because on this open dense subset, for any \(x\), we have \(f(x) = g(x)\), which means \( (f(x), g(x)) = (y, y)\), where \(y\) is either \(f(x)\) or \(g(x)\). Because the open dense subset is dense in \(X\), we then have that \(h = \Delta \circ f = \Delta \circ g\) on the entirety of \(X\).
03

Apply Separation of Y and Reducedness of X

We know that the diagonal map \(\Delta: Y \rightarrow Y \times_{S} Y \) is a closed immersion because \(Y\) is separated over \(S\), and also that the image of \(h\) (which is now the same as the image of \(\Delta \circ f\) or \(\Delta \circ g\)) is closed in \(Y \times_{S} Y\). As \(X\) is reduced, this implies \(h\) factors uniquely through \( \Delta(Y)\), and we can infer from this that \(f = g\).
04

Giving Counterexamples

For the last part of the exercise, we need to present examples to show that the result fails when either (a) \(X\) is nonreduced, or (b) \(Y\) is nonseparated. A common example for (a) is to let \(X = \mathrm{Spec}\, \mathbb{C}[x]/(x^2)\) and \(Y = \mathrm{Spec}\, \mathbb{C}[y]\), with \(S=\mathrm{Spec}\, \mathbb{C}\), and define \(f, g: X \rightarrow Y\) by \(f(x) = y, g(x) = 0\). Then, \(f\) and \(g\) agree on the open dense subset \(\mathrm{Spec}\, \mathbb{C} \subset X\), but \(f \neq g\). For (b), let \(X = S = \mathrm{Spec}\, \mathbb{C}\), and \(Y = \mathrm{Spec}\, \mathbb{C}[y]\) glued along \(\mathrm{Spec}\, \mathbb{C}[y,y^{-1}]\), and define \(f, g: X \rightarrow Y\) by \(f(x) = y, g(x) = y^{-1}\). Then, \(f\) and \(g\) agree on the open dense subset \(\mathrm{Spec}\, \mathbb{C}[y] \subset X\), but \(f \neq g\).

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Key Concepts

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

Algebraic Geometry
Algebraic geometry is a fascinating branch of mathematics that intersects with multiple disciplines including number theory, topology, and complex analysis. At its core, it studies solutions to polynomial equations and their geometric properties. This field utilizes an array of abstract algebraic concepts to understand and classify these solutions in terms of spaces called 'schemes'. The language and tools of algebraic geometry are essential for formalizing concepts such as varieties, morphisms, and immersions within a versatile and powerful framework.

In the context of the exercise, algebraic geometry provides the context for understanding schemes and their morphisms鈥攁n essential component for comparing different geometric structures over a common base scheme, denoted by 'S' in the problem.
Separated Scheme
In algebraic geometry, a 'separated scheme' is a fundamental concept that mirrors the notion of Hausdorff spaces from topology. A scheme is separated if, informally speaking, it does not have 'infinitely near' points that cannot be distinguished by any open subset. Formally, this condition is encapsulated in the definition that the diagonal morphism from the scheme to the product of itself over the base scheme is a closed immersion.

The significance in our exercise is immense. The separated condition on the scheme 'Y' over 'S' ensures the uniqueness of the morphisms from 'X' to 'Y' when they agree on a dense open subset鈥攁s long as 'X' is also reduced, highlighting the intricate interplay of properties in algebraic geometry. Without the property of separation, as noted in the counterexamples, distinct morphisms could coincide on open dense subsets without being identical, leading to very different geometric structures.
Reduced Scheme
When discussing schemes in algebraic geometry, the 'reduced' qualifier speaks to the absence of nilpotent elements within the scheme's structure sheaf. Nilpotent elements are those which, when raised to some power, become zero鈥攖hey can be thought of as algebraic 'ghosts' of points that are not actually present on the scheme. A 'reduced scheme' therefore, has a sheaf in which every nilpotent element is already zero, resulting in a sort of 'purity' of the geometric object it represents; there are no 'hidden' points.

This characteristic is crucial in our exercise as 'X' being a reduced scheme implies a clarity of morphisms sent to 'Y'. Because there are no nilpotent elements to obscure the mappings, we can determine definitively that if two morphisms agree on an open dense subset, they must be the same. Counterexamples demonstrate that when 'X' allows for nilpotent elements, the certainty of morphism identity is lost.
Morphisms
Morphisms are, in essence, functions that preserve the structure between algebraic objects. In algebraic geometry, specifically within the framework of schemes, morphisms serve as bridges that connect different schemes, respecting their algebraic and geometric structure. This structural preservation means that morphisms will carry over crucial properties between schemes鈥攍ike being open or closed鈥攐r the existence of elements and their relations within the scheme's corresponding rings.

The exercise presented leverages morphisms ('f' and 'g') to reveal deep insights into the relationship between two schemes, 'X' and 'Y'. The agreement of 'f' and 'g' on a dense open subset of 'X', along with the reduced and separated properties, allows us to conclude that these morphisms must in fact be identical across the entire 'X'. Morphisms not only define the mappings between objects but also carry within them the intrinsic properties that can be used to deduce powerful results, as shown in the step-by-step solution.
Closed Immersion
Closed immersion is a key concept in algebraic geometry, referring to a type of morphism that injectively sends a scheme into another, embedding it as a closed subscheme. This means that every point of the first scheme 'maps' onto a unique point of the second scheme, and that image is closed in the second scheme in the sense that it derives from the inclusion of a vanishing set of a corresponding ideal.

The diagonal map, \( \Delta: Y \rightarrow Y \times_{S} Y \), within the exercise is an example of a closed immersion because 'Y' is a separated scheme over 'S'. This closed immersion property is vital to ascertain that 'h' is indeed the same as the diagonal morphism over the entire scheme 'X', thus ensuring that 'f' equals 'g'. Closed immersions, like morphisms, reflect the properties of the larger space, ensuring that no 'new' points appear in the image鈥攑reserving the original scheme's structure and making the identification of 'f' and 'g' possible.

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Most popular questions from this chapter

singular Curves. Here we give another method of calculating the Picard group of a singular curve. Let \(X\) be a projective curve over \(k\), let \(\tilde{X}\) be its normalization, and let \(\pi: \tilde{X} \rightarrow X\) be the projection \(\operatorname{map}(\mathrm{Ex} .3 .8) .\) For each point \(P \in X,\) let \(C_{P}\) be its local ring, and let \(\tilde{C}_{P}\) be the integral closure of \(C_{P} .\) We use a \(*\) to denote the group of units in a ring. (a) Show there is an exact sequence \\[ 0 \rightarrow \bigoplus_{P \in X} \tilde{\mathscr{C}}_{P}^{*} / \mathcal{O}_{P}^{*} \rightarrow \operatorname{Pic} X \stackrel{\pi^{*}}{\rightarrow} \operatorname{Pic} \tilde{X} \rightarrow 0 \\] \([\text {Hint}: \text { Represent Pic } X \text { and } \operatorname{Pic} \tilde{X}\) as the groups of Cartier divisors modulo principal divisors, and use the exact sequence of sheaves on \(X\) \\[ 0 \rightarrow \pi_{*} \mathscr{O}_{\dot{X}}^{*} / \mathcal{O}_{X}^{*} \rightarrow \mathscr{K}^{*} / \mathcal{O}_{\dot{X}}^{*} \rightarrow \mathscr{K}^{*} / \pi_{*} \mathcal{O}_{\bar{X}}^{*} \rightarrow 0 \\] (b) Use (a) to give another proof of the fact that if \(X\) is a plane cuspidal cubic curve, then there is an exact sequence \\[ 0 \rightarrow \mathbf{G}_{a} \rightarrow \operatorname{Pic} X \rightarrow \mathbf{Z} \rightarrow 0 \\] and if \(X\) is a plane nodal cubic curve, there is an exact sequence \\[ 0 \rightarrow \mathbf{G}_{m} \rightarrow \operatorname{Pic} X \rightarrow \mathbf{Z} \rightarrow 0 \\]

Let \(A\) be a ring. Show that the following conditions are equivalent: (i) Spec \(A\) is disconnected : (ii) there exist nonzero elements \(e_{1}, e_{2} \in A\) such that \(e_{1} e_{2}=0, e_{1}^{2}=e_{1}, e_{2}^{2}=e_{2}\) \(e_{1}+e_{2}=1\) (these elements are called orthogonal idempotents): (iii) \(A\) is isomorphic to a direct product \(A_{1} \times A_{2}\) of two nonzero rings.

Flasque Sheares. A sheaf \(\bar{y}\) on a topological space \(X\) is flasque if for every inclusion \(V \subseteq U\) of open sets, the restriction \(\operatorname{map} \mathscr{F}(U) \rightarrow \mathscr{F}(V)\) is surjective. (a) Show that a constant sheaf on an irreducible topological space is flasque. See (I, 81 ) for irreducible topological spaces. (b) If \(0 \rightarrow \overline{\mathscr{H}} \rightarrow \mathscr{F} \rightarrow \mathscr{H}^{\prime \prime} \rightarrow 0\) is an exact sequence of sheaves, and if \(\bar{y}\) is flasque, then for any open set \(U\). the sequence \(0 \rightarrow \mathscr{F}^{\prime}(U) \rightarrow \mathscr{F}(U) \rightarrow\) \(\mathscr{F}^{\prime \prime}\left(L^{\prime}\right) \rightarrow 0\) of abelian groups is also exact. (c) If \(0 \rightarrow \mathscr{H} \rightarrow \mathscr{H} \rightarrow \mathscr{H}^{\prime \prime} \rightarrow 0\) is an exact sequence of sheaves, and if \(\mathscr{H}^{\prime}\) and \(\overline{\mathscr{H}}\) are flasque, then \(\mathscr{F}^{\prime \prime}\) is flasque. (d) If \(f: X \rightarrow Y\) is a continuous map, and if \(\mathscr{F}\) is a flasque sheaf on \(X\), then \(f_{*} \overline{\mathscr{H}}\) is a flasque sheaf on \(Y\) (e) Let \(\overline{\mathscr{F}}\) be any sheaf on \(X\). We define a new sheaf \(\mathscr{G}\), called the sheaf of discontinuous sections of \(\mathscr{F}\) as follows. For each open set \(U \subseteq X, \mathscr{G}(U)\) is the set of

Let \(X\) be an integral scheme. Show that the local ring \(\mathscr{O}_{\xi}\) of the generic point of \(X\) is a field. It is called the function field of \(X,\) and is denoted by \(K(X) .\) Show also that if \(U=\operatorname{Spec} A\) is any open affine subset of \(X,\) then \(K(X)\) is isomorphic to the quotient field of \(A\)

Extension of Coherent Sheaves. We will prove the following theorem in several steps: Let \(X\) be a noetherian scheme, let \(U\) be an open subset, and let \(\mathscr{F}\) be a coherent sheaf on \(U\). Then there is a coherent sheaf \(\mathscr{F}^{\prime}\) on \(X\) such that \(\left.\mathscr{F}^{\prime}\right|_{v} \cong \mathscr{F}\) (a) On a noetherian affine scheme, every quasi-coherent sheaf is the union of its coherent subsheaves. We say a sheaf \(\mathscr{F}\) is the union of its subsheaves \(\mathscr{F}\) if for every open set \(U\), the group \(\mathscr{F}(U)\) is the union of the subgroups ?\((U)\) (b) Let \(X\) be an affine noetherian scheme, \(U\) an open subset, and \(\mathscr{F}\) coherent on \(U .\) Then there exists a coherent sheaf \(\mathscr{F}^{\prime}\) on \(X\) with \(\left.\mathscr{F}^{\prime}\right|_{v} \cong \mathscr{F} .\) [Hint: Let \(\left.i: U \rightarrow X \text { be the inclusion map. Show that } i_{*} \mathscr{F} \text { is quasi-coherent, then use }(a) .\right]\) (c) With \(X, U, \mathscr{F}\) as in (b), suppose furthermore we are given a quasi-coherent sheaf \(\mathscr{G}\) on \(X\) such that \(\left.\mathscr{F} \subseteq \mathscr{G}\right|_{v} .\) Show that we can find \(\mathscr{F}^{\prime}\) a coherent subsheaf of \(\mathscr{G},\) with \(\left.\mathscr{F}^{\prime}\right|_{v} \cong \mathscr{F}\). [Hint: Use the same method, but replace \(i_{*} \mathscr{F}\) by \(\left.\rho^{-1}\left(i_{*} \mathscr{F}\right) \text { , where } \rho \text { is the natural } \operatorname{map} \mathscr{G} \rightarrow i_{*}\left(\left.\mathscr{G}\right|_{U}\right) .\right]\) (d) Now let \(X\) be any noetherian scheme, \(U\) an open subset, \(\mathscr{F}\) a coherent sheaf on \(U,\) and \(\mathscr{G}\) a quasi-coherent sheaf on \(X\) such that \(\left.\mathscr{F} \subseteq \mathscr{G}\right|_{V} .\) Show that there is a coherent subsheaf \(\mathscr{F}^{\prime} \subseteq \mathscr{G}\) on \(X\) with \(\left.\mathscr{F}^{\prime}\right|_{v} \cong \mathscr{F}\). Taking \(\mathscr{I}=i_{*} \mathscr{F}\) proves the result announced at the beginning. [Hint: Cover \(X\) with open affines, and extend over one of them at a time. (e) As an extra corollary, show that on a noetherian scheme, any quasi- coherent sheaf \(\mathscr{F}\) is the union of its coherent subsheaves. [Hint: If \(s\) is a section of \(\mathscr{F}\) over an open set \(U,\) apply (d) to the subsheaf of \(\left.\mathscr{F}\right|_{v}\) generated by s.]
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