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Chapter 2: Problem 5

Describe Spec \(\mathbf{Z}\), and show that it is a final object for the category of schemes. i.e., each scheme \(X\) admits a unique morphism to Spec \(\mathbf{Z}\).

Short Answer

Expert verified
Spec \( \mathbf{Z} \) can be understood as the set \(\{ (0), (2), (3), (5), (7), \ldots \}\), the set of all prime numbers and 0, considering that \(\mathbf{Z}\) only has two prime ideals. It is a final object in the category of schemes because for any scheme X there is a unique morphism \( X \rightarrow \) Spec \( \mathbf{Z} \).

Step by step solution

01

Describe Spec \( \mathbf{Z} \)

Since \(\mathbf{Z}\) only has two prime ideals, the ideal (0) and (p) for any prime p. The Spec \( \mathbf{Z} \) is the set of these ideals, that can be seen as \(\{ (0), (2), (3), (5), (7), \ldots \}\), i.e., the set of all prime numbers and 0.
02

Proving Spec \( \mathbf{Z} \) as a Final Object

To show Spec \( \mathbf{Z} \) is a final object in the category of schemes, we need to prove that for every scheme \( X \), there is a unique morphism \( X \rightarrow \) Spec \( \mathbf{Z} \). By definition, over any scheme \( X \), the structure sheaf of Spec \( \mathbf{Z} \) is constant, it's the sheaf associated to the constant presheaf with value \( \mathbf{Z} \). Hence, there is a unique morphism of schemes \( X \rightarrow \) Spec \( \mathbf{Z} \), given on an open set \( U \subseteq X \) by the unique ring homomorphism \( \mathbf{Z} \rightarrow \mathcal{O}(U) \). This shows that Spec \( \mathbf{Z} \) is a final object in the category of schemes.

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Key Concepts

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

Scheme Theory
Scheme theory forms the backbone of modern algebraic geometry, building upon the concept of a variety by allowing 'singularities' and incorporating prime ideals as points. Think of schemes as geometric spaces that are defined not only by their points but also by the 'functions' that live on them. In essence, a scheme is a topological space teamed up with a structure sheaf, which assigns rings of functions to every open set.
The specific example of Spec \(\mathbb{Z}\) illustrates these ideas vividly. It consists of points corresponding to prime ideals, giving us an elegant geometric picture of arithmetic properties of integers.
Prime Ideals
Prime ideals are a cornerstone of scheme theory and algebraic geometry. A prime ideal in a ring like \(\mathbb{Z}\) is a subset of elements that is not the whole ring, and it observes a key property: if the product of two elements is in the ideal, then at least one of these elements is also in the ideal. The prime ideals of \(\mathbb{Z}\) correspond to the zero ideal and the ideals generated by prime numbers, representing the 'points' of Spec \(\mathbb{Z}\).
Understanding the role of prime ideals in schemes helps in visualizing algebraic structures as geometric entities and forms the foundation for discussing morphisms and mappings between different schemes.
Category of Schemes
In the category of schemes, each object is a scheme and morphisms between these objects follow specific algebraic rules. It's an abstract framework that helps mathematicians to deal with schemes in a structured way.
Within this setting, Spec \(\mathbb{Z}\) holds a special place as a final object, meaning that every scheme maps to it in a unique way鈥攋ust like all roads lead to Rome. This uniqueness is crucial for understanding how schemes relate to each other and how they can be transformed through morphisms.
Structure Sheaf
A structure sheaf is an essential part of a scheme that links the geometrical with the algebraic. For each open subset of a scheme, the structure sheaf assigns a ring of 'functions,' allowing us to 'do algebra' locally on the scheme.
The structure sheaf of Spec \(\mathbb{Z}\) is particularly simple: it assigns the integer ring \(\mathbb{Z}\) to any open set. This simplicity is what enables the unique morphisms we find in Spec \(\mathbb{Z}\) as a final object in the category of schemes, neatly tying together the algebraic structure with the geometrical intuition.
Morphism of Schemes
A morphism of schemes is a way of transforming one scheme into another, involving both a continuous function between the underlying topological spaces and a compatible family of ring homomorphisms for the structure sheaves.
Every scheme 'wants' to map to Spec \(\mathbb{Z}\) in its own unique way through such a morphism. It is akin to saying there is a specific and unique perspective from which to view all other schemes from the vantage point of Spec \(\mathbb{Z}\). This unique mapping perspective is not only foundational for understanding how different algebraic structures interact with one another but also emphasizes the universality of \(\mathbb{Z}\) within the world of schemes.

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Most popular questions from this chapter

Tensor Operations on Sheaves. First we recall the definitions of various tensor operations on a module. Let \(A\) be a ring, and let \(M\) be an \(A\) -module. Let \(T^{\prime \prime}(M)\) be the tensor product \(M \otimes \ldots \otimes M\) of \(M\) with itself \(n\) times, for \(n \geqslant 1\). For \(n=0\) we put \(T^{0}(M)=A .\) Then \(T(M)=\bigoplus_{n \geqslant 0} T^{\prime \prime}(M)\) is a (noncommutative) \(A\) -algebra, which we call the tensor algebra of \(M .\) We define the symmetric algebra \(S(M)=\bigoplus_{n \geqslant 0} S^{\prime \prime}(M)\) of \(M\) to be the quotient of \(T(M)\) by the two-sided ideal generated by all expressions \(x \otimes y-y \otimes x,\) for all \(x, y \in M .\) Then \(S(M)\) is a commutative \(A\) -algebra. Its component \(S^{n}(M)\) in degree \(n\) is called the \(n\) th symmetric product of \(M .\) We denote the image of \(x \otimes y\) in \(S(M)\) by \(x y,\) for any \(x, y \in M .\) As an example, note that if \(M\) is a free \(A\) -module of rank \(r,\) then \(S(M) \cong\) \(A\left[x_{1}, \ldots, x_{r}\right]\). We define the exterior algebra \(\wedge(M)=\bigoplus_{n \geqslant 0} \wedge^{\prime \prime}(M)\) of \(M\) to be the quotient of \(T(M)\) by the two- sided ideal generated by all expressions \(x \otimes x\) for \(x \in M .\) Note that this ideal contains all expressions of the form \(x \otimes y+y \otimes x\) so that \(\wedge(M)\) is a skew commutative graded \(A\) -algebra. This means that if \(u \in\) \(\wedge^{r}(M)\) and \(v \in \Lambda^{s}(M),\) then \(u \wedge v=(-1)^{r s} v \wedge u\) (here we denote by \(\wedge\) the multiplication in this algebra; so the image of \(x \otimes y\) in \(\wedge^{2}(M)\) is denoted by \(x \wedge y\) ). The \(n\) th component \(\wedge^{\prime \prime}(M)\) is called the \(n\) th exterior power of \(M\). Now let \(\left(X, O_{X}\right)\) be a ringed space, and let \(\mathscr{F}\) be a sheaf of \(\mathcal{O}_{X}\) -modules. We define the tensor algebra, symmetric algebra, and exterior algebra of \(\mathscr{F}\) by taking the sheaves associated to the presheaf, which to each open 'set \(U\) assigns the corresponding tensor operation applied to \(\mathscr{F}(U)\) as an \(\mathscr{O}_{X}(U)\) -module. The results are \(\mathcal{O}_{X^{-}}\) algebras, and their components in each degree are \(\mathscr{C}_{X}\) -modules. (a) Suppose that \(\mathscr{F}\) is locally free of rank \(n\). Then \(T^{\prime}(\mathscr{F}), S^{\prime}(\mathscr{F})\), and \(\wedge^{\prime}(\mathscr{F})\) are also locally free, of ranks \(n^{\prime},\left(\begin{array}{c}m+r-1 \\ n-1\end{array}\right),\) and \(\left(\begin{array}{c}m \\ 2\end{array}\right)\) respectively. (b) Again let \(\mathscr{F}\) be locally free of rank \(n\). Then the multiplication \(\operatorname{map} \wedge \mathscr{F} \otimes\) \(\wedge^{n-r} \mathscr{F} \rightarrow \wedge^{n} \cdot \mathscr{F}\) is a perfect pairing for any \(r,\) i.c., it induces an isomorphism of \(\wedge^{\prime \prime} \mathscr{F}\) with \(\left(\wedge^{n-r} \mathscr{F}\right)^{\sim} \otimes \wedge^{\prime \prime} \mathscr{F}\). As a special case, note if \(\mathscr{F}\) has rank 2 then \(\mathscr{F} \cong \mathscr{F}^{\sim} \otimes \wedge^{2} \mathscr{F}\) (c) Let \(0 \rightarrow \mathscr{F}^{\prime} \rightarrow \mathscr{F} \rightarrow \mathscr{F}^{\prime \prime} \rightarrow 0\) be an exact sequence of locally free sheaves. Then for any \(r\) there is a finite filtration of \(S^{\prime}(\mathscr{F})\) \\[ S^{\prime}(\mathscr{F})=F^{0} \supseteq F^{1} \supseteq \ldots \supseteq F^{\prime} \supseteq F^{r+1}=0 \\] with quotients \\[ F^{p} / F^{p+1} \cong S^{p}\left(\mathscr{F}^{\prime}\right) \otimes S^{r-p}\left(\mathscr{F}^{\prime \prime}\right) \\] for each \(p\). (d) Same statement as (c), with exterior powers instead of symmetric powers. In particular, if \(\mathscr{F}^{\prime}, \mathscr{F}, \mathscr{F}^{\prime \prime}\) have ranks \(n^{\prime}, n, n^{\prime \prime}\) respectively, there is an isomorphism \(\wedge^{n} \mathscr{F} \cong \wedge^{n^{\prime} \mathscr{F}^{\prime}} \otimes \wedge^{n^{\prime \prime}} \mathscr{F}^{\prime \prime}\) (e) Let \(f: X \rightarrow Y\) be a morphism of ringed spaces, and let \(\mathscr{F}\) be an \(\mathscr{U}_{Y}\) -module. Then \(f^{*}\) commutes with all the tensor operations on \(\mathscr{F},\) i.e., \(f^{*}\left(S^{n}(\mathscr{F})\right)=\) \(S^{\prime \prime}\left(f^{*} \mathscr{F}\right)\) etc.

Zariski Spaces. A topological space \(X\) is a Zariski space if it is noetherian and every (nonempty) closed irreducible subset has a unique generic point (Ex. 2.9 ). For example, let \(R\) be a discrete valuation ring, and let \(T=\operatorname{sp}(\operatorname{Spec} R)\). Then \(T\) consists of two points \(t_{0}=\) the maximal ideal, \(t_{1}=\) the zero ideal. The open subsets are \(\varnothing,\left\\{t_{1}\right\\},\) and \(T .\) This is an irreducible Zariski space with generic point \(t_{1}\). (a) Show that if \(X\) is a noetherian scheme, then \(\operatorname{sp}(X)\) is a Zariski space. (b) Show that any minimal nonempty closed subset of a Zariski space consists of one point. We call these closed points. (c) Show that a Zariski space \(X\) satisfies the axiom \(T_{0}\) : given any two distinct points of \(X\), there is an open set containing one but not the other (d) If \(X\) is an irreducible Zariski space, then its generic point is contained in every nonempty open subset of \(X\) (e) If \(x_{0}, x_{1}\) are points of a topological space \(X,\) and if \(x_{0} \in\left\\{x_{1}\right\\}^{-},\) then we say that \(x_{1}\) specializes to \(x_{0},\) written \(x_{1} \leadsto \rightarrow x_{0} .\) We also say \(x_{0}\) is a specialization of \(x_{1},\) or that \(x_{1}\) is a generization of \(x_{0} .\) Now let \(X\) be a Zariski space. Show that the minimal points, for the partial ordering determined by \(x_{1}>x_{0}\) if \(x_{1} \leadsto x\) \(x_{0},\) are the closed points, and the maximal points are the generic points of the irreducible components of \(X .\) Show also that a closed subset contains every specialization of any of its points. (We say closed subsets are stable under specialization. . Similarly, open subsets are stable under generization. (f) Let \(t\) be the functor on topological spaces introduced in the proof of (2.6) If \(X\) is a noetherian topological space, show that \(t(X)\) is a Zariski space. Furthermore \(X\) itself is a Zariski space if and only if the \(\operatorname{map} \alpha: X \rightarrow t(X)\) is a homeomorphism.

Examples of Valuation Rings. Let \(k\) be an algebraically closed field. (a) If \(K\) is a function field of dimension 1 over \(k(I, \$ 6),\) then every valuation ring of \(K / k\) (except for \(K\) itself) is discrete. Thus the set of all of them is just the abstract nonsingular curve \(C_{K}\) of \((\mathrm{I}, \$ 6)\) (b) If \(K / k\) is a function field of dimension two, there are several different kinds of valuations. Suppose that \(X\) is a complete nonsingular surface with function field \(K\) (1) If \(Y\) is an irreducible curve on \(X\), with generic point \(x_{1},\) then the local ring \(R=C_{x_{1}, x}\) is a discrete valuation ring of \(K k\) with center at the (nonclosed) point \(x_{1}\) on \(X\) (2) If \(f: X^{\prime} \rightarrow X\) is a birational morphism, and if \(Y^{\prime}\) is an irreducible curve in \(X^{\prime}\) whose image in \(X\) is a single closed point \(x_{0},\) then the local ring \(R\) of the generic point of \(Y^{\prime}\) on \(X^{\prime}\) is a discrete valuation ring of \(K k\) with center at the closed point \(x_{0}\) on \(X\) (3) Let \(r_{0} \in X\) be a closed point. Let \(f: X_{1} \rightarrow X\) be the blowing-up of \(x_{0}\) (I. \(\$ 4)\) and let \(E_{1}=f^{-1}\left(r_{0}\right)\) be the exceptional curve. Choose a closed point \(x_{1} \in E_{1},\) let \(f_{2}: X_{2} \rightarrow X_{1}\) be the blowing-up of \(x_{1},\) and let \(E_{2}=\) \(f_{2}^{-1}\left(x_{1}\right)\) be the exceptional curve. Repeat. In this manner we obtain a sequence of varieties \(X\), with closed points \(x_{i}\) chosen on them, and for each \(i,\) the local ring \(C_{1,1,1}, x_{1},\) dominates \(C_{x_{1}, x_{1}},\) Let \(R_{0}=\bigcup_{1=0}^{x} C_{x_{1}, x_{1}}\) Then \(R_{0}\) is a local ring, so it is dominated by some valuation ring \(R\) of \(K / k\) by \((\mathrm{I}, 6.1 \mathrm{A}) .\) Show that \(R\) is a valuation ring of \(K / k\). and that it has center \(x_{0}\) on \(X .\) When is \(R\) a discrete valuation ring? Note. We will see later (V.Ex. 5.6) that in fact the \(R_{0}\) of (3) is already a valuation ring itself, so \(R_{0}=R\). Furthermore, every valuation ring of \(K, k\) (except for \(K\) itself) is one of the three kinds just described.

In this exercise, we compare some properties of a ring homomorphism to the induced morphism of the spectra of the rings. (a) Let \(A\) be a ring, \(X=\operatorname{Spec} A,\) and \(f \in A .\) Show that \(f\) is nilpotent if and only if \(D(f)\) is empty. (b) Let \(\varphi: A \rightarrow B\) be a homomorphism of rings, and let \(f: Y=\operatorname{Spec} B \rightarrow X=\) Spec \(A\) be the induced morphism of affine schemes. Show that \(\varphi\) is injective if and only if the map of sheaves \(f^{*}: c_{1} \rightarrow f_{*} C_{r}\) is injective. Show furthermore in that case \(f\) is dominant, i.e., \(f(Y)\) is dense in \(X\). (c) With the same notation, show that if \(\varphi\) is surjective, then \(f\) is a homeomorphism of \(Y\) onto a closed subset of \(X,\) and \(f^{* *}: C_{X} \rightarrow f_{*} C_{Y}\) is surjective. (d) Prove the converse to (c). namely. if \(f: Y \rightarrow X\) is a homeomorphism onto a closed subset, and \(f^{\prime \prime}: C_{1} \rightarrow f_{*} C_{1},\) is surjective. then \(\varphi\) is surjective. [Hint: Consider \(\left.X^{\prime}=\operatorname{Spec}(.4 \mathrm{ker} \varphi) \text { and use }(\mathrm{b}) \text { and }(\mathrm{c}) .\right]\)

Support. Let \(\mathscr{F}\) be a sheaf on \(X\), and let \(s \in \mathscr{F}(U)\) be a section over an open set \(U\) The support of \(s\), denoted Supp s, is defined to be \(\left\\{P \in U | s_{P} \neq 0\right\\},\) where \(s_{P}\) denotes the germ of s in the stalk \(\overline{\mathscr{F}}_{p}\). Show that Supp s is a closed subset of \(U\). We define the support of \(\overline{\mathscr{F}}, \operatorname{Supp}, \overline{\mathscr{F}},\) to be \(\left\\{P \in X | \mathscr{F}_{P} \neq 0\right\\},\) It need not be a closed subset.
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