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Chapter 2: Problem 15

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.

Short Answer

Expert verified
For proving geometric irreducibility, we show that: (i) irreducibility over the algebraic closure implies irreducibility over the separable closure; (ii) irreducibility over the separable closure implies irreducibility over all extensions; (iii) irreducibility over all extensions implies irreducibility over the algebraic closure. For geometrically reduced, the same logic can be applied but with the perfect closure: (i) reducedness over the algebraic closure implies reducedness over the perfect closure; (ii) reducedness over the perfect closure implies reducedness over all extensions; (iii) reducedness over all extensions implies reducedness over the algebraic closure. An integral scheme that is neither geometrically irreducible nor geometrically reduced is the scheme \(X = Spec(k[x]/(x^{2}+1))\) over the field of real numbers.

Step by step solution

01

Prove equivalence for being geometrically irreducible

To prove these conditions are equivalent requires only demonstrating that one condition leads to another, as follows:i \(\rightarrow\) ii: If \(X \times_{k} \bar{k}\) is irreducible, then \(X \times_{k} k_{s}\) is also irreducible, as \(k_{s}\) is an intermediate field of \(\bar{k}\) over \(k\).ii \(\rightarrow\) iii: If \(X \times_{k} k_{s}\) is irreducible for the separable closure, then for any field extension \(K\) of \(k\), \(X \times_{k} K\) is also irreducible. This is by definition of irreducible: any non-empty open subset of \(X \times_{k} K\) is dense in \(X \times_{k} K\), and \(X \times_{k} k_s\) is such a subset.iii \(\rightarrow\) i: If \(X \times_{k} K\) is irreducible for every extension field \(K\) of \(k\), then especially it is for \(\overline{k}\), the algebraically closed field. So \(X \times_{k} \overline{k}\) is irreducible.
02

Prove equivalence for being geometrically reduced

Again, it's crucial to show that one condition implies the next:i \(\rightarrow\) ii: If \(X \times_{k} \overline{k}\) is reduced, then \(X \times_{k} k_{p}\) (the perfect closure of \(k\)) is also reduced, as \(k_{p}\) is an intermediate field of \(\overline{k}\) over \(k\).ii \(\rightarrow\) iii: If \(X \times_{k} k_{p}\) is reduced for the perfect closure, then for any field extension \(K\) of \(k\), \(X \times_{k} K\) is also reduced. This follows from the fact that any scheme is reduced if and only if the stalk of each of its points is reduced, which is the case for \(k_{p}\), hence it holds for all its extensions.iii \(\rightarrow\) i: If \(X \times_{k} K\) is reduced for every extension field \(K\) of \(k\), then it is in particular for \(\overline{k}\), the algebraically closed field. So \(X \times_{k} \overline{k}\) is reduced.
03

Provide examples of integral schemes not geometrically irreducible or reduced

An example of an integral scheme which is not geometrically irreducible nor geometrically reduced is \(X = Spec(k[x]/(x^{2}+1))\) over \(k = \mathbb{R}\), the field of real numbers. It is integral because it is defined by a single (irreducible) polynomial over \(k\).Yet, over \(\overline{k} = \mathbb{C}\), it's not reduced: \(x^{2}+1\) factors as \((x+i)(x-i)\) in \(\mathbb{C}[x]\), hence \(X \times_{k} \overline{k}\) is not irreducible. It's also not reduced, because extensions of \(k\) including \(i\) will have non-trivial nilpotents.

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Key Concepts

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

Geometrically Reduced Schemes
In algebraic geometry, a geometrically reduced scheme is a vital concept to grasp. Essentially, a scheme is called geometrically reduced when it remains reduced even when base-changed to an algebraic closure of its field. A scheme is "reduced" if it doesn't have any "nilpotent" elements, which are elements that become zero when raised to some power.

When we talk about base change, we mean that we're considering the scheme over a new field. Usually, this involves using the algebraic closure, the largest field extension with all algebraic numbers over the original field. This means we take the scheme defined over a field and check it over all the possible numbers that might fit into the solution.

To say it in simple terms:
  • A geometrically reduced scheme stays "clean" without nilpotent elements even in the broadest possible context of numbers.
Ensuring a scheme is geometrically reduced is useful for understanding its behavior across different fields.
Integral Schemes
Integral schemes are another cornerstone in understanding algebraic structures. These schemes are both "irreducible" and "reduced." This means:
  • An integral scheme is irreducible: it cannot be split into smaller parts (sub-schemes).
  • It's also reduced: it contains no weird elements that disappear when multiplied by themselves multiple times (no nilpotents).
When you mix these qualities, an integral scheme becomes a powerful and clean entity that doesn't decompose easily and doesn't have any hidden "ghost" elements (nilpotents).

Imagine an integral scheme as a strong house built from solid bricks without any hidden cavities. No half-measures are sneaking around inside its structure. Understanding integral schemes helps lay the groundwork for more complex algebraic constructs, especially in coming to grips with powerhouses in algebraic geometry.
Algebraic Closure
The term "algebraic closure" refers to an incredibly comprehensive expansion of a field. In the realm of algebra, it's about taking a field (let's say, the field of rational numbers, \( \mathbb{Q} \)) and adding in all possible roots of polynomials from that field.

In simpler terms:
  • The algebraic closure contains every possible solution (or root) to a polynomial, regardless of whether or not those solutions could originally exist in the field.
For example, the algebraic closure of the real numbers \( \mathbb{R} \) is the complex numbers \( \mathbb{C} \) because it includes roots for equations like \( x^2 + 1 = 0 \).

Working with algebraic closures allows mathematicians to handle mathematical objects in the broadest capacity, ensuring that every potential outcome or solution is accounted for. This concept is crucial for examining schemes in cases where fields may not initially contain all potential roots, thereby enabling the consideration of schemes under the most generalized conditions.

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Most popular questions from this chapter

A topological space is quasi-compact if every open cover has a finite subcover. (a) Show that a topological space is noetherian (I, \(\$ 1)\) if and only if every open subset is quasi-compact. (b) If \(X\) is an affine scheme. show that \(\operatorname{sp}(X)\) is quasi- compact. but not in general noetherian. We say a scheme \(X\) is quati-ciompact if \(\operatorname{sp}(X)\) is. (c) If \(A\) is a noetherian ring. show that spiSpec 1 ) is a nocthcrian topological space. (d) Give an example to show that sp(Spec \(A\) ) can be noetherian even when \(A\) is not.

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 \(\left.X^{\prime}=\operatorname{Spec}(.4 \mathrm{ker} \varphi) \text { and use }(\mathrm{b}) \text { and }(\mathrm{c}) .\right]\)

Closed Subschemes. (a) Closed immersions are stable under base extension: if \(f: Y \rightarrow X\) is a closed immersion, and if \(X^{\prime} \rightarrow X\) is any morphism, then \(f^{\prime}: Y \times_{X} X^{\prime} \rightarrow X^{\prime}\) is also a closed immersion. (b) If \(Y\) is a closed subscheme of an affine scheme \(X=\operatorname{Spec} A\), then \(Y\) is also affine, and in fact \(Y\) is the closed subscheme determined by a suitable ideal \(\mathfrak{a} \subseteq A\) as the image of the closed immersion \(\operatorname{Spec} A / \mathfrak{a} \rightarrow \operatorname{Spec} A\). [Hints: First show that \(Y\) can be covered by a finite number of open affine subsets of the form \(D\left(f_{i}\right) \cap Y,\) with \(f_{i} \in A .\) By adding some more \(f_{i}\) with \(D\left(f_{i}\right) \cap Y=\varnothing\) if necessary, show that we may assume that the \(D\left(f_{i}\right)\) cover \(X .\) Next show that \(f_{1}, \ldots, f_{r}\) generate the unit ideal of \(A .\) Then use (Ex. 2.17 b) to show that \(Y\) is affine, and (Ex. \(2.18 \mathrm{d}\) ) to show that \(Y\) comes from an ideal \(\mathfrak{a} \subseteq\) A. .] Note: We will give another proof of this result using sheaves of ideals later (5.10). (c) Let \(Y\) be a closed subset of a scheme \(X\), and give \(Y\) the reduced induced subscheme structure. If \(Y^{\prime}\) is any other closed subscheme of \(X\) with the same underlying topological space, show that the closed immersion \(Y \rightarrow X\) factors through \(Y^{\prime} .\) We express this property by saying that the reduced induced structure is the smallest subscheme structure on a closed subset. (d) Let \(f: Z \rightarrow X\) be a morphism. Then there is a unique closed subscheme \(Y\) of \(X\) with the following property: the morphism \(f\) factors through \(Y\), and if \(Y^{\prime}\) is any other closed subscheme of \(X\) through which \(f\) factors, then \(Y \rightarrow X\) factors through \(Y^{\prime}\) also. We call \(Y\) the scheme-theoretic image of \(f\). If \(Z\) is a reduced scheme, then \(Y\) is just the reduced induced structure on the closure of the image \(f(Z)\)

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.

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.
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