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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 morphism \(f: X \rightarrow Y\) of schemes is quasi-compact if there is a cover of \(Y\) by open affines \(V_{i}\) such that \(f^{-1}\left(V_{i}\right)\) is quasi-compact for each \(i .\) Show that \(f\) is quasicompact if and only if for every open affine subset \(V \subseteq Y, f^{-1}(V)\) is quasi-compact.

The Infinitesimal Lifting Property. The following result is very important in studying deformations of nonsingular varieties. Let \(k\) be an algebraically closed field, let \(A\) be a finitely generated \(k\) -algebra such that Spec \(A\) is a nonsingular variety over \(k .\) Let \(0 \rightarrow I \rightarrow B^{\prime} \rightarrow B \rightarrow 0\) be an exact sequence, where \(B^{\prime}\) is a \(k\) -algebra, and \(I\) is an ideal with \(I^{2}=0 .\) Finally suppose given a \(k\) -algebra homomorphism \(f: A \rightarrow B .\) Then there exists a \(k\) -algebra homomorphism \(g: A \rightarrow B^{\prime}\) making a commutative diagram We call this result the infinitesimal liffing property for \(A .\) We prove this result in several steps. (a) First suppose that \(g: A \rightarrow B^{\prime}\) is a given homomorphism lifting \(f\). If \(g^{\prime}: A \rightarrow B^{\prime}\) is another such homomorphism, show that \(\theta=g-g^{\prime}\) is a \(k\) -derivation of \(A\) into \(I,\) which we can consider as an element of \(\mathrm{Hom}_{A}\left(\Omega_{A / k}, I\right) .\) Note that since \(I^{2}=0, I\) has a natural structure of \(B\) -module and hence also of \(A\) -module. Conversely, for any \(\theta \in \operatorname{Hom}_{A}\left(\Omega_{A / k}, I\right), g^{\prime}=g+\theta\) is another homomorphism lifting \(f .\) (For this step, you do not need the hypothesis about Spec \(A\) being nonsingular.) (b) Now let \(P=k\left[x_{1}, \ldots, x_{n}\right]\) be a polynomial ring over \(k\) of which \(A\) is a quotient, and let \(J\) be the kernel. Show that there does exist a homomorphism \(h: P \rightarrow B^{\prime}\) making a commutative diagram, and show that \(h\) induces an \(A\) -linear map \(\hbar: J / J^{2} \rightarrow I\) (c) Now use the hypothesis Spec \(A\) nonsingular and (8.17) to obtain an exact sequence \\[ 0 \rightarrow J / J^{2} \rightarrow \Omega_{P / k} \otimes A \rightarrow \Omega_{A / k} \rightarrow 0 \\] Show furthermore that applying the functor Hom \(_{A}(\cdot, I)\) gives an exact sequence \\[ 0 \rightarrow \operatorname{Hom}_{A}\left(\Omega_{A / k}, I\right) \rightarrow \operatorname{Hom}_{P}\left(\Omega_{P | k}, I\right) \rightarrow \operatorname{Hom}_{A}\left(J / J^{2}, I\right) \rightarrow 0 \\] Let \(\theta \in \mathrm{Hom}_{P}\left(\Omega_{P / k}, I\right)\) be an element whose image gives \(\bar{h} \in \mathrm{Hom}_{A}\left(J / J^{2}, I\right)\) Consider \(\theta\) as a derivation of \(P\) to \(B^{\prime}\). Then let \(h^{\prime}=h-\theta\), and show that \(h^{\prime}\) is a homomorphism of \(P \rightarrow B^{\prime}\) such that \(h^{\prime}(J)=0 .\) Thus \(h^{\prime}\) induces the desired homomorphism \(g: A \rightarrow B^{\prime}\).

(a) Show that a morphism \(f: X \rightarrow Y\) is of finite type if and only if it is locally of finite type and quasi-compact. (b) Conclude from this that \(f\) is of finite type if and only if for every open affine subset \(V=\operatorname{Spec} B\) of \(Y, f^{-1}(V)\) can be covered by a finite number of open affines \(U_{i}=\operatorname{Spec} A_{i},\) where each \(A_{i}\) is a finitely generated \(B\) -algebra. (c) Show also if \(f\) is of finite type, then for every open affine subset \(V=\operatorname{Spec} B \subseteq\) \(Y,\) and for every open affine subset \(U=\operatorname{Spec} A \subseteq f^{-1}(V), A\) is a finitely gener- ated \(B\) -algebra.

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 \(f(Z)\)

Espace Etale of a Presheuf. (This exercise is included only to establish the connection between our definition of a sheaf and another definition often found in the literature. See for example Godement [1. Ch. II, \$1.2].) Given a presheaf \(\mathscr{F}\) un \(X\), we define a topological space Sp茅( \(\bar{y}\) ), called the espuce 茅ral茅 of \(\mathscr{F},\) as follows. As a set. Sp茅(. \(\overline{\mathscr{F}})=\cup_{P e}, x^{-} \overline{\mathscr{H}}_{P} .\) We define a projection map \(\pi: \operatorname{Sp茅}(\mathscr{F}) \rightarrow X\) by sending \(s \in \overline{\mathscr{H}}_{p}\) to \(P\). For each open set \(U \subseteq X\) and each section \(s \in \overline{\mathscr{F}}(\mathcal{L}),\) we obtain a \(\operatorname{map} \bar{\Im}: L \rightarrow \operatorname{Spei}(\mathscr{F})\) by sending \(P \mapsto s_{P},\) its germ at \(P .\) This map has the property that \(\pi \quad \bar{s}=\) id \(_{l},\) in other words, it is a "section" of \(\pi\) over \(U\). We now make Sp茅(. \(\overline{\mathscr{H}}\) ) into a topological space by giving it the strongest topology such that show that the sheaf \(\bar{y}^{+}\) associated to \(\bar{y}\) can be described as follows: for any open set \(l \subseteq X, \overline{\mathscr{H}}^{+}(U)\) is the set of continuous sections of \(\operatorname{Spei}(\mathscr{F})\) over \(U\). In particular, the original presheaf \(\mathscr{I}\) was a sheaf if and only if for each \(U, \mathscr{F}(U)\) is equal to the set of all continuous sections of Sp茅(. \(\overline{\mathcal{F}}\) ) over \(U\)
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