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Chapter 2: Problem 1

Let \(X\) be an affine variety and \(K\) a finite extension of \(k(X)\). Prove that there exists an affine variety \(Y\) and a map \(f: Y \rightarrow X\) with the properties (1) \(f\) is finite; (2) \(Y\) is normal; (3) \(k(Y)=K\) with \(f^{*}: k(X) \hookrightarrow k(Y)=K\) the given inclusion. Prove that \(Y\) is uniquely determined by these properties. It is called the normalisation of \(X\) in \(K\).

Short Answer

Expert verified
Construct \(Y = \text{Spec}(B)\) where \(B\) is the integral closure of \(A\) in \(K\). Prove \(f\) is finite, \(Y\) is normal, \(k(Y) = K\), and show that \(Y\) is uniquely determined.

Step by step solution

01

- Understand the Problem

We need to show that for an affine variety \(X\) and a finite extension \(K\) of \(k(X)\), there exists an affine variety \(Y\) with corresponding map \(f: Y \rightarrow X\) such that \(f\) is finite, \(Y\) is normal, \(k(Y)=K\), and the inclusion \(f^{*}: k(X) \rightarrow k(Y)\) holds. Finally, we need to prove that \(Y\) is uniquely determined by these properties.
02

- Constructing the Variety \(Y\)

Consider \(X = \text{Spec}(A)\) for some affine coordinate ring \(A\). Similarly, let \(K\) be a finite extension of the fraction field \(k(X) = \text{Frac}(A)\). Let \(B\) be the integral closure of \(A\) in \(K\). This ring \(B\) is called the normalization of \(A\). Then \(Y = \text{Spec}(B)\) is our candidate variety.
03

- Proving the Finite Map

Since \(K\) is a finite extension of \(k(X)\), and \(B\) is integral over \(A\), the map \(f: Y \rightarrow X\) induced by the inclusion \(A \rightarrow B\) is finite in the sense that \(B\) is a finitely generated \(A\)-module. This ensures that \(f\) is a finite map.
04

- Ensuring Normality

By construction, \(B\) is the integral closure of \(A\) in \(K\). The integral closure of a ring in a field is normal, so \(Y = \text{Spec}(B)\) is a normal variety.
05

- Field of Functions

The fraction field of \(B\) is \(K\), and since \(B\) is the integral closure of \(A\) in \(K\), we have \(k(Y) = K\). The map \(f^{*} : k(X) \rightarrow k(Y)\) is the inclusion map and holds because of the way \(K\) was defined as extending \(k(X)\). Hence, the required property \(k(Y) = K\) with inclusion \(f^{*}\) is satisfied.
06

- Uniqueness

Suppose there is another variety \(Y' = \text{Spec}(B')\) with the same properties where \(B'\) is also an integral closure of \(A\) in \(K\). Any other integral closure of \(A\) in \(K\) must coincide with \(B\), thus \(Y' = Y\). This proves the uniqueness of \(Y\).

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Key Concepts

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

Affine Variety
An affine variety is a fundamental concept in algebraic geometry. It is defined as the set of solutions to a collection of polynomial equations within an affine space. Affine varieties bridge algebra and geometry by representing geometric structures through algebraic equations.
To understand affine varieties, start by considering a polynomial ring over a field. The solutions to the system of polynomial equations from this ring in an affine space form our affine variety. These varieties essentially bring together the concepts of polynomial functions and geometric objects.
Finite Extension
In field theory, a finite extension is a field extension that has a finite degree. This means the larger field can be expressed as a finite-dimensional vector space over the smaller field.
Given two fields, where one is an extension of another, we can measure the degree of this extension by counting the number of basis elements needed to express elements of the extended field.
When dealing with affine varieties, understanding finite extensions helps in grasping how one set of rational functions can extend another set's structure.
Integral Closure
The integral closure is a crucial concept when studying rings and fields. It refers to the process of finding all elements in an extension ring that are roots of monic polynomials with coefficients from the base ring.
If you take a ring and extend it to include solutions to polynomial equations, the new ring can be more comprehensive in scope. This extended ring where all elements are roots of such polynomials is the integral closure of the original ring.
For affine varieties, considering the integral closure of the coordinate ring within a function field helps identify 'normal' varieties, which are varieties with no singularities in their structure.
Finite Map
A finite map between varieties is one where the induced map between their coordinate rings results in a finite module. Essentially, for map-induced module relationships, if the module is finite, the map is considered finite.
This concept helps ensure that the map from one variety to another remains controlled and limited in complexity. When dealing with affine varieties, a finite map indicates that the target variety has a structured, non-infinite relationship with the source.
Normal Variety
A normal variety is a type of affine variety characterized by having no singularities in its structure. Specifically, a variety is normal if its coordinate ring is integrally closed in its function field.
This concept roots back to algebra, where integral closure ensures that every element satisfies a certain polynomial equation. For a normal variety, this means that locally at every point, the ring structure is as simple and 'normal' as possible, implying smoothness and the absence of pathological points.
Understanding normal varieties is essential when working through affine varieties and their normalizations, as it helps us ensure neat and clean algebraic structures without irregularities.

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Most popular questions from this chapter

Consider the rational map \(\varphi: \mathbb{P}^{2} \rightarrow \mathbb{P}^{4}\) given by $$ \varphi\left(x_{0}: x_{1}: x_{2}\right)=\left(x_{0} x_{1}: x_{0} x_{2}: x_{1}^{2}: x_{1} x_{2}: x_{2}^{2}\right) $$ Prove that \(\varphi\) is a birational map to a surface \(\overline{\varphi\left(\mathbb{P}^{2}\right)}\), and that the inverse map \(\overline{\varphi\left(\mathbb{P}^{2}\right)} \rightarrow \mathbb{P}^{2}\) coincides with the blowup.

We use the following assertion in subsequent exercises. Let \(X \subset \mathbb{A}^{n}\) be an affine variety and \(x \in X\). Suppose that \(\mathfrak{a}_{X}=\left(f_{1}, \ldots, f_{m}\right)\). Prove that $$ \widehat{\mathcal{O}}_{x}=k\left[\left[T_{1}, \ldots, T_{n}\right]\right] / \overline{\mathfrak{a}}_{X}, \quad \text { where } \overline{\mathfrak{a}}_{X}=\left(\tau\left(f_{1}\right), \ldots, \tau\left(f_{m}\right)\right) $$ and \(\tau\left(f_{i}\right)\) is the Taylor series of \(f_{i}\) as in Section 2.2. [Hint: Use the results of Atiyah and Macdonald [8, Chapter 10].]

Determine the branch locus of the map \(X \rightarrow \mathbb{P}^{n}\), where \(X\) is the normalisation of \(\mathbb{P}^{n}\) in the quadratic extension of \(k\left(\mathbb{P}^{n}\right)=k\left(x_{1}, \ldots, x_{n}\right)\) defined by the equation \(y^{2}=\) \(f\left(x_{1}, \ldots, x_{n}\right)\), where \(f\) is a polynomial of degree \(m\). [Hint: The answer depends on the parity of \(m\).]

Let \(x\) be a point of a variety \(X\) and \(\mathfrak{m}_{x} \subset \mathcal{O}_{x}\) the local ring at \(x\) and its maximal ideal. Prove that for every integer \(t>0\), the module \(\mathfrak{m}_{x}^{t} / \mathfrak{m}_{x}^{t+1}\) is a finite dimensional vector space over \(k\).

Prove that if \(X \subset \mathbb{P}^{n}\) is a nonsingular hypersurface and not a hyperplane, then as \(x\) runs through \(X\), the tangent hyperplanes \(\Theta_{x}\) form a hypersurface in the dual space \(\mathbb{P}^{n *}\)
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