Question

(a) Decompose \(V\left(X^{2}+Y^{2}-1, X^{2}-Z^{2}-1\right) \subset A^{3}(\mathbb{C})\) into irreducible components. (b) Let \(V=\left\\{\left(t, t^{2}, t^{3}\right) \in \mathbb{A}^{3}(\mathbb{C}) \mid t \in \mathbb{C}\right\\} .\) Find \(I(V)\), and show that \(V\) is irreducible.

Short answer

In summary, for the variety \(V\left(X^{2}+Y^{2}-1, X^{2}-Z^{2}-1\right) \subset A^{3}(\mathbb{C})\), we found four irreducible components: 1. The variety formed by \(X = 0\), \(Y = 1\), and \(Z = 0\) 2. The variety formed by \(X = 0\), \(Y = -1\), and \(Z = 0\) 3. The variety formed by \(X^{2}+Y^{2}-1 = 0\) and \(Z = Y\) 4. The variety formed by \(X^{2}+Y^{2}-1 = 0\) and \(Z = -Y\) For the variety \(V=\left\\{\left(t, t^{2}, t^{3}\right) \in \mathbb{A}^{3}(\mathbb{C}) \mid t \in \mathbb{C}\right\\}\), we found its ideal \(I(V) = \left\langle Y - X^2, Z - X^3 \right\rangle\), and we showed that the variety is irreducible because \(I(V)\) is a prime ideal.

Step-by-step solution

Part (a) - Decomposing the variety V into irreducible components

To find the irreducible components of the variety \(V\left(X^{2}+Y^{2}-1, X^{2}-Z^{2}-1\right)\), we will start by examining the given polynomial equations and then finding the common solutions to those equations. The variety is given by the zero set of the following system of polynomial equations: 1. \(X^{2}+Y^{2}-1 = 0\) 2. \(X^{2}-Z^{2}-1 = 0\) We will now consider the following cases for the variables \(X\), \(Y\), and \(Z\).

Case 1: X = 0

If \(X = 0\), we have the equation \(Y^{2} = 1\), which gives us two solutions: \(Y = 1\) and \(Y = -1\). Since \(X^{2} = Z^{2}\), we can conclude that either \(Z = 0\) or \(Z = 0\) in this case. Considering the variety formed by \(X = 0\), \(Y = 1\), and \(Z = 0\), it is an irreducible variety. Similarly, considering the variety formed by \(X = 0\), \(Y = -1\), and \(Z = 0\), it is another irreducible variety.

Case 2: X ≠ 0

Now, if \(X ≠ 0\), we can rewrite the second equation as \(Z^{2} = X^{2} - 1\). From the first equation, we have \(Y^{2}=1-X^2\). So we can rewrite the second equation using the first equation as \(Z^{2}=Y^{2}\). Therefore, \(Z=Y\) or \(Z=-Y\). Considering the variety formed by the equations \(X^{2}+Y^{2}-1 = 0\) and \(Z-Y=0\), it is an irreducible variety. Similarly, considering the variety formed by the equations \(X^{2}+Y^{2}-1 = 0\) and \(Z+Y=0\), it is another irreducible variety. Finally, we can conclude that the variety \(V\left(X^{2}+Y^{2}-1, X^{2}-Z^{2}-1\right) \subset A^{3}(\mathbb{C})\) has four irreducible components: 1. The variety formed by \(X = 0\), \(Y = 1\), and \(Z = 0\) 2. The variety formed by \(X = 0\), \(Y = -1\), and \(Z = 0\) 3. The variety formed by \(X^{2}+Y^{2}-1 = 0\) and \(Z = Y\) 4. The variety formed by \(X^{2}+Y^{2}-1 = 0\) and \(Z = -Y\)

Part (b) - Finding I(V) and showing that V is irreducible

The given variety is defined as \(V=\left\\{\left(t, t^{2}, t^{3}\right) \in \mathbb{A}^{3}(\mathbb{C}) \mid t \in \mathbb{C}\right\\}\). We need to find its ideal \(I(V)\) and show that the variety is irreducible. First, we find the parametrized representation of this variety. Let \(t=X\): 1. \(Y = t^2 = X^2\) 2. \(Z = t^3 = X^3\) Now, we will find the polynomials which vanish on this variety by eliminating the parameter \(t=X\). We begin by taking the difference between the first two equations: 1. \(Y - X^2 = 0\) Now we take the difference between the second and third equations: 2. \(Z - X^3 = 0\) So the ideal \(I(V)\) is given by: \(I(V) = \left\langle Y - X^2, Z - X^3 \right\rangle\) To show that the variety is irreducible, we need to show that \(I(V)\) is a prime ideal, i.e., if \(fg \in I(V)\), then either \(f \in I(V)\) or \(g \in I(V)\). Suppose \(fg \in \left\langle Y - X^2, Z - X^3 \right\rangle\). Then, \(fg = h_1(Y - X^2) + h_2(Z - X^3)\) for some polynomials \(h_1\) and \(h_2\). Substituting the parametrized representation \((t, t^2, t^3)\) for \((X, Y, Z)\), we have \(f(t)g(t) = h_1(t)(t^2 - t^2) + h_2(t)(t^3 - t^3) = 0\) for all \(t \in \mathbb{C}\). Since \(f(t)g(t) = 0\) for all \(t \in \mathbb{C}\), either \(f(t) = 0\) or \(g(t) = 0\) for all \(t \in \mathbb{C}\). Thus, either \(f \in I(V)\) or \(g \in I(V)\). This means that \(I(V)\) is a prime ideal, and therefore, the variety \(V\) is irreducible.

Key concepts

Algebraic Geometry

Algebraic geometry is a branch of mathematics that studies geometric structures through algebraic expressions like polynomial equations. It blends concepts from both algebra and geometry to explore entities known as algebraic varieties. These varieties are essentially the solution sets of systems of polynomial equations.

In simpler terms, you can imagine graphs of equations like circles, lines, and parabolas, but in higher dimensions. Instead of just considering real numbers, algebraic geometry considers solutions in different fields, such as the complex numbers (noted as \(\mathbb{C}\)). This allows for a richer set of tools and insights.

• This field helps to understand the shapes defined by polynomial equations.
• Uses the language of rings, fields, and ideals from abstract algebra.

By translating geometric questions into algebraic ones, algebraic geometry allows mathematicians to solve complex problems using algebraic techniques. This exercise, for example, looks at finding the irreducible components of a variety, which means identifying the most basic building blocks of a geometric object.

Variety Decomposition

Variety decomposition is a technique used in algebraic geometry to break down an algebraic variety into simpler, non-decomposable pieces called irreducible components. Think of an algebraic variety as a puzzle, and variety decomposition is the process of identifying and extracting the individual puzzle pieces that can't be broken down further.

In our example, a variety is defined by a set of polynomial equations with complex solutions forming a geometric figure in \(\mathbb{A}^3(\mathbb{C})\). To find its irreducible components, we look at cases where certain variables set other equations to a simpler form - essentially keeping the most basic configurations intact.

• Irreducible components are key because they represent the fundamental parts of the variety.
• The decomposition helps in understanding the structure and properties of varieties.

Knowing the irreducible components of a variety is vital as they help in studying and predicting the behavior of the variety under various mathematical operations and transformations.

Prime Ideal

A prime ideal is an important concept in algebraic geometry and abstract algebra, serving as a bridge between the algebraic and geometric perspectives. In the context of polynomials, an ideal is a set of polynomials that "behave together." A prime ideal has a special property where if the product of two polynomials belongs to the ideal, then at least one of those polynomials is in the ideal as well.

For example, in the exercise, we found that the ideal \(I(V)\) is generated by the polynomials \(Y - X^2\) and \(Z - X^3\). This ideal is prime because, given any product of polynomials within it, at least one factor must also be in the ideal.

• Prime ideals are integral to defining irreducible varieties.
• They ensure that the variety represented by the ideal cannot be broken down into simpler pieces.

Understanding prime ideals helps in understanding the structure and "indivisibility" of the algebraic variety they define.

Polynomial Equations

Polynomial equations are crucial in algebraic geometry as they form the cornerstone for defining algebraic varieties. A polynomial is an expression involving variables raised to various powers, like \(X^2 + Y^2 - 1\). These equations describe curves, surfaces, and other geometric objects in various dimensions.

In the context of the exercise, polynomial equations are used to define the variety which we decomposed into irreducible components. For example, the equations \(X^2 + Y^2 - 1 = 0\) and \(X^2 - Z^2 - 1 = 0\) lead to a complex geometric structure in three-dimensional space which we break down using algebraic techniques.

• They are foundational in representing and solving geometric problems.
• Equations can describe both simple and complex geometries with infinite solutions.

By solving polynomial equations, either individually or simultaneously, we model geometric objects and explore their properties, which is a central pursuit in algebraic geometry.