Question

(a) Find the irreducible components of \(V\left(Y^{2}-X Y-X^{2} Y+X^{3}\right)\) in \(A^{2}(\mathbb{R})\), and also in \(\mathrm{A}^{2}\) (C). (b) Do the same for \(V\left(Y^{2}-X\left(X^{2}-1\right)\right)\), and for \(V\left(X^{3}+X-X^{2} Y-Y\right)\).

Short answer

Question: Find the irreducible components of the following algebraic sets: 1. \(V(Y^2 - XY - X^2Y + X^3)\) in \(A^2(\mathbb{R})\) and \(A^2(\mathbb{C})\) 2. \(V\left(Y^{2}-X\left(X^{2}-1\right)\right)\) in \(A^2(\mathbb{R})\) and \(A^2(\mathbb{C})\) 3. \(V\left(X^{3}+X-X^{2}Y-Y\right)\) in \(A^2(\mathbb{R})\) and \(A^2(\mathbb{C})\) Answer: 1. \(V(Y^2 - XY - X^2Y + X^3)\): - In \(A^2(\mathbb{R})\): \(V(X^2+X)\) and \(V(Y-X)\) - In \(A^2(\mathbb{C})\): \(V(X)\), \(V(X+1)\), and \(V(Y-X)\) 2. \(V\left(Y^{2}-X\left(X^{2}-1\right)\right)\): - In \(A^2(\mathbb{R})\): \(V(Y^2)\), \(V(X)\), \(V(X-1)\), and \(V(X+1)\) - In \(A^2(\mathbb{C})\): \(V(Y^2)\), \(V(X)\), \(V(X-1)\), and \(V(X+1)\) 3. \(V\left(X^{3}+X-X^{2}Y-Y\right)\): - In \(A^2(\mathbb{R})\): \(V(X-Y)\) and \(V(X^2 + 1)\) - In \(A^2(\mathbb{C})\): \(V(X-Y)\) and \(V(X^2 + 1)\)

Step-by-step solution

Factor the polynomial \(Y^2 - XY - X^2Y + X^3 = X^2(Y-X) + X(Y-X) = (X^2+X)(Y-X)\). #Step 2: Determine the algebraic sets of factors#

We have two factors: \((X^2+X)\) and \((Y-X)\). Their algebraic sets are as follows: - \(V(X^2+X) = \{(x,y) \in \mathbb{R}^2 \ | \ x^2+x=0\} = \{(x, y) \in \mathbb{R}^2 \ | \ x(x+1)=0\}\) - \(V(Y-X) = \{(x,y) \in \mathbb{R}^2 \ | \ y-x=0\}\) #Step 3: Identify irreducible components in \(A^2(\mathbb{R})\)#

Since both factors are distinct irreducible polynomials, their algebraic sets (computed in the previous step) are irreducible components in \(A^2(\mathbb{R})\). Thus, the irreducible components are \(V(X^2+X)\) and \(V(Y-X)\). Now, determine the irreducible components in \(A^2(\mathbb{C})\). The steps are essentially the same as for the real case, but we need to check for any further factorization of polynomials over the complex numbers. #Step 3: Identify irreducible components in \(A^2(\mathbb{C})\)#

Note that the polynomial \(X^2+X\) can be further factored as \(X(X+1)\) over the complex numbers. Thus, we have three irreducible components in \(A^2(\mathbb{C})\): - \(V(X)\) - \(V(X+1)\) - \(V(Y-X)\) (b) Do the same for \(V\left(Y^{2}-X\left(X^{2}-1\right)\right)\), and for \(V\left(X^{3}+X-X^{2}Y-Y\right)\). We will follow the same steps for each polynomial equation: 1. For \(V\left(Y^{2}-X\left(X^{2}-1\right)\right)\): - Factor the polynomial: \(Y^2-(X^3-X) = Y^2 - X(X^2-1) = Y^2 - X(X-1)(X+1)\) - Identify irreducible components in \(A^2(\mathbb{R})\) and \(A^2(\mathbb{C})\): - \(V(Y^2)\) - \(V(X)\) - \(V(X-1)\) - \(V(X+1)\) 2. For \(V\left(X^{3}+X-X^{2}Y-Y\right)\): - Factor the polynomial: \(X^3+X-X^2Y-Y = X(X^2+1) - Y(X^2+1) = (X-Y)(X^2+1)\) - Identify irreducible components in \(A^2(\mathbb{R})\): - \(V(X-Y)\) - \(V(X^2 + 1)\) - Identify irreducible components in \(A^2(\mathbb{C})\): - \(V(X-Y)\) - \(V(X^2 + 1)\) (Since \(X^2 + 1\) has no further factorization over \(\mathbb{C}\))

Key concepts

Algebraic Sets

Algebraic sets play a fundamental role in algebraic geometry. They are solution sets of polynomial equations. Imagine you have a polynomial in two variables, like \(f(x, y)\). The set of all points \((x, y)\) in the plane \(A^2(\mathbb{R})\) or \(A^2(\mathbb{C})\), such that \(f(x, y) = 0\), forms an algebraic set. These sets help us understand the geometric structure of equations.

For example, given the polynomial \(Y^2 - XY - X^2Y + X^3\), after factoring, we get separate polynomial equations, such as \((X^2+X)(Y-X)\). Each of these factors leads to an algebraic set. The solutions to \(X^2+X=0\) are points where this equation becomes zero, leading to the algebraic set \(V(X^2+X)\). Similarly, \(V(Y-X)\) describes points where \(y-x=0\), essentially the line \(y=x\).

Algebraic sets are often made up of simpler sets, known as irreducible components. Understanding these components is key to studying the solutions of polynomial equations.

Polynomial Factorization

Polynomial factorization involves expressing a polynomial as a product of simpler polynomials, ideally making the structure of solutions more explicit. Understanding how to factor polynomials allows us to break them down into irreducible parts.

Consider the polynomial \(Y^2 - X Y - X^2 Y + X^3\). This can be factored into \((X^2+X)(Y-X)\). The factorization process reveals the individual shapes and lines associated with the polynomial in geometric terms. Each factor, like \(X^2+X\) and \(Y-X\), represents an algebraic set.

Factoring a polynomial makes it easier to find solutions, especially when analyzing over different fields like real or complex numbers. For instance, over the complex numbers, \(X^2 + X\) can further be split into \((X)(X+1)\), revealing more irreducible components. Factorization is not only about finding simpler forms but also about revealing deeper geometric insights.

Complex Numbers in Algebraic Geometry

Complex numbers provide a richer field for exploring algebraic geometry because they allow us to see additional structures that might not be apparent over the reals. When factoring polynomials over \(\mathbb{C}\), they can often be split into more components than when factored over \(\mathbb{R}\).

Take the example of the polynomial \(X^2+X\). Over the real numbers, this polynomial has roots \(-1\) and \(0\), resulting in factors \((X)(X+1)\). However, over the complex numbers, every real factorization remains unchanged, unlike some polynomials which gain additional components such as complex roots which can affect the irreducible components.

The complex plane in algebraic geometry lets us treat all roots symmetrically, providing a complete picture of a polynomial's behavior. This view uncovers new geometrical and topological properties that are inaccessible when limiting ourselves to the real numbers.

Real Numbers in Algebraic Geometry

In algebraic geometry, when dealing with real numbers, some polynomials present limitations. Real algebraic sets can be seen as slices of the broader complex algebraic sets, giving us a constricted view unless fully analyzed.

When factoring polynomials like \(X^2+X\), the solutions in real numbers are often fewer than their complex counterparts. Crucially, the real set obtained from this factorization is \(V(X(x+1)) = \{ (0, y), (-1, y) \} \), each representing a distinct solution component, contributing to understanding the polynomial's structure over \(\mathbb{R}\).

However, many polynomials have critical points or behavior that only fully emerge when looked at through the lens of the real number line. Understanding these can help bridge the intuitive gaps between real and complex algebraic geometry. This perspective encourages exploring transitions between fields, using real solutions as a foundational understanding of more complex structures.