Question

Show that each of the following sets is not algebraic: (a) \(\left\\{(x, y) \in \mathrm{A}^{2}(\mathbb{R}) \mid y=\sin (x)\right\\}\). (b) \(\left\\{\left.(z, w) \in \mathbb{A}^{2}(\mathrm{C})|| z\right|^{2}+|w|^{2}=1\right\\}\), where \(|x+i y|^{2}=x^{2}+y^{2}\) for \(x, y \in \mathbb{R}\). (c) \(\left\\{(\cos (t), \sin (t), t) \in \mathbb{A}^{3}(\mathbb{R}) \mid t \in \mathbb{R}\right\\} .\)

Short answer

To conclude, we have shown that the three given sets are not algebraic by demonstrating that there is no polynomial equation that satisfies their elements and only their elements. The first set involved the sine function, which led to an infinite number of zeroes and the contradiction with a polynomial function's properties. The second set in complex coordinates represented the surface of a four-dimensional unit sphere, which is not smooth and contradicts the properties of complex polynomials. The third set involved the unit circle, which is known not to be algebraic, and the contradiction in the projection of their polynomial equation.

Step-by-step solution

Set (a)

For the first set, \((x, y) \in \mathrm{A}^{2}(\mathbb{R}) \mid y=\sin (x)\), we need to show that there is no polynomial equation that is satisfied by its elements and only by them. Suppose that there exists a polynomial equation \(P(x, y) = 0\), such that \((x, y) \in \mathrm{A}^{2}(\mathbb{R}) \mid y=\sin (x)\) if and only if \(P(x, y)=0\). Consider the function \(f(x) = P(x, \sin x)\). Since \(P(x, y)\) is a polynomial, \(f(x)\) is a composition of a polynomial function and a continuous function, so \(f(x)\) is continuous in \(\mathbb{R}\). However, we know that \(\sin x\) is periodic with period \(2\pi\). So, for any solution \((x, y)\) of \(P(x, y)=0\), \((x+2n\pi, y)\) is also a solution, where \(n \in \mathbb{Z}\). This means that the function \(f(x)\) has infinitely many zeroes, which contradicts the fact that polynomial functions have a finite number of zeroes. Therefore, such a polynomial equation does not exist, and the set is not algebraic.

Set (b)

For the second set, \(\left\\{(z, w) \in \mathbb{A}^{2}(\mathrm{C})||z|^2+|w|^2=1\right\\}\), we need to show that there is no polynomial equation with complex coefficients that is satisfied by its elements and only by them. Let us consider the set in polar coordinates, \((r_1e^{i\theta_1},r_2e^{i\theta_2})=(p_1+ip_2,p_3+ip_4)\). Then we have the equation \(r_{1}^{2}+r_{2}^{2}=1\) with \(r_1=\sqrt{p_1^2+p_2^2}\) and \(r_2=\sqrt{p_3^2+p_4^2}\) for each point in the set. Now assume there exists a polynomial equation \(Q(z, w) = 0\) in complex variables, such that \((z, w) \in \mathbb{A}^{2}(\mathbb{C}) \mid |z|^2+|w|^2=1\) if and only if \(Q(z, w) = 0\). Complex polynomials are smooth functions, with derivatives of all orders existing. The equation \(Q(z, w) = 0\) represents a smooth surface in four-dimensional space with variables \((p_1, p_2, p_3, p_4)\) (where \(z=p_1 + ip_2\) and \(w = p_3 + ip_4\)). However, the equation \(r_{1}^2+r_{2}^2=1\) represents the surface of a four-dimensional unit sphere (since each \((r_1, r_2)\) pair represents a point in a 2D plane). The surface of this set is not smooth, as it has a border (e.g., think of a three-dimensional sphere, which has a border in two-dimensional space). This contradiction implies that such a polynomial equation does not exist and the set is not algebraic.

Set (c)

For the third set, \(\left\\{(\cos (t), \sin (t), t) \in \mathbb{A}^{3}(\mathbb{R}) \mid t \in \mathbb{R}\right\\}\), we need to show that there is no polynomial equation with real coefficients that is satisfied by its elements and only by them. Assume there exists a polynomial equation \(R(x, y, z) = 0\) in real variables, such that \((\cos (t), \sin (t), t) \in \mathbb{A}^{3}(\mathbb{R})\) if and only if \(R(x, y, z) = 0\). Let us consider the projection of the curve defined by the equation \(R(x, y, z) = 0\) onto the \(xy\)-plane. This projection is given by the equation \(R(x, y, 0) = 0\). This equation should be satisfied by all points \((\cos(t), \sin(t)) \in \mathbb{R}^2\) for \(t \in \mathbb{R}\). However, we know that the points \((\cos(t), \sin(t))\) form a unit circle in the plane, and we know that the unit circle is not an algebraic set, as we showed in Step 1. This contradiction implies the set described in (c) is also not algebraic.

Key concepts

Non-algebraic sets

Algebraic geometry deals with sets defined by polynomial equations. When a set cannot be described this way, it is termed a non-algebraic set. Understanding what constitutes a non-algebraic set is crucial for students studying algebraic geometry as it differentiates algebraic and non-algebraic objects.
For example, let's look at the set where \(y = \sin(x)\). In algebraic terms, a set is algebraic if there is a polynomial equation that its points satisfy. However, due to the periodic nature of functions like sine and cosine, the conditions for forming polynomial equations are not met.
Non-algebraic sets often involve transcendental functions like trigonometric or exponential functions, which have properties such as periodicity that are fundamentally incompatible with polynomial behavior.

Polynomial equations

Polynomial equations form the backbone of algebraic geometry. A polynomial is a mathematical expression involving a sum of powers in one or more variables, like \(P(x, y) = ax^n + bx^{n-1}y + \ldots + k\). Algebraic sets are defined as those that can satisfy polynomial equations.
However, some sets such as the unit circle, defined by a trigonometric parameterization like \((\cos(t), \sin(t))\), cannot be expressed solely using polynomial functions, especially when complex dependencies or periodicity are introduced.
Importantly, polynomial equations are finite, smooth, and do not accommodate the concept of infinity well, which limits their capacity to represent sets generated by periodic functions fully.

Periodic functions

Periodic functions, such as sine and cosine, repeat values at regular intervals. This uniqueness makes periodic functions incompatible with being expressed solely through polynomial equations. Polynomials can have only a finite number of roots, meaning it's nearly impossible for a polynomial to equate periodic solutions repeating an infinite number of times.
For example, the sine function repeats every \(2\pi\), so the set \(\{(x, \sin(x))\}\) includes points that keep looping on the same values as \(x\) increases. This inherent infinite repetition prevents a polynomial from being an exact representation of a periodic function.
Understanding this behavior is important in algebraic geometry as it highlights why and how periodic functions lead to non-algebraic sets.

Complex variables

In algebraic geometry, complex variables are used when dealing with equations involving complex numbers. A complex number includes a real and an imaginary part, expressed as \(z = a + bi\), where \(i\) is the square root of \(-1\).
Sets that include conditions such as \(|z|^2 + |w|^2 = 1\) are non-algebraic despite seeming to have a simple geometric representation — the unit sphere in four-dimensional space defined by complex numbers. The smooth nature of polynomials over complex domains limits their ability to describe more intricate boundaries such as those seen in these equations.
This understanding is critical because it illustrates the relationship and constraints between complex variables and algebraic sets, further highlighting limitations within polynomial-defined frameworks.