Question

Let \(F\) be a nonconstant polynomial in \(k\left[X_{1}, \ldots, X_{n}\right], k\) algebraically closed. Show that \(\mathrm{A}^{n}(k) \backslash V(F)\) is infinite if \(n \geq 1\), and \(V(F)\) is infinite if \(n \geq 2 .\) Conclude that the complement of any proper algebraic set is infinite. (Hint: See Problem 1.4.)

Short answer

Question: Show that the complement of any proper algebraic set is infinite. Answer: We proved that the complement of any proper algebraic set, denoted as \(\mathrm{A}^{n}(k) \backslash V(F)\), is infinite if \(n \geq 1\), and that \(V(F)\) is infinite if \(n \geq 2\). Thus, the complement of any proper algebraic set is indeed infinite.

Step-by-step solution

Understanding the problem

We are given a nonconstant polynomial \(F \in k\left[X_{1}, \ldots, X_{n}\right]\) where \(k\) is an algebraically closed field. We need to show that the complement of any proper algebraic set is infinite by showing that \(\mathrm{A}^{n}(k) \backslash V(F)\) is infinite if \(n \geq 1\), and \(V(F)\) is infinite if \(n \geq 2\).

Proving \(\mathrm{A}^{n}(k) \backslash V(F)\) is infinite if \(n \geq 1\)

As the polynomial \(F\) is nonconstant, it has at least one variable. So for any \(a \in k\), if we consider the 1-dimensional slice \(X_1 = a\), we have a polynomial \(G \in k[X_2, \ldots, X_n]\). Since \(k\) is algebraically closed, \(G\) has a root \((b_2, \dots, b_n)\) by Problem 1.4. So \((a, b_2, \ldots, b_n)\) is a point in the affine space \(\mathrm A^n(k)\) but not in \(V(F)\). Since this can be done for every \(a \in k\), there must be infinitely many points in \(\mathrm{A}^{n}(k) \backslash V(F)\). Hence, the claim is proved.

Proving \(V(F)\) is infinite if \(n \geq 2\)

Let \(n \geq 2\). As \(F\) is nonconstant, there exist a monomial, say \(X_1^{d_1} X_2^{d_2} \cdots X_n^{d_n}\) in \(F\) with \(d_i \geq 1\) for some \(i\). Without loss of generality, let's assume \(i=1\). Now for each \(y \in k\), the polynomial \(F_y = F(y, x_2, \dots, x_n)\) is a nonconstant polynomial in \(x_1\) (as the polynomial contains the monomial \(x_1^{d_1}\)). By Problem 1.4, \(F_y\) also has infinitely many roots in \(k\). As \(y\) can be any element in \(k\), we see that the set \(V(F)\) has infinitely many points in \(\mathrm A^n(k)\).

Concluding the result

We showed that \(\mathrm{A}^{n}(k) \backslash V(F)\) is infinite if \(n \geq 1\) and also \(V(F)\) is infinite if \(n \geq 2\). Therefore, the complement of any proper algebraic set is infinite.

Key concepts

Algebraically Closed Field

Understanding the concept of an algebraically closed field is pivotal when delving into algebraic geometry. An algebraically closed field, denoted as 'k' in the given problem, is a field in which every non-constant polynomial in one variable has at least one root. This means that there are no 'loopholes' for polynomial equations; every polynomial equation has a solution within this field.

Imagine playing a game where for every challenge there's at least one guaranteed solution—this is exactly what an algebraically closed field provides for polynomial equations. The most common example of such a field is the set of complex numbers. Being algebraically closed is crucial for the proof in the exercise since it ensures that nonconstant polynomials have roots that are used to construct infinite points not lying in the algebraic set defined by the polynomial.

Affine Space

The concept of affine space, represented as \( \mathrm{A}^{n}(k) \) in the exercise, can be viewed much like our intuitive understanding of geometric space. In algebraic geometry, it's an n-dimensional space over an algebraically closed field, 'k'. Like a stage set for a play, affine space is where all the action takes place—polynomials graph out their zeros, and shapes called algebraic sets emerge.

To visualize, picture a coordinate grid, but instead of just flat paper, extend this into higher dimensions where each point is determined by n-coordinates. It's a setting void of origin or axes, focusing purely on the relationships between points. In the context of the problem, understanding affine space is a foundation to figuring out where the polynomial equation does or does not have solutions.

Algebraic Set

An algebraic set is a collection of points in an affine space that are the solutions to a given set of polynomial equations. Remember, within the realm of algebraic geometry, we observe the universe through the lens of polynomials. An equation like \( F = 0 \) carves out a shape in affine space, marking the boundary of where the equation holds true. This shape is our algebraic set, denoted as \( V(F) \) in the exercise, resembling a 'no trespassing' sign for all the possible solutions of the polynomial.

Picture an algebraic set as a sort of 'forbidden city' where all the points inside follow the strict rules laid out by the polynomial. The beauty lies in its diversity—algebraic sets can be points, curves, surfaces, or more complex higher-dimensional objects. The exercise illuminates the abundant nature of these sets, proving that under certain conditions, they have infinitely many points.

Polynomials in Algebraic Geometry

In the realm of polynomials in algebraic geometry, we wield polynomials as tools to sculpt shapes in the affine space. These polynomials are not just mere equations; they're like a script that describes a geometric narrative in 'k', the algebraically closed field. With each polynomial, we can describe an algebraic set—these are the tangible manifestations of the polynomial's zeros in the multidimensional space.

Furthermore, the behavior of these polynomials determines the complexity and properties of the algebraic sets they define. Taking a nonconstant polynomial in the exercise, it introduces at least one variable allowing for infinite nuances and an infinite set of solutions in affine space. These solutions are the very backbone of algebraic geometry, revealing an intricate dance between algebraic expressions and their geometric interpretations.