Question

If \(k\) is a finite field, show that every subset of \(A^{n}(k)\) is algebraic.

Short answer

Question: Show that every subset of \(A^n(k)\) is algebraic, where \(k\) is a finite field and \(A^n(k)\) is the affine space over the field. Answer: We showed that any subset \(V\) of \(A^n(k)\) is algebraic by considering an arbitrary point \(\textbf{a}\) in \(V\). We then constructed a set of polynomials in \(k[x_1,...,x_n]\) that have \(\textbf{a}\) as a common zero, and verified that these polynomials do not have any common zero in \(V\) other than the point \(\textbf{a}\). Thus, every subset of \(A^n(k)\) is algebraic.

Step-by-step solution

Consider an arbitrary subset of \(A^n(k)\)

Let \(V\) be any subset of \(A^n(k)\). Since \(A^n(k)\) has \(q^n\) points, \(V\) also has a finite number of points, say \(m\) points.

Pick an arbitrary point in \(V\)

Let \(\textbf{a} = (a_1, a_2, ..., a_n)\) be a point in \(V\). Our goal is to find a set of polynomials in \(k[x_1,...,x_n]\) that have \(\textbf{a}\) as a common zero.

Construct the set of polynomials

We can easily construct a set of polynomials that have \(\textbf{a}\) as a common zero: $$\{x_i - a_i: 1 \le i \le n\}$$ This is a set of \(n\) polynomials in \(k[x_1,...,x_n]\), and they all have \(\textbf{a}\) as a common zero, because $$ f_i(\textbf{a}) = (a_i - a_i) = 0 $$ for every \(1 \le i \le n\).

Verify that the polynomials have only one common zero in \(V\)

We need to ensure that the constructed polynomials have only one common zero in \(V\), which is the point \(\textbf{a}\). To see this, consider any point \(\textbf{a'} = (a_1', a_2', ..., a_n') \neq \textbf{a}\) in \(V\). Then there exists at least one \(j\) such that \(a_j' \neq a_j\). Since \(f_j(\textbf{a'}) = (a_j' - a_j) \neq 0\), we deduced that the constructed polynomials do not have any common zero in \(V\) other than the point \(\textbf{a}\).

Conclusion

Since we showed that for any arbitrary point \(\textbf{a} \in V\), there exists a set of polynomials in \(k[x_1,...,x_n]\) that have that point as a common zero, we conclude that every subset of \(A^n(k)\) is algebraic.

Key concepts

Algebraic Sets

When we talk about algebraic sets, we're referring to particular collections of points that satisfy a given set of polynomial equations. Imagine you have a list of conditions defined by polynomials, and you're searching for all the possible points that make these equations true. In simpler terms, if you have a list of equations, the solutions to these equations collected in a set form what we call an algebraic set.

In the context of finite fields, the situation simplifies further because you're dealing with a limited number of possible points. Every possible point in our space can potentially be part of an algebraic set if it can be described by polynomial conditions. Therefore, any subset of our space could be turned into an algebraic set by identifying the right polynomials that have zero values at all points in that subset. This is why in finite fields, every subset can be seen as an algebraic set; we can always find a polynomial, or set of polynomials, to describe any subset precisely.

• Algebraic sets are defined by zero sets of polynomials.
• In finite fields, always possible to pick polynomials for any subset.
• Finite nature of field means limited solutions, aiding in algebraic set formation.

Polynomials

Polynomials are central to understanding algebraic structures in mathematics. Essentially, a polynomial in several variables is an expression made up of sums and products of variables raised to whole-number exponents and coefficients from a field. For example, in two variables, a polynomial might look like this: \(3x^2 + 2xy - 7y^3 + x - 5\).

In our specific problem, we construct polynomials to treat each variable as a potential solution point within the finite field. This means that for each point \(\textbf{a} = (a_1, a_2, \ldots, a_n)\), we can create a set of simple linear polynomials: \(\{x_i - a_i: 1 \le i \le n\}\). All these polynomials have the point \(\textbf{a}\) as a zero, meaning when evaluated at \textbf{a}, these polynomials yield zero, aligning perfectly with the defining characteristic of algebraic sets.

• Polynomials have terms consisting of variables, exponents, and coefficients.
• Key role in creating algebraic sets for any finite field.
• Used to identify zero points in solution spaces.

Common Zero

In the realm of finite fields and algebraic geometry, the idea of a common zero is pivotal. A common zero is a point at which multiple polynomials simultaneously evaluate to zero. When dealing with systems of polynomial equations, a common zero refers to a solution that satisfies all equations at once.

In our exercise, we explore the use of polynomials to find a common zero within a set \(V\) by constructing polynomials that specifically nullify at a given point \(\textbf{a}\). Crucially, it's not just about finding one zero but ensuring that a set of polynomials only allows that specific point, \(\textbf{a}\), to satisfy all conditions. This ensures the uniqueness of the solution for our constructed set of polynomials, verifying there are no other common zeros in \(V\).

• Common zeros satisfy all polynomials in the system.
• Key for identifying unique solutions in finite field problems.
• Ensures proper algebraic structure representation in sets.