Exercise 111.15.1. Prove that an algebraic set can always be written as the zero locus of finitely many polynomials.
111.15 Constructible sets
Let k be an algebraically closed field, for example the field \mathbf{C} of complex numbers. Let n \geq 0. A polynomial f \in k[x_1, \ldots , x_ n] gives a function f : k^ n \to k by evaluation. A subset Z \subset k^ n is called an algebraic set if it is the common vanishing set of a collection of polynomials.
With notation as above a subset E \subset k^ n is called constructible if it is a finite union of sets of the form Z \cap \{ f \not= 0\} where f is a polynomial.
Exercise 111.15.2. Show the following
the complement of a constructible set is a constructible set,
a finite union of constructible sets is a constructible set,
a finite intersection of constructible sets is a constructible set, and
any constructible set E can be written as a finite disjoint union E = \coprod E_ i with each E_ i of the form Z \cap \{ f \not= 0\} where Z is an algebraic set and f is a polynomial.
Exercise 111.15.3. Let R be a ring. Let f = a_ d x^ d + a_{d - 1} x^{d - 1} + \ldots + a_0 \in R[x]. (As usual this notation means a_0, \ldots , a_ d \in R.) Let g \in R[x]. Prove that we can find N \geq 0 and r, q \in R[x] such that
a_ d^ N g = q f + r
with \deg (r) < d, i.e., for some c_ i \in R we have r = c_0 + c_1 x + \ldots + c_{d - 1}x^{d - 1}.
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