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

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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