---
tag: 1XYZNC2E6G
kind: definition
name: Algebraic set
msc: [14A10, 14A05]
defines: VHFMQQYSTN
uses: [04C5ZBG1WT, 0SDC934R52, 096Z9MP2ME, 0YKA9J3XEE, 8SQK7RTB63]
sources:
  - url: https://en.wikipedia.org/wiki/Algebraic_variety
---

## Definition

Let $k$ be a [[field|04C5ZBG1WT]], let $n \ge 0$ be an integer, and let $\mathbb{A}^n = k^n$ be the affine $n$-space over $k$.
Let $k[x_1, \dots, x_n]$ be the [[polynomial ring|0SDC934R52]] in $n$ variables over $k$.
For a subset $S \subseteq k[x_1, \dots, x_n]$, its zero set is
$$V(S) = \set{a \in k^n : f(a) = 0 \text{ for all } f \in S}.$$
A subset $X \subseteq k^n$ is an *algebraic set* (or affine algebraic set) if $X = V(S)$ for some subset $S \subseteq k[x_1, \dots, x_n]$.

## Remarks

Example: in $\R^2$, the unit circle is $V(x^2 + y^2 - 1)$, so it is an algebraic set.
Each point $a = (a_1, \dots, a_n)$ is algebraic, since $\set{a} = V(x_1 - a_1, \dots, x_n - a_n)$.

Non-example: the half-line $H = \set{(t, 0) : t \ge 0}$ in $\R^2$ is not an algebraic set.
If a polynomial $f$ vanishes on $H$, then $t \mapsto f(t, 0)$ is a polynomial in one variable with infinitely many roots, so it is zero.
Then $f$ also vanishes at $(-1, 0)$, so every $V(S)$ that contains $H$ contains the whole $x$-axis.

The zero set only depends on the [[ideal|096Z9MP2ME]] $I$ generated by $S$: $V(S) = V(I)$.
By [[Hilbert's basis theorem|0YKA9J3XEE]], $I$ is finitely generated, so every algebraic set is the zero set of finitely many polynomials.
The rules $V(I) \cup V(J) = V(IJ)$ and $\bigcap_i V(I_i) = V\left(\sum_i I_i\right)$ show that the algebraic sets are the closed sets of a topology on $k^n$.
This topology is the classical form of the [[Zariski topology|8SQK7RTB63]].