Affine Algebraic Sets

Recall from the Affine n-Space over a Field page that if $K$ is a field then we denote the affine n-space over $K$ to be the set:

(1)

\begin{align} \quad \mathbb{A}^n(K) = \underbrace{K \times K \times ... \times K}_{n \: \mathrm{times}} = K^n \end{align}

Furthermore, $K[x_1, x_2, ..., x_n]$ denotes the ring of polynomials in the variables $x_1, x_2, ..., x_n$ and if $F \in K[x_1, x_2, ..., x_n]$ we say that a point $\mathbf{p} = (p_1, p_2, ..., p_n) \in \mathbb{A}^n(K)$ is a zero or root of $F$ if $F(\mathbf{p}) = 0$.

We are now ready to define what it means for a set to be an affine algebraic set.

Definition: Let $K$ be a field and let $S \subseteq K[x_1, x_2, ..., x_n]$. The Zero Locus of $S$ or Vanishing Set of $S$ is the set $V(S) = \{ \mathbf{p} \in \mathbb{A}^n(K) : F(\mathbf{p}) = 0, \: \forall F \in S \}$, that is, $V(S)$ is the set of all points in the affine $n$-space $\mathbb{A}^n(K)$ such that every function $f \in S$ vanishes at $\mathbf{p}$. A set $X \subseteq \mathbb{A}^n(K)$ is said to be an Affine Algebraic Set if there exists an $S \subseteq K[x_1, x_2, ..., x_n]$ such that $X = V(S)$.

If $S$ is a finite set of functions in $K[x_1, x_2, ..., x_n]$, say $S = \{ F_1, F_2, ..., F_k \}$, then we can write "$V(F_1, F_2, ..., F_k)$" instead of "$V(\{ F_1, F_2, ..., F_k \})$".

For example, let $K = \mathbb{R}$ and consider the affine $2$-space $\mathbb{A}^2(\mathbb{R})$. Consider the polynomial $F \in \mathbb{R}[x, y]$ given by:

(2)

\begin{align} \quad F(x, y) = x^2 - y \end{align}

Then $V(F)$ is simply the set of points $(x, y) \in \mathbb{A}^2(\mathbb{R})$ such that $F(x, y) = 0$, i.e., $0 = x^2 - y$:

(3)

\begin{align} \quad V(F) = V(x^2 - y) = \{ (x, y) \in \mathbb{A}^2(\mathbb{R}) : y = x^2 \} \end{align}

We can graph this affine algebraic set in the Cartesian plane:

For another example, consider the polynomials $F, G \in \mathbb{R}[x, y]$ where $F$ is given as above and $G$ is given by:

(4)

\begin{align} \quad G(x, y) = y^2 - x \end{align}

We graph $F$ and $G$ simultaneously in the Cartesian plane:

So the set of points $(x, y) \in \mathbb{A}^2(\mathbb{R})$ such that $F(x, y) = 0$ and $G(x, y) = 0$ is precisely:

(5)

\begin{align} \quad V(F, G) = V(x^2 - y, y^2 - x) = \{ (x, y) \in \mathbb{A}^2(\mathbb{R}) : y = x^2 \: \mathrm{and} \: y^2 = x \} = \{ (0, 0), (1, 1) \} \end{align}

Therefore $\{ (0, 0), (1, 1) \}$ is an affine algebraic set in the affine $2$-space $\mathbb{A}^2(\mathbb{R})$.