For a given field $ {\mathbb{F}}$ and positive integer $ n$, the $ n$-dimensional affine space over $ {\mathbb{F}}$ is the set

$\displaystyle {\mathbb{F}}^n = \{(c_1,\ldots,c_n) \;\vert\; c_1, \ldots, c_n \in {\mathbb{F}}\} .$ (4.51)

For our purposes in this section, an affine space can be considered as a vector space (for an exact definition, see [438]). Thus, $ {\mathbb{F}}^n$ is like a vector version of the scalar field $ {\mathbb{F}}$. Familiar examples of this are $ {\mathbb{Q}}^n$, $ {\mathbb{R}}^n$, and $ {\mathbb{C}}^n$.

A polynomial in $ f \in {\mathbb{F}}[x_1,\ldots,x_n]$ can be converted into a function,

$\displaystyle f : {\mathbb{F}}^n \rightarrow {\mathbb{F}},$ (4.52)

by substituting elements of $ {\mathbb{F}}$ for each variable and evaluating the expression using the field operations. This can be written as $ f(a_1,\ldots,a_n) \in {\mathbb{F}}$, in which each $ a_i$ denotes an element of $ {\mathbb{F}}$ that is substituted for the variable $ x_i$.

We now arrive at an interesting question. For a given $ f$, what are the elements of $ {\mathbb{F}}^n$ such that $ f(a_1,\ldots,a_n) = 0$? We could also ask the question for some nonzero element, but notice that this is not necessary because the polynomial may be redefined to formulate the question using 0. For example, what are the elements of $ {\mathbb{R}}^2$ such that $ x^2 + y^2 = 1$? This familiar equation for $ {\mathbb{S}}^1$ can be reformulated to yield: What are the elements of $ {\mathbb{R}}^2$ such that $ x^2
+ y^2 - 1 = 0$?

Let $ {\mathbb{F}}$ be a field and let $ \{f_1,\ldots,f_k\}$ be a set of polynomials in $ {\mathbb{F}}[x_1,\ldots, x_n]$. The set

$\displaystyle V(f_1,\ldots,f_k) = \{ (a_1,\ldots,a_n) \in {\mathbb{F}}\;\vert\; f_i(a_1,\ldots,a_n) = 0$    for all $\displaystyle 1 \leq i \leq k \}$ (4.53)

is called the (affine) variety defined by $ f_1,\ldots,f_k$. One interesting fact is that unions and intersections of varieties are varieties. Therefore, they behave like the semi-algebraic sets from Section 3.1.2, but for varieties only equality constraints are allowed. Consider the varieties $ V(f_1,\ldots,f_k)$ and $ V(g_1,\ldots,g_l)$. Their intersection is given by

$\displaystyle V(f_1,\ldots,f_k) \cap V(g_1,\ldots,g_l) = V(f_1,\ldots,f_k,g_1,\ldots,g_l) ,$ (4.54)

because each element of $ {\mathbb{F}}^n$ must produce a 0 value for each of the polynomials in $ \{f_1,\ldots,f_k, g_1,\ldots,g_l\}$.

To obtain unions, the polynomials simply need to be multiplied. For example, consider the varieties $ V_1, V_2 \subset {\mathbb{F}}$ defined as

$\displaystyle V_1 = \{ (a_1,\ldots,a_n) \in {\mathbb{F}}\;\vert\; f_1(a_1,\ldots,a_n) = 0 \}$ (4.55)


$\displaystyle V_2 = \{ (a_1,\ldots,a_n) \in {\mathbb{F}}\;\vert\; f_2(a_1,\ldots,a_n) = 0 \} .$ (4.56)

The set $ V_1 \cup V_2 \subset {\mathbb{F}}$ is obtained by forming the polynomial $ f = f_1 f_2$. Note that $ f(a_1,\ldots,a_n) = 0$ if either $ f_1(a_1,\ldots,a_n) = 0$ or $ f_2(a_1,\ldots,a_n) = 0$. Therefore, $ V_1 \cup V_2$ is a variety. The varieties $ V_1$ and $ V_2$ were defined using a single polynomial, but the same idea applies to any variety. All pairs of the form $ f_i g_j$ must appear in the argument of $ V(\cdot)$ if there are multiple polynomials.

Steven M LaValle 2012-04-20