Varieties

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

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

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

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

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

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

and

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