Definition nb. Let \(\mathbb{F}\)F be a field and \(n\)n a positive integer. The \(n\)n-dimensional affine space over \(\mathbb{F}\)F is the set \[\mathbb{F}^n = \{(a_1, \ldots, a_n) \mid a_1, \ldots, a_n \in \mathbb{F}\}.\]
Remark nn. A polynomial \(f \in \mathbb{F}[x_1, \ldots, x_n]\)f F[x 1, , x n] can be regarded as a function \(f : \mathbb{F}^n \to \mathbb{F}\)f : F n F that takes points in the affine space \(\mathbb{F}^n\)F n and produces elements of the field \(\mathbb{F}\)F. A zero point or root of \(f\)f is a point \(\mathbf{a} = (a_1, \ldots, a_n) \in \mathbb{F}^n\)a = (a 1, , a n) \mathbbF^n such that \(f(\mathbf{a}) = 0\)f(a) = 0.
Definition nd. Let \(\{f_1, \ldots, f_s\} \subset \mathbb{F}[x_1, \ldots, x_n]\)\f 1, \ldots, f s\ \subset \mathbbF[x_1, \ldots, x_n] be a set of polynomials and \(\mathbb{F}^n\)F n an affine space. The affine variety \(V(f_1, \ldots, f_s)\)V(f 1, , f s) defined by \(\{f_1, \ldots, f_s\}\)\f 1, \ldots, f s\ is the set \[V(f_1, \ldots, f_s) = \{(a_1, \ldots, a_n) \in \mathbb{F}^n \mid f_i(a_1, \ldots, a_n) = 0 \text{ for all } 1 \le i \le s\}\] of all common zero points of the polynomials in \(\{f_1, \ldots, f_s\}\)\f 1, \ldots, f s\.
Solving an equation that can be expressed as a polynomial in multiple variables can be seen as finding the zero points of the corresponding polynomial. Affine varieties generalize this notion to systems of polynomial equations.
Example dq. Consider the real coordinate space \(\mathbb{R}^2\)R 2 and the polynomial \(f(x, y) = x^2 + y^2 - 1\)f(x, y) = x 2 + y 2 - 1. The variety \(V(f)\)V(f) is the unit circle centered at the origin.
Proposition nr. If \(\{f_1, \ldots, f_s\}\)\f 1, \ldots, f s\ and \(\{g_1, \ldots, g_t\}\)\g 1, \ldots, g t\ are two bases of the same ideal in \(\mathbb{F}[x_1, \ldots, x_n]\)F[x 1, , x n], so that \(\langle f_1, \ldots, f_s \rangle = \langle g_1, \ldots, g_t \rangle\) f 1, , f s = \langle g_1, \ldots, g_t \rangle, then \(V(f_1, \ldots, f_s) = V(g_1, \ldots, g_t)\)V(f 1, , f s) = V(g_1, , g_t).
Proof. Choose any \((a_1, \ldots, a_n) \in V(f_1, \ldots, f_s)\)(a 1, , a n) V(f_1, , f_s). All polynomials in \(\{f_1, \ldots, f_s\}\)\f 1, \ldots, f s\ are zero at \((a_1, \ldots, a_n)\)(a 1, , a n). Now choose any \(g \in \langle g_1, \ldots, g_t \rangle\)g g 1, , g t \rangle. Since \(\langle g_1, \ldots, g_t \rangle = \langle f_1, \ldots, f_s \rangle\) g 1, , g t = \langle f_1, \ldots, f_s \rangle, we can write \(g = \sum_{i=1}^s h_i f_i\)g = i=1 s h_i f_i, \(h_i \in \mathbb{F}[x_1, \ldots, x_n]\)h i F[x 1, , x_n]. Then \(g(a_1, \ldots, a_n) = \sum_{i=1}^s h_i(a_1, \ldots, a_n) \cdot f_i(a_1, \ldots, a_n) = 0\)g(a 1, , a n) = _i=1^s h_i(a_1, , a_n) \cdot f_i(a_1, \ldots, a_n) = 0, which shows that \(V(f_1, \ldots, f_s) \subseteq V(g_1, \ldots, g_t)\)V(f 1, , f s) V(g_1, , g_t). The opposite inclusion follows by the same argument.
Example dp shows that an ideal may have multiple different bases while Proposition nr reveals that a variety is determined by the ideal generated by its basis and not by the basis itself.
Definition dx. Let \(I \subseteq \mathbb{F}[x_1, \ldots, x_n]\)I F[x 1, , x n] be an ideal. We denote by \(V(I)\)V(I) the set \[V(I) = \{(a_1, \ldots, a_n) \in \mathbb{F}^n \mid f(a_1, \ldots, a_n) = 0 \text{ for all } f \in I\}.\]
Proposition do. \(V(I)\)V(I) is an affine variety. In particular, if \(I = \langle f_1, \ldots, f_m \rangle\)I = f 1, , f m , then \(V(I) = V(f_1, \ldots, f_m)\)V(I) = V(f 1, , f m) (Cox 2015).
In combination with the Hilbert Basis Theorem (Theorem n4), Proposition do shows that even though a nonzero ideal contains infinitely many polynomials, its variety can still be defined by a finite set of polynomial equations. Proposition do is a generalization of Proposition nr.
Definition nf. Let \(V \subseteq \mathbb{F}^n\)V F n be an affine variety. The ideal of \(V\)V is defined as \[I(V) = \{f \in \mathbb{F}[x_1, \ldots, x_n] \mid f(a_1, \ldots, a_n) = 0 \text{ for all } (a_1, \ldots, a_n) \in V\}.\]
Proposition ng. If \(V \subseteq \mathbb{F}^n\)V F n is an affine variety, then \(I(V) \subseteq \mathbb{F}[x_1, \ldots, x_n]\)I(V) F[x 1, , x n] is an ideal.
Proof. We have \(0 \in I(V)\)0 I(V) since the zero polynomial vanishes on all points in \(\mathbb{F}^n\)F n. Now let \(f, g \in I(V)\)f, g I(V), \(h \in \mathbb{F}[x_1, \ldots, x_n]\)h F[x 1, , x n], and let \((a_1, \ldots, a_n)\)(a 1, , a n) be an arbitrary point in \(V\)V. We get \(f(a_1, \ldots, a_n) + g(a_1, \ldots, a_n) = 0 + 0 = 0\)f(a 1, , a n) + g(a_1, , a_n) = 0 + 0 = 0 and \(h(a_1, \ldots, a_n) \cdot f(a_1, \ldots, a_n) = h(a_1, \ldots, a_n) \cdot 0 = 0\)h(a 1, , a n) f(a_1, , a_n) = h(a_1, \ldots, a_n) \cdot 0 = 0, which shows that \(I(V)\)I(V) is an ideal.