Algebraic set

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 $I$ generated by $S$: $V(S) = V(I)$. By Hilbert's basis theorem, $I$ 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.