Ideal of an algebraic set

Remarks

$I(X)$ is an ideal, and it equals its own radical: if $f^m$ vanishes at a point $a$, then $f(a)^m = 0$, so $f(a) = 0$.

Example: for the parabola $X = V(y - x^2) \subseteq k^2$ with $k$ infinite, $I(X) = (y - x^2)$. Check: dividing $f(x, y)$ by $y - x^2$ as a polynomial in $y$ gives $f = (y - x^2) q + r(x)$; if $f$ vanishes on $X$, then $r(t) = 0$ for all $t \in k$, so $r = 0$.

Non-example: the ideal $(x^2)$ in $k[x]$ is not the ideal of any set. It is not radical, since $x \notin (x^2)$ while $x^2 \in (x^2)$; in fact $I(V(x^2)) = I(\set{0}) = (x)$.

For an algebraic set $X$ one has $V(I(X)) = X$, and $X \subseteq Y$ if and only if $I(Y) \subseteq I(X)$. If $k$ is algebraically closed, Hilbert's Nullstellensatz gives $I(V(J)) = \sqrt{J}$ for every ideal $J$. So $X \mapsto I(X)$ is then an inclusion-reversing bijection from algebraic sets in $k^n$ to radical ideals of $k[x_1, \dots, x_n]$.