Ideal of affine algebraic variety is radical In remark 1.15(c) of these notes by Gathman, it is said that for a ring $R$, an ideal of an affine variety is radical, for if $f\in \sqrt I$ then $f^k|_X=0$ and hence $f|_X=0$ meaning $f\in I$. I'm confused. If "ideal of an affine algebraic variety $X$" just means an ideal of $R$ which is a subset of $X$, then why would it be radical if $R$ is not reduced?
 A: If the affine variety $X \subset \Bbb{A}^n$ is defined by an ideal $J \subset k[x_1 , \dots , x_n] $, then the ideal of $X$ is not necessarily $J$.
The ideal of $X$ is by definition something else, namely
$$I(X)= \{ f \in k[x_1 , \dots , x_n] : f|_X = 0\}$$
which turns out to be an ideal containing the ideal $J$. The fact that $I(X)$ is a radical ideal does not mean that $J$ is a radical ideal.
EDIT: Here is the proof. Let $f^k \in I(V(J))$: we want to prove that $f \in I(V(J))$. At every point $p \in V(J)$ you have
$$f^k(p)=(f(p))^k=0$$
but $k$ is a field, in particular it is a reduced ring. Hence
$$f(p)=0$$
by arbitrarity of $p$ you have that $f \in I(V(J))$.
A: Let $X \subset A^n$ be an affine variety. Then the ideal of $X$, denoted $I_X$, is its vanishing ideal, i.e. the set of all polynomials that vanish on $X$. This ideal is radical: Suppose $f \in \sqrt{I_X}$. Suppose for the sake of contradiction that $f \not\in I_X$. Hence there exists some $x \in X$ such that $f(x) \neq 0$. On the other hand $f^\ell \in I_X$ for some $\ell$, and so $f(x)^\ell = 0$. Now $f(x) \in k$, $k$ being the underlying field. Since any field is an integral domain, induction gives $f(x)=0$, which is a contradiction. Hence $f \in I_X$.
By the context of Remark 1.15(c) in the referred notes, i suppose that Gathman means that $R$ is the polynomial ring, since $I$ is the ideal of an affine variety and such an ideal lives (up to isomorphism) in a polynomial ring.
