Q:

Ideal of variety in polynomial ring

Let $k$ be a field, let $A$ be an ideal of $k[x_1,\ldots,x_n]$, and let $B$ be an ideal of elements $f\in k[x_1,\ldots,x_n]$ such that $f^n\in A$ for some $n$. I want to show that $B\subseteq I(V(A))$. This means showing that for any polynomial $f\in B$, we have $f(a_1,a_2,\ldots,a_n)=0$ for all $(a_1,\ldots,a_n)\in V(A)$. How to show this?

A:

The ideal $B$ is called radical of $A$. If $f\in B$ then $f^n\in A$ for some $n$, so $f^n(a)=0$ for all $a\in V(A)$. But $f^n(a)=(f(a))^n$, and it is $0$ exactly when $f(a)=0$, because $f(a)\in k$ which is a field.

