Let $R=k[x_1,...,x_n]$. I want to prove that $$\sqrt I\subset \sqrt J\implies Z(J)\subset Z(I),$$ where $I,J$ are ideals of $R$ and $Z(I)$ and $Z(J)$ varieties. So let $(c_1,...,c_n)\in Z(J)$. Then, $f(c_1,...,c_n)=0$ for all $f\in J$. Let $g\in I$. Then, $g\in \sqrt I$, and thus $g\in \sqrt J$. Therefore, there is $m$ s.t. $g^m\in J$. Then, $g^m(c_1,...,c_n)=0$ and thus $g(c_1,...,c_n)=0$.
Question : I just don't understand the last implication, i.e. "$g^m(c_1,...,c_n)=0$ and thus $g(c_1,...,c_n)=0$". Why do we have this result ?
It is simply that $g^m(c_1,\ldots, c_n)=g(c_1,\ldots, c_n)^m$ by definition, and since $k$ is a field, any element $x$ of $k$ such that $x^m=0$ must be such that $x=0$.