I'm having trouble recalling one direction of the following bi-implication.
Suppose $S,T$ are subsets of the polynomial ring $k[X_1,\dots,X_n]$ over an algebraically closed field. We have $V(S)\subseteq V(T)\iff T\subseteq\sqrt{\langle S\rangle}$.
For one direction, I suppose $T\subseteq\sqrt{\langle S\rangle}$. If $f\in T$, then $f^m\in\langle S\rangle$ for some $m$. Let $(a_i)$ be an $n$-tuple in $V(S)=V(\langle S\rangle)$. Then $f^m((a_i))=(f((a_i)))^m=0$, hence $f((a_i))=0$. Then we get $(a_i)\in V(T)$, so $V(S)\subseteq V(T)$.
What's the way to get the other implication? Thanks.
The implication $\implies$ is false for a non-algebraically closed field, like $\mathbb R$: $\emptyset =V(X^2+1)\subset \emptyset =V(X^2+2)$ but $X^2+2\not\subset \sqrt {(X^2+1)}=(X^2+1)$.
This suggests using the Nullstellensatz, which indeed immediately proves $\implies$ .