Inclusion containing V and radicals Let $A$ be a commutative ring with unit and $I,J$ be two ideals of $A$. Also, denote $V(I):=\{\mathfrak{p}\in\operatorname{Spec}A\mid I\subset\mathfrak{p}\}$. Why is it true that if $J\subseteq \sqrt{I}$ then $V(I)\subseteq V(J)$? I guess it has something to do with the fact that the radical $\sqrt{I}$ is the intersection of the prime ideals $\mathfrak{p}\in V(I)$.
 A: Suppose $J \subseteq \sqrt I$. We want to show $V(I) \subseteq V(J)$.
If you have the following result at your disposal: $\sqrt I = \cap \{\mathfrak p$ prime $ : I \subseteq \mathfrak p\}$, then your reasoning works:
Let $\mathfrak p \in V(I)$, then $I \subseteq \mathfrak p$. But since $J \subseteq \sqrt I = \cap \{\mathfrak p $ prime $: I \subseteq \mathfrak p\}$, it follows that $J \subseteq \mathfrak p$ and thus $\mathfrak p \in V(J)$. So your reasoning was correct.
If you don't have this result, Olivier Bégassat's reasoning works nicely:
Observe that $V(I) = V(\sqrt I)$ since $I \subseteq \mathfrak p$ if and only if $\sqrt I \subseteq \mathfrak p$:


*

*If $\sqrt I \subseteq \mathfrak p$, and we know $I \subseteq \sqrt I$, then $I \subseteq \mathfrak p$.

*If $I \subseteq \mathfrak p$, let $x \in \sqrt I$, then there exists $n > 0$ such that $x^n \in I \subseteq \mathfrak p$ which means $x \in \mathfrak p$ since $\mathfrak p$ is prime. Hence $\sqrt I \subseteq \mathfrak p$.


If $J \subseteq \sqrt I$, then $V(I) = V(\sqrt I) \subseteq V(J)$ which was what you wanted.
