Property of the Zariski Topology $V(J) \subset V(I)$

Question

I am just starting to learn about the Zariski topology, and there is one thing that keeps coming up in notes (I have been looking through many online) without proof. It is stated as a "remark", but I cannot see why the property follows so logically. Maybe someone could explain it to me?

It is this:

Given two ideals $I, J$, such that $I \subset J$, then $V(J) \subset V(I)$.

(Where we define $V(I) = \big\{p \in SpecA : I \subset p\big\}$)

I must be missing something very obvious, as I have only been able to find this property given as a remark. It would be helpful and useful to see a documented proof, regardless how trivial it may actually be!

Answer 1

Suppose $I\subseteq J$ and $p\in V(J)$. By definition of $V(J)$, this means $J\subseteq p$. We now have $I\subseteq J$ and $J\subseteq p$, and so $I\subseteq p$. By definition, this means $p\in V(I)$.

Answer 2

Hint: $P\in V(J)$ implies $I\subset J\subset P$.