I have the following problem:
Let $R$ be a ring and $a\subset R$. $V_R(a):=\{p\in Spec \ R \ | \ a\subset p\}\subset Spec \ R$
Show: Let $(a_i)_{i\in I}$ be a family of subsets of $R$ then the following is true $$ \bigcap_{i\in I}V_R(a_i)=V_R(\bigcup_{i\in I}a_i)$$
I see why this equation has to be true, but in our script it is not proven, because it should be obvious. My problem is that I don't know how to proof it. Is it really that simple to show?
Working with the Zariski topology can be tricky at first. In this case, the result follows directly from the definitions: $$ \begin{align*} \bigcap_{i \in I} V_R(a_i) &= \bigcap_{i \in I}\{p \in \text{Spec}R; a_i \subseteq p\}\\ &= \{p \in \text{Spec}R; a_i \subseteq p \ \forall i \in I\}\\ &= \{p\in \text{Spec} R; \cup_{i\in I}a_i \subseteq p\} = V_R\left(\bigcup_{i \in I}a_i\right). \end{align*} $$
It follows from this that sets of the form $V_R(a)$ are closed under arbitrary intersection. Together with the facts that these sets are closed under finite union (exercise), $V_R(\emptyset) = \text{Spec}R$ and $V_R(R) = \emptyset,$ this shows that sets of the form $V(a)$ may be considered as the closed sets in a topology on $\text{Spec}A.$ This topology is called the Zariski topology.
A friend recently shared this with me, and I found it to be a good heuristic motivation for the definition of the Zariski topology: https://blogs.ams.org/mathgradblog/2017/07/16/idea-scheme/.