Is the closure of $\{p\}$ equal to $V(p)$ in Spec $A$? Let $A$ be a commutative ring with unity and consider Spec$A$. Does the closure of $\{p\}$, $p \in$ Spec $A$, equal to $V(p)$ which is defined to be the set of all primes ideals containing $p$? Thank you. 
 A: That's right: $\mathfrak q\in\overline{\{\,\mathfrak p\,\}}$ means that  any elementary open set $D(f)$ for the Zariski topology which contains $\mathfrak q$ contains $\mathfrak p$ – in other words, if an element $f\notin \mathfrak q$, then  $f\notin \mathfrak p$.
Indeed, suppose  $\mathfrak p\not\subseteq \mathfrak q$. This means there exists an element $a\in\mathfrak p$ which does not belong to $\mathfrak q$. Then $D_a$ in an open neighbourhood of $\mathfrak q$, whereas $\mathfrak p\in V(a)$.
A: Another way to think of it is: the Zariski topology is defined so that the closed sets are $V(I)$ where $I$ is an ideal (it's also enough just to look at radical ideals). Therefore $\overline{\{p\}}$, which is a closed set, is equal to $V(I)$ for some ideal $I$. To say that $p \in V(I)$ means that $p \supseteq I$.
In fact, if $I$ is any ideal such that $p \in V(I)$ then $p \supseteq I$. A closure is necessarily the smallest closed set containing $p$. Thus, we want $p = V(I)$ with $V(I)$ as small as possible. How do we achieve this? We make $I$ as big as possible because $V$ reverses inclusions. So what's the biggest ideal $I$ such that $p \supseteq I$? Clearly $I = p$. Thus $V(p)$ is the smallest closed set containing $p$.
