Atiyah-Macdonald has the following question:
If $x$ is a prime ideal, then $\overline{\{x\}}=V(p_x)$, where $V(x)$ is the set of prime ideals that contains $x$.
Why can we not have a smaller closed set containing $x$? Why can $x$ itself not be a closed set, even if it is not maximal?
That is because the closed sets in the Zariski topology have the $V(I)$, where $I$ is some ideal of the ring $R$. If $\mathfrak p_x\in V(I)$, it means that $I\subset \mathfrak p_x$, whence $$V(\mathfrak p_x)\subset V(I).$$
$x$ is closed if and only if $\,\overline{\!\{x\}\!}\,=\{x\}$, i.e. $V{\mathfrak p_x}=\{\mathfrak p_x\}$, or the only prime ideal containing $\mathfrak p_x$ is $\mathfrak p_x$ itself. Doesn't this mean $\mathfrak p_x$ is maximal?