How to prove the correspondence between closed points of the spectrum of a ring and its maximal ideals? Let $R$ be an entire ring.
How does one prove $\mathfrak{p}$ is a closed point of $\text{Spec} R$ if and only if $\mathfrak{p}$ is a maximal ideal of $R$?
 A: Let $\mathfrak{p}$ be a prime ideal of $R$. The required preliminary result is to know the closure $\overline{\lbrace \mathfrak{p} \rbrace}$. Can you guess? Do you see a closed subset wrt the Zariski topology that obviously contains $\lbrace \mathfrak{p} \rbrace$?
Actually, the set of all prime ideals of $R$ containing $\mathfrak{p}$, usually denoted $V(\mathfrak{p})$, is the closure of $\lbrace \mathfrak{p} \rbrace$.
Let $\mathfrak{p}$ be a closed point, hence $\lbrace \mathfrak{p} \rbrace = \overline{\lbrace \mathfrak{p} \rbrace}$. Recall there exists a maximal ideal $\mathfrak{m}$ such that $\mathfrak{p} \subseteq \mathfrak{m}$ and from our preliminary result one has
$$
V(\mathfrak{p}) = \overline{\lbrace \mathfrak{p} \rbrace} = \lbrace \mathfrak{p} \rbrace.
$$
Since a maximal ideal is prime, one concludes $\mathfrak{m} = \mathfrak{p}$.
Conversely, let $\mathfrak{p}$ be a maximal ideal. Then, one has $V(\mathfrak{p}) = \lbrace \mathfrak{p} \rbrace$. From our preliminary result one concludes $\lbrace \mathfrak{p} \rbrace = \overline{\lbrace \mathfrak{p} \rbrace}$, i.e. $\mathfrak{p}$ is a closed point.
A: By definition $\mathfrak{p}$ is a closed point iff there is no other prime ideals containing $\mathfrak{p}$ other than itself. But every prime ideal (in fact, every ideal) must be contained in some maximal ideal, and maximal ideals are prime.
