Prime Number iff Generates Principal Maximal Ideal

Theorem
Let $\Z_{>0}$ be the set of strictly positive integers.

Let $p \in \Z_{>0}$.

Let $\left({p}\right)$ be the principal ideal of $\Z$ generated by $p$.

Then $p$ is prime $\left({p}\right)$ is a maximal ideal of $\Z$.

Proof
First, note that $\Z$ is a principal ideal domain, so all ideals are principal.

Suppose $\left({p}\right)$ is prime.

From Integer Divisor is Equivalent to Subset of Ideal, $m \mathrel \backslash n \iff \left({n}\right) \subseteq \left({m}\right)$.

But as $p$ is prime, the only divisors of $p$ are $1$ and $p$ itself.

By Natural Numbers Set Equivalent to Ideals of Integers, it follows that if $p$ is prime, then $\left({p}\right)$ must be a maximal ideal.

Conversely, let $p \in \Z_{>0}$ such that $\left({p}\right)$ is maximal.

Then if $\left({p}\right) \subseteq \left({q}\right)$ for some $q \in \Z_{>0}$, we have:
 * $\left({q}\right) = \left({p}\right) \implies q = p$

or
 * $\left({q}\right) = \left({1}\right) \implies q = 1$

Hence, as $p \in \left({q}\right)$, we have:
 * $p \in q \iff q = 1 \text { or } q = p$

and:
 * $p \in \left({q}\right) \implies q \mathrel \backslash p$

Hence $p$ is prime, so $\left({p}\right)$ is a prime ideal.