I learnt of this characterization from benedict gross's lectures, lecture 31 .

We usually define a number $p \in R$ as prime iff the ideal generated by $p$, $(p)$ is prime. Formally, for all $a, b \in R$, if $ab \in (p)$ then $a \in (p)$ or $b \in (p)$.

This can be thought of as saying that among all principal ideals, the ideal $(p)$ is maximal: no other principal ideal $(a)$ contains it.

§ Element based proof

• So we are saying that if $(p) \subseteq (a)$ then either $(p) = (a)$
• Since $(p) \subseteq (a)$ we can write $p = ar$. Since $(p)$ is prime, and $ar = p \in (p)$, we have that either $a \in (p) \lor r \in (p)$.
• Case 1: If $a \in (p)$ then we get $(a) \subseteq (p)$. This gives $(a) \subseteq (p) \subseteq (a)$, or $(a) = (p)$.
• Case 2: Hence, we assume $a \not \in (p)$, and $r \in (p)$. Since $r \in (p)$, we can write $r = r'p$ for some $r' \in R$. This gives us $p = ar$ and $p = a(r'p)$. Hence $ar' = 1$. Thus $a$is a unit, therefore $(a) = R$.