## § Prime numbers as maximal among principal ideals

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 $(a) = R$.