A principal ideal domain if and only if proof

Question

Show an ideal $(p)$ in a principal ideal domain in a maximal ideal if and only if $p$ is irreducible. This is a new concept i do not know how to go about this.

Answer 1

For the $\impliedby$ direction:

Suppose $p$ is irreducible and $(p)$ is not maximal. Then there exists a proper ideal $(t)$ such that $(p) \subset(t)$.
So $p \in (t) \implies \exists$ a non-unit $s$ in $A$ such that $p=ts$, else $(p)=(t)$.
$t$ is not a unit, otherwise $(t)=A$, and $s$ is not a unit by assumption.Thus $p$ is not irreducible. Contradiction.
Hence $(p)$ is a maximal ideal.

Answer 2

Hint $\ $ For principal ideals: $\,\ \rm\color{#0a0}{contains} =\color{#c00}{divides},\,$ i.e. $\, (a)\supset (b)\iff a\mid b.\ $ Therefore

$$\begin{align} (p)\text{ is maximal} \iff&\, (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (a)\\ \iff&\ \ \,p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ a\\ \iff &\ \ \,p\ \, \text{ is irreducible} \end{align}\qquad$$