Proving $\pi$ is irreducible/prime in $R.$

Can I please get feedback on my proof or help proving the following problem I am working on? Thank you for your time and help.

Denote $R$ as the ring of algebraic integers in an imaginary quadratic field and consider $R$ a UFD. Let $P$ be a maximal/prime ideal in $R$ and let $\pi \in P$ be an element of minimal norm. I am trying to prove $\pi$ is irreducible/prime in $R.$

$\textit{Proof.}$ Suppose $P$ is prime ideal, then $P$ is a maximal ideal as every prime ideal in a ring of algebraic integers is maximal. Then $\pi$ is irreducible and $\pi$ is prime because in UFD, $\alpha$ is prime if and only if $\alpha$ is irreducible.

Solution 1:

That's the right idea. More generally, using the same idea, in any domain $D$, suppose $\,0\neq b\in P$ prime, and suppose $\,b\,$ has a factorization $\,a_1\cdots a_n\,$ into atoms (irreds) $\,a_i.\,$ Since $P$ is prime some $\,a_i \in P.\,$ Thus if further $\,b\,$ is an element of $P$ minimal wrt to divisibility then $\,a\,$ must be an atom (so prime if $D$ is a UFD). So $D$ a UFD $\Rightarrow$ prime ideals can be generated by primes. The converse is also true, i.e. UFDs are exactly those domains whose prime ideals can be generated by primes.