I have one question but first I will give some context:
For the field $\mathbb{Q}(\sqrt{m})$ I consider the case where the ring of algebraic integers is $\mathcal{O}_{\mathbb{Q}(\sqrt{m})}=\mathbb{Z}[\omega_{m}]=\{a+\omega_{m}b\;|\; a,b\in\mathbb{Z}\}$ where $$\omega_{m}=\sqrt{m}.$$
Therefore for $\alpha=a+\omega_m b$, we have that the norm
$$N(\alpha):= \alpha\cdot\overline{\alpha}=a^{2}-mb^{2}$$
where $\overline{\alpha}=a-\omega_m b$ is the conjugate of $\alpha$
Now I am considering the quadratic imaginary case; i.e. $m<0$ and $\mathbb{Z}[\omega_{m}]$ is a PID.
If $\alpha=a+\omega_m b$ is a prime element in $\mathbb{Z}[\omega_{m}]$ such that $N(\alpha)=p$ is a prime number.
I know that $p$ can not divide $a$ and $b$ at the same time, otherwise from the definition of norm we have that
$$p=N(\alpha)= \alpha\cdot\overline{\alpha}=p\alpha\beta$$
for some $\beta\in\mathbb{Z}[\omega_{m}]$. It follows that $\alpha$ is a unit contradicting that is a prime element in $\mathbb{Z}[\omega_{m}]$.
So my question is the following: Is it possible that $p$ divides only $a$?.
I am thinking that the answer is no, since if $p$ divides $a$ we have that $N(\alpha)>p$ (since $m<0$ and therefore $a^{2}-mb^{2}$ is not negative)
With a similar argument we can answer negatively the question Is it possible that $p$ divides only $b$?
So I am confused about my reasoning, so any idea or correction it would be appreciated.
Thank you!