TITLE: Are all prime ideals in $\mathbb Z[\sqrt{-5}]$ of the form $\langle c, a + b \sqrt{-5} \rangle$?
QUESTION [5 upvotes]: Where $a, b, c \in \mathbb Z$?
I know that if in an UFD, $\langle c, a + b \sqrt{d} \rangle$ would boil down to a principal ideal. But it seems to me that in $\mathbb Z[\sqrt{-5}]$, for any purely real integer $c$ that is irreducible, the ideal $\langle c \rangle$ can be properly contained in some suitable choice of $\langle c, a + b \sqrt{-5} \rangle$ that is itself properly contained within the ring.

REPLY [5 votes]: No, it is not the case that for any irreducible integer $c$ the principal ideal $(c)$ is properly contained in an $(c, a + b \sqrt{-5})$ that is properly contained in the full ring.
If $p$ is a prime number number that is inert in $\mathbb Z[\sqrt{-5}]$ then essentially by definition $(p)$ is a prime ideal in  $\mathbb Z[\sqrt{-5}]$. 
Now, as a Dedekind domain $\mathbb Z[\sqrt{-5}]$ is one-dimensional and thus each non-zero prime ideal is maximal. Thus for an inert prime $p$ the ideal $(p)$ is not properly contained in any ideal but the full ring. 
As explained in other answers it is possible to write every ideal in the form $(c, a + b \sqrt{-5})$, and in particular $(p)$, in the form $(p, a + b \sqrt{-5})$, but in the case I mention there is no proper containment.
It is also true that every prime ideal of $\mathbb Z[\sqrt{-5}]$ contains some prime number and thus the principal ideal $(p)$ for some prime number $p$. Thus, if you drop the condition that $(p)$ is properly contained in $(p, a + b \sqrt{-5})$ then (but only then) what you say seems to be true is in fact true.