Are all prime ideals in $\mathbb Z[\sqrt{-5}]$ of the form $\langle c, a + b \sqrt{-5} \rangle$?

Question

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.

Answer 1

A Dedekind domain $\,D\,$ satisfies what's known as the $\,1\ 1/2\,$ (one and a half) generator property: given an ideal $\,I\subset D\,$ and $\,0\neq i\in I\,$ then there is some $\,j\in I\,$ such that $\,I = (i,j).\,$

For example, if $\, (n) = I\cap \Bbb Z\,$ then $\, I = (n,w)\,$ for some $\,w,\, $ as you seek.

Answer 2

Yes. Let $I$ be an ideal of $\mathbb{Z}[\sqrt{-5}]$. Then $I \cap \mathbb{Z}$ is an ideal of $\mathbb{Z}$ and hence principal, take $I \cap \mathbb{Z} = (c)$. Then, as abelian groups, $I/c \mathbb{Z}$ injects into $\mathbb{Z}[\sqrt{-5}]/\mathbb{Z} \cong \mathbb{Z}$. Let the image be generated by the coset $\mathbb{Z}+b \sqrt{-5}$ and lift this coset to $a+b \sqrt{-5}$ in $I$. Then $a+b \sqrt{-5}$ and $c$ generate $I$ as an abelian group and, therefore, as an ideal.

Answer 3

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.