Ideals in quadratic integers

Question

I am studying for an algebra exam and have stumbled upon these two questions:

1. Show that $(2, \varepsilon)$ is not a principal ideal in $\mathbb{Z}[\sqrt{0}]$
2. Show that $(2, \varepsilon)$ is a principal ideal in $\mathbb{Z}[\sqrt{-1}]$

$\varepsilon$ is not specified, but I suppose $\varepsilon = \sqrt{0}$ and $\varepsilon = \sqrt{-1}$ respectively. Here are my thoughts:

1. Isn’t $\mathbb{Z}[\sqrt{0}] = \mathbb{Z}$ and hence a PID?
2. I know that $\mathbb{Z}[\sqrt{-1}] = \mathbb{Z}[i]$ and that the Gaussian integers are a PID. How could I prove the statement without this fact?

Edit: I have found this definition in the lecture notes: Let $A$ be a (commutative) ring, $d \in A$, and $\varepsilon$ a “square root” of $d$ that we add to $A$, then $A[\sqrt{d}] = \{a + b\varepsilon, a, b \in A\}$. That’s why I think $\mathbb{Z}[\sqrt{0}] = \{a + b\sqrt{0}, a, b \in \mathbb{Z}\} = \mathbb{Z}$.

Answer 1

After sorting out the notational difficulties with Daniel Fisher, I am now able to answer my own question:

1. $\mathbb{Z}[\sqrt{0}]$: If $(2, \varepsilon)$ was a principal ideal, it would be $(\operatorname{GCD}(2, \varepsilon))$ with $\operatorname{GCD}(2, \varepsilon)$ being a non-unit. However, 2 and $\varepsilon$ are relatively prime in $\mathbb{Z}[\sqrt{0}]$ and hence $(2, \varepsilon)$ cannot be principal.
2. $\mathbb{Z}[\sqrt{-1}]$ = $\mathbb{Z}[i]$: Here, $(2, \varepsilon) = (2, i)$. $i$ is a unit in the Gaussian integers and hence $(2, i) = (1) = \mathbb{Z}[i]$ and a principal ideal.