Let $R:=\mathbb{Z}[\sqrt{-6}]$. Prove that $R$ is not a Unique Factorization Domain 
Question: Let $R:=\mathbb{Z}[\sqrt{-6}]$.
(a) Prove that $R$ is not a Unique Factorization Domain
(b) Find an irreducible element of $R$ that is not prime
(c) Find a non-principal ideal in $R$
(d) Find two elements of $R$ with no $\text{GCD}$.

I just don't see how to tackle this problem. Throughout Dummit and Foote, the example $\mathbb{Z}[\sqrt{-5}]$ is used for similar questions. For example, $3\in\mathbb{Z}[\sqrt{-5}]$ is an irreducible element that is not prime:
$$
(2+\sqrt{-5})(2-\sqrt{-5})=9=3^2.
$$
But why is $3$ irreducible? I know the units in $\mathbb{Z}[\sqrt{-5}]$ are $\pm1$. They also take the ideal $I:=(3,2+\sqrt{-5})$, and prove it is not principal. I don't see how to remedy this for $R$.
 A: You can think $\mathbb{Z}[\sqrt{-6}]$ as $\mathbb{Z}+\mathbb{Z}\sqrt{-6}$. The units of $\mathbb{Z}[\sqrt{-6}]$ are $\pm 1$ because $N(a+b \sqrt{-6})=1 \Rightarrow a^2+6b^2=1$ has only solutions $a =\pm 1$, $b=0$, where $N$ is the norm function.
$(a)$ To show $\mathbb{Z}[\sqrt{-6}]$ is not UFD, it is sufficient to show some at least one element of $\mathbb{Z}[\sqrt{-6}]$ admits different factorization into irreducibles. For $$2 \cdot 3=6=(-1) \cdot (\sqrt{-6})^2.$$
$(b)$ $2$ is an irreducible element in $\mathbb{Z}[\sqrt{-6}]$ because you can not find an element in $\mathbb{Z}[\sqrt{-6}]$ whose norm is $2$. Similarly, $3$ is also an irreducible element in $\mathbb{Z}[\sqrt{-6}]$. By the same reason $\sqrt{-6}$ is also an irreducible element in $\mathbb{Z}[\sqrt{-6}]$.
But $2$ is not prime because $2 | (2+\sqrt{-6})(2-\sqrt{-6})$ but $2$ divides neither of the factors.
$(c)$ Let $p$ be a prime in $\mathbb{Z}$ or we can say irreducible in $\mathbb{Z}[\sqrt{-6}]$ and consider the ideal $I=(p, a+b \sqrt{-6})$ for some $a,b \in \mathbb{Z}$. Then we realize that $I \cdot \bar I=p \mathbb{Z}[\sqrt{-6}]$, where $\bar I=a-b \sqrt{-6}$. For, $I \cdot \bar I$ is generated by $p^2, p(a\pm b \sqrt{-6}), a^2+6b^2$. The first two elements are clear elements of $p \mathbb{Z}[\sqrt{-6}]$ while the third element $a^2+6b^2$ is the norm, and hence an element of $p \mathbb{Z}[\sqrt{-6}]$. The conclusion is $$p \in I \cdot \bar I.$$
Next we will show that if $I$ is principal then $p=a^2+6b^2$ for some $a,b \in \mathbb{Z}$. For, suppose $I$ is principal, say $I=(a+b \sqrt{-6})$ for some $a,b \in \mathbb{Z}$. Then $I \cdot \bar I=(a^2+6b^2)$. Now by the above argument, $p \in I cdot \bar I$ and therefore $a^2+6b^2$ divides $p$. But since $p$ is prime is $\mathbb{Z}$, we have $$p=a^2+6b^2.$$
This gives us the answer. Take a prime $p \in \mathbb{Z}$ which is not of the form $a^2+6b^2$. Clearly, $2 \neq a^2+6b^2$  for any $a,b \in \mathbb{Z}$. Thus $$I=(2, 1+\sqrt{-6})$$
is a non-principal ideal.
