Decompose $33+11\sqrt{-7}$ into irreducible integral elements of $\mathbb{Q}(\sqrt{-7})$ This is exercise I.3.1 in Neukrich's Algebraic Number Theory. Writing
$$ 33+11\sqrt{-7} = 11(3+\sqrt{-7}), $$
I have managed (by trying small values) to find
$$ 11 = (2+\sqrt{-7})(2-\sqrt{-7}) $$
and
$$ 3+\sqrt{-7} = 2 \left(\frac{3+\sqrt{-7}}{2}\right) = \left(\frac{1+\sqrt{-7}}{2}\right)\left(\frac{1-\sqrt{-7}}{2}\right)\left(\frac{3+\sqrt{-7}}{2}\right). $$
This shows that $(2\pm\sqrt{-7})$ and $\frac12(1\pm\sqrt{-7})$ are irreducible (because they are prime elements?) but I have a hard time continuing.
How does one determine whether an element (e.g. $\frac12(3+\sqrt{-7})$ here) is irreducible? 
It seems that showing that an element is prime is easy, but what about irreducible elements that are not prime?
Also, how does one find the factorization of a reducible element in general (other than trial and error)?
 A: I find that simplifying the notation helps with the calculations. This is what I did:
Since $-7 \equiv 1 \hspace{2pt}\mathrm{mod}\hspace{2pt} 4$ the ring of integers is $\mathbf{Z}[\alpha]$, where $\alpha :=\tfrac{1}{2}(1 + \sqrt{-7})$, and we know that $\{1, \alpha\}$ is an integral basis and that $\alpha^2 -\alpha +2 =0$. The norm form is $$\mathrm{Nm}(x+y\alpha) = x^2 +xy + 2y^2.$$ Starting with $33 + 11\sqrt{-7} = 11\cdot(3 + \sqrt{-7})$ we find that $1 + 2\alpha$ has norm equal to $11$, so in particular it is irreducible. Thus $11 = (1+2\alpha)\cdot(3-2\alpha)$. On the other hand, $3 + \sqrt{-7} = 2 + 1 + \sqrt{-7} = 2 + 2\alpha = 2\cdot(1+\alpha)$. We observe that $\alpha$ has norm $2$, so that $2 = \alpha \cdot (1-\alpha)$. The norm of $1 + \alpha$ is $4$, so we check if one of the elements with norm $2$ is a factor. Luckily, $1 + \alpha = -(1-\alpha)^2$. Summarizing:
\begin{equation*}
    33 + 11\sqrt{-7} = -(1 + 2\alpha)\cdot(3-2\alpha)\cdot (\alpha)\cdot (1-\alpha)^3 .
  \end{equation*}
