is it true that in $\mathbb{Z}[\sqrt{-11}]$, there's only one way to factorize 22 into irreducibles?(ignoring permutation and associates) for a specific example $22$, it seems to me that the only way to factorize it into irreducibles is $(-2)(\sqrt{-11})(\sqrt{-11})$ and indeed these are irreducibles. And I can manage to show that there's no way we can factorize 22 into products of the form $(a+b\sqrt{-11})(a-b\sqrt{-11})$ since there's no integers $a,b \in \mathbb{Z}$ satisfy this equation. But then it got me thinking that if there are any other ways to factorize $22$ into products of three,four or even more.
1.for $22$ is my derivation correct? (there's only one way to factorize $22$ into irreducibles in this ring if we ignoring permutations and associates)
2.how do we formalize the proof that there's no other ways $22$ can be factorized into irreducibles?
Thanks in advance.
 A: Yes, you are correct that this is essentially the only factorization. The usual way of dealing with these questions ad hoc is to use the norm map.
Consider the multiplicative norm map,
$$N\colon\mathbb{Z}[\sqrt{-11}]\to\mathbb{Z}$$
given by $N(a+b\sqrt{-11}) = a^2+11b^2$, with $a,b\in\mathbb{Z}$. This map satisfies $N(xy)=N(x)N(y)$, and that $x$ is a unit if and only if $N(x)=1$.
In particular, we have that if $x,y\in\mathbb{Z}[\sqrt{-11}]$ are such that $x$ divides $y$, then $N(x)$ must divide $N(y)$ in $\mathbb{Z}$.
Since $N(22) = 22^2 = 2^2\times 11^2$, any factorization of $22$ in $\mathbb{Z}[\sqrt{-11}]$ must involve factors of norms $2$, $4$, $11$, or $11^2$.
The only solutions to $a^2+11b^2 = 11$ with $a,b\in\mathbb{Z}$ are $a=0$, $b=\pm 1$, so the only elements with norm $11$ are $\pm\sqrt{-11}$.
There are no solutions to $a^2+11b^2=2$. The only solutions to $a^2+11b^2=4$ are $a=\pm 2$, $b=0$.
The only solutions to $a^2+11b^2=11^2$ are $a=\pm 11$, $b=0$; for $b=\pm1,\pm 2,\pm3$ we have $11^2-11b^2$ is not a square, and $|b|\gt 3$ is impossible.
Thus any nontrivial factorization of $22$ would have to involve a factor of norm $4$, and either two factors of norm $11$ or one factor of norm $11^2$. But $\pm 11$ is not irreducible, since $\pm11 = \mp(\sqrt{-11})^2$. So a factorization into irreducibles must be a product of $\pm 2$ and the square of $\sqrt{-11}$, with appropriate signs.

Aside: As it happens, the rings of integers of $\mathbb{Q}(\sqrt{d})$ with $d=-1$, $-2$, $-3$, $-7$, $-11$, $-19$, $-43$, $-67$, and $-163$, are UFDs, and in fact are the only quadratic imaginary number fields with that property. The ring of integers of $\mathbb{Q}(\sqrt{-11})$, however, is $\mathbb{Z}[\frac{1+\sqrt{-11}}{2}]$ and not $\mathbb{Z}[\sqrt{-11}]$, which is why you cannot invoke unique factorization directly above. However, your factorization happens to be the unique factorization of $22$ in $\mathbb{Z}[\frac{1+\sqrt{-11}}{2}]$, so any other factorization in $\mathbb{Z}[\sqrt{-11}]$ is likewise impossible.
