TITLE: Finding units in quadratic integer rings
QUESTION [9 upvotes]: I want to find the units in $\mathbb{Z}[\alpha]$, where $\alpha=\frac{1+\sqrt{-11}}{2}$.
One can of course use norms to find the units in quadratic integer rings of the form $\mathbb{Z}[\sqrt{D}]$ for some squarefree integer $D$. This is my attempt at generalizing this method...
Let $r_1,r_2,s_1,s_2\in \mathbb{Z}$ such that 
$(r_1+r_2\alpha)(s_1+s_2\alpha)=1$. Then
$$4 = (2r_1+r_2(2\alpha))(2s_1+s_2(2\alpha))=(2r_1+r_2+r_2\sqrt{-11})(2s_1+s_2+s_2\sqrt{-11}).$$
Taking norms, we have
$$16 = ((2r_1+r_2)^2 + 11r_2^2)((2s_1+s_2)^2 + 11s_2^2) .$$
We can write $16$ as the product of two positive integers in three different ways: $16=4\times 4,8\times 2, 16\times 1$. The latter two factorizations present no solutions while the former presents the trivial solutions $(r_1+r_2\alpha,s_1+s_2\alpha)=(\pm 1, \pm 1)$. 
The units in $\mathbb{Z}[\alpha]$ are therefore $\pm 1$.

REPLY [7 votes]: The much simpler way, which you can generalize to all imaginary quadratic integer rings, is to remember that negative norms are impossible in these rings. Also remember that as these numbers get further away from $0$, they have larger norms.
If $D \equiv 1 \pmod 4$ (which is the case with $D = -11$), then you need to consider $N(z) = 1$ and $N(z) = 4$ ($z$ is an algebraic integer in the relevant domain). Otherwise, you only need to consider $N(z) = 1$.
The purely imaginary algebraic integers $z$ that are nontrivially closest to $0$ have $N(z) = |D|$. In the $D = -11$ case, that means $N(z) = 11$, which is clearly greater than $1$.
With $D \equiv 1 \pmod 4$, we also have $$N\left( \frac{1}{2} + \frac{\sqrt{D}}{2} \right) = \frac{|D| + 1}{4}.$$ In the $D = -11$ case, that means $3 > 1$. We conclude $1, -1$ are the only units in this particular doma