I was trying to comprehend a simple exercise from my elementary number theory class.
Let $\theta=\frac{1+\sqrt{-19}}{2}$, and let $R$ be the ring $\mathbb{Z}[\theta] .$ Show that the units of $R$ are $\pm 1$, and that 2,3, and $\theta$ are irreducible in $\mathbb{Z}[\theta]$.
The official solution:
If $z=a+b \theta$ is a unit of $\mathbb{Z}[\theta]$ then $z \bar{z}=1$, so we have $a^{2}+a b+5 b^{2}=1$ from which we conclude $a=\pm 1$, so the units are $\pm 1$. If $z=a+b \theta$ is a nonunit, non-associate divisor of 2 then $z \bar{z}=2$, but there are no solutions to $a^{2}+a b+5 b^{2}=2$. The arguments for 3 and $\theta$ are similar.
This solution is much too concise for me.
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Firstly, why does $z=a+b \theta$ being a unit of $\mathbb{Z}[\theta]$ imply $z \bar{z}=1$? At this point in time, the norm for $\mathbb{Z}[\theta]$ has not been defined, so cannot be used. I was hypothesizing that for $z_1,z_2 \in \mathbb{Z}[\theta]$, $z_1 z_2 \in \mathbb{Z}$ implied (by comparing imaginary parts) that $z_1 = \lambda \overline{z_2}$ for some $\lambda \in \mathbb{Q}$, but this also is not easy to prove. I even failed to brute force this, I couldn't continue the calculation.
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Why does $z=a+b \theta$ being a nonunit, non-associate divisor of 2 imply that $z \bar{z}=2$?
Any partial answers (answering only one of the questions) are very welcome! I will upvote them too. If the above problem has other solutions, not necessarily following the same reasoning as the official solution, then those are also very welcome!
Best Answer
Denote $z\bar z$ by $N(z)$. Then
Proof:
If $z$ is a unit, then there is a $w\in\Bbb Z[\theta]$ such that $zw=1$. Then $N(z)N(w)=1$. So $N(z)=1$
If $N(z)=1$, then $z\bar z=1$, so $z$ is a unit.
The units in $\Bbb Z[\theta]$ satisfy $N(z)=1$. To find the units notice that $$a^2+ab+5b^2=1\\(2a+b)^2+19b^2=4$$ gives $b=0$, and so $a=\pm 1$. Therefore the units of $\Bbb Z[\theta]$ are $\pm 1$.
Let $z$ be a non-unit, and a non-associate divisor of $2$. Then $zw=2$, and so $N(z)N(w)=4$.
Therefore $N(z)=2$.