Description about the set of quadratic integer with prime norm.

Question

I have a guess that, if $p$ is a prime number, then $$ \text{if }\exists a\in\mathbb{Z}[\sqrt{k}] \text{ such that } N(a)=p, \\ \text{ then }\left\{z\in \mathbb{Z}[\sqrt{k}]:N(z)=p\right\}= \left\{ a,\overline{a} \right\}\cdot \mathbb{Z}[\sqrt{k}]^* $$ In short, all the element with norm $p$ is actually associated with $a$ or $\overline{a}$, where $\overline{a}$ means $a$'s conjugate. Is it true? (I seems have proved this for cases of $k\equiv2,3 (\text{mod } 4)$, while the case for $k\equiv1 (\text{mod } 4)$ seems rather complicated)

Answer 1

Let $d > 1$ be a squarefree positive integer, and let $R=\mathbb{Z}[\sqrt{d}]$.

Fix a prime $p$.

Suppose $a\in R$ is such that $N(a)=p$, and let $\overline{a}$ denote the conjugate of $a$.

Claim:$\;$If $b\in R$ is such that $N(b)=p$, then either $b/a$ or $b/\overline{a}$ is an element of $R$ (and hence is a unit of $R$).

Proof:

Assume the hypothesis.

Then we get $$ \left\lbrace \begin{align*} \frac{b}{a}&=\frac{b\overline{a}}{p}\\[4pt] \frac{b}{\overline{a}}&=\frac{ba}{p}\\[4pt] \end{align*} \right. $$ Wrting $ba=g_0+g_1\sqrt{d}$ and $b\overline{a}=h_0+h_1\sqrt{d}$, our goal is to show that either $g_0,g_1$ are both divisible by $p$, or $h_0,h_1$ are both divisible by $p$.

Write $a=x+y\sqrt{k}$ and $b=u+v\sqrt{k}$.

Expanding $ba$ yields $$ \left\lbrace \begin{align*} g_0&=ux+vdy\\[4pt] g_1&=vx+uy\\[4pt] \end{align*} \right. $$ and expanding $b\overline{a}$ yields $$ \left\lbrace \begin{align*} h_0&=ux-vdy\\[4pt] h_1&=vx-uy\\[4pt] \end{align*} \right. $$ Then we have \begin{align*} g_1h_1 &= (vx+uy)(vx-uy) \\[4pt] &= v^2x^2-u^2y^2 \\[4pt] &= v^2(p+dy^2)-(p+dv^2)y^2 \\[4pt] &= p(v^2-y^2) \\[4pt] \end{align*} so $p{\,\mid\,}g_1$ or $p{\,\mid\,}h_1$.

Without loss of generality, we can assume $p{\,\mid\,}g_1$ (the case $p{\,\mid\,}h_1$ is analagous).

Then we get $$ g_0^2-dg_1^2=N(ba)=N(b)N(a)=p^2 $$ hence from $p{\,\mid\,}g_1$, we get $p{\,\mid\,}g_0$.

Thus $g_0,g_1$ are both divisible by $p$, which completes the proof.