TITLE: Do units in $\mathbb{Z} [(1+\sqrt{d})/2] $ have norm $\pm1$?
QUESTION [6 upvotes]: Do units in $\mathbb{Z} [(1+\sqrt{d})/2] $ have norm $\pm1$? As I understood the norm of units of integer rings of $\mathbb{Q} (\sqrt{d} ) $ have norm $\pm 1$ and at the same time $N( \frac{a + b \sqrt{d}}{2} ) = \frac{1}{4} (a^2 - d b^2)$ in my understanding. Is that right?

REPLY [5 votes]: It's important to distinguish between $d$ positive and negative and $d \equiv 1 \bmod 4$ and $d \not\equiv 1 \bmod 4$. Maybe you do have those distinctions in mind, but didn't write them in your question here. If that's the case, I might be simple reiterating what you already know very well.
If $d \equiv 1 \bmod 4$, then the ring of integers $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ includes numbers of the form $$\frac{a + b \sqrt{d}}{2}$$ with $a$ and $b$ both odd, in addition to numbers of the form $a + b \sqrt{d}$ with $a$ and $b$ of whatever parity combination. Then, with $a$ and $b$ both odd, we have $$N\left(\frac{a + b \sqrt{d}}{2}\right) = \frac{a^2 - b^2 d}{4}.$$ Since $a$ is odd, $a^2 \equiv 1 \bmod 4$, and since $b$ is odd, then $b^2 \equiv 1 \bmod 4$ as well. Then $a^2 - b^2 d$ is a multiple of $4$. Then that's why you care to find $N(x) = -4$ or $4$.
With two exceptions, if $d < 0$, then there are precisely only two units in $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$: $-1$ and $1$, and they both have norm $1$. In the two exceptions ($\mathbb{Z}[i]$ and $\mathbb{Z}[\omega]$), there are a few more units, but they all also have norm $1$. If $d > 0$, then there could be units of norm $-1$, but that's not guaranteed. Look up "fundamental unit" on this site for a more thorough discussion than I have time to write here.