= Solution
{c}
For $z=a+b\sqrt d$, let $\bar z=a-b\sqrt d$ denote the <algebraic conjugate>, not necessarily ordinary complex conjugation when $d>0$. Conjugation respects multiplication, and $z\bar z=a^2-db^2$. Therefore
$$
N(zz')=|(zz')\overline{zz'}|
=|z\bar z|\,|z'\bar z'|
=\boxed{N(z)N(z')}.
$$
If $z$ is a <unit in a ring>, choose its inverse $w$ in the same <ring>. Then $N(z)N(w)=N(1)=1$, and both norms are nonnegative <integers>, forcing $N(z)=1$. Conversely, if $N(z)=1$, then $z\bar z=\pm1$, so $z^{-1}=\pm\bar z$ belongs to the <ring>. Thus \b[$z$ is a <unit in a ring> exactly when $N(z)=1$].
For negative $d$, the equation becomes $a^2+|d|b^2=1$. If $d=-1$, its <integer> solutions give the four <units> $\pm1,\pm i$. If $d<-1$, the term $|d|b^2$ forces $b=0$, leaving $a=\pm1$. Hence
$$
\boxed{\mathbb Z[\sqrt d]^\times=
\begin{cases}\{1,-1,i,-i\},&d=-1,\\
\{1,-1\},&d<-1.\end{cases}}
$$
For $d=2$, the element $u=1+\sqrt2$ has <field norm> $|1-2|=1$. Its powers $u^n$, $n\geq0$, are all <units> by <multiplicative function> and are distinct, since their positive real values strictly increase with $n$. Thus $\mathbb Z[\sqrt2]$ has infinitely many <units>. This argument concerns the stated integer-coefficient <ring>; it does not replace it by a larger <ring> of algebraic <integers>.
Back to article page
