Consider $\mathbb Z[\sqrt {d}]$, where $d$ is any non - square integer, define $$N(a + \sqrt d b) = a^2 - db^2 = (a + \sqrt d b)(a - \sqrt d b)$$ as $\mathbb Z \subseteq \mathbb Z[\sqrt {d}]$, so from this definition, it is clear that $x|N(x), \forall x \in \mathbb Z[\sqrt {d}]$ now if for some $u \in \mathbb Z[\sqrt {d}] $, $|N(u)| = 1$, then, $u|N(u)$ implies $u||N(u)| = 1$ in $\mathbb Z[\sqrt {d}]$, so $1 = uv = vu$ for some $v \in \mathbb Z[\sqrt {d}]$, thus $u$ is a unit . Now in general let $R$ be an integral domain with unity and suppose there is some function $N:R \to \mathbb Z$ such that $N(1) \ne 0$, $N(ab) = N(a)N(b), \forall a,b \in R$, it is clear that, if $u$ is a unit in $R$, then $|N(u)| = 1$.
My question is: When does the converse hold? i.e. Is there any sufficient condition(s) on $R$ or $N$ which grantees $|N(u)| = 1$ implies $u$ is a unit, or also what are the necessary conditions (properties) that $R$, $N$ should have if $|N(u)| = 1$ implies $u$ is a unit ? I know that this converse holds if $N$ is a Euclidean evaluation, but when else does it hold ? Please help . Thanks in advance ( If it helps, also assume that $N(x) = 0$ only when $x = 0$ )
There is only one possible condition to put on $R$ for the result to be guaranteed (and a non-interesting one at that), namely that $R$ is a field. You always have the trivial map $N: R \to \mathbf{Z}$ given by $N(a) = 1$ if $a \neq 0$, $N(0) = 0$. Hence if $R$ has any non-unit, the condition is not satisfied.
On the other hand, this condition that $N(a)$ is a unit if and only if $a$ is a unit in $R$, is exactly the condition that defines a norm map. This is equivalent to saying $N^{-1}(\mathbf{Z}^\times) = R^\times$.