Norms on general integral domains.

Question

I know that on $\mathbb Z[\sqrt n]$, $|N(a)|=1$ iff $a$ is a unit in $\mathbb Z[\sqrt n]$.

Note: $N$ is the norm function i.e., $N(0)=0$ and $N(a)N(b)= N(ab)$

However, does this property still hold for any norm on any Integral Domain?

Edit: A "norm" $N$ on any integral domain $D$ is a function $N:D\rightarrow \mathbb Z$ that satisfies the following properties: $N(0)=0$ and $N(a)N(b)= N(ab)$

Answer 1

No. For instance, on any integral domain you can define a norm by $N(0)=0$ and $N(x)=1$ for all $x\neq 0$. This will not satisfy the property you mention unless the integral domain is actually a field.

Answer 2

Arbitrary integral domains don't come equipped with a norm.

One can do the following: consider the class of (commutative) rings $R$ which are infinite, not fields, and such that if $I$ is a nonzero ideal then $R/I$ is finite (Pete Clark calls these abstract number rings). Then we can define the norm $N(I)$ to be the cardinality $|R/I|$, and we can define the norm $N(r), r \in R$ to be $N((r)) = |R/(r)|$. Then it's straightforward to verify that $N(I) = 1$ iff $I$ is the unit ideal, hence that $N(r) = 1$ iff $(r)$ is the unit ideal iff $r$ is a unit.

However, it's no longer true that $N(IJ) = N(I) N(J)$ in general.