Let $D$ be an integral domain and $N:D\rightarrow \mathbb Z$ s.t $N(0)=0$ and $N(ab)=N(a)N(b)$.
I have shown that if $u$ is a unit in $D$ then $|N(u)|=1$. However, the converse might not be true. i.e., let $D=\mathbb Z$ and $N(0)=0$, $N(x)=1$
Also if we suppose that N has the property that $|N(u)|=1$ iff $u$ is a unit, we can show that if $N(a)=p$ where $p$ is a prime, $a$ is irreducible. i.e., if $a=bc$, $N(a)=N(bc)=N(b)N(c)=p$. Therefore, $N(b)=1$ or $N(c)=1$. Hence, either $c$ or $b$ is a unit which implies $a$ is irreducible.
However, if we remove the assumption: $|N(u)|=1$ iff $u$ is a unit - Is the statement still true? i.e., if $N(a)=p$ then a is irreducible.
I don't think it is but I can't find any counter examples.
No, because if $N(x) = 1$ and $N(a) = p$ then $N(ax) = p$ and $ax$ can only be irreducible if $x$ is a unit.