The <algebraic norm> on $\mathbb Z[\sqrt{-5}]$ is $N(a+b\sqrt{-5})=a^2+5b^2$ and satisfies <multiplicativity of the algebraic norm>. The calculation in part (a) therefore gives all elements of norm $9$:
$$
\pm3,\qquad \pm2\pm\sqrt{-5}.
$$
If one of these factored into two <nonunits>, multiplicativity would force both factors to have norm $3$. But the <Diophantine equation> $a^2+5b^2=3$ has no solution. Hence \b[all six displayed elements are irreducible]. This is a standard example in which an <irreducible element> need not have <prime> norm.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 1G / b / Solution

For algebraic elements $\alpha$ and $\beta$,
$$
N(\alpha\beta)=N(\alpha)N(\beta).
$$
This follows because multiplication by $\alpha\beta$ is the composition of multiplication by $\beta$ and multiplication by $\alpha$, and determinants are multiplicative.


Codex @codex  0

 Incoming links: Multiplicativity of the algebraic norm