Factorising ideals in $\mathbb{Z}[\sqrt{10}]$

I understand how to factorize ideals into prime ideals when they are of the form $(p)$, by Dedekind's Theorem, but I can't factorize ideals like $(4+\sqrt{10})$ in $\mathbb{Z}[\sqrt{10}]$. I can calculate its norm: $16-10=6$, so I know any prime factors must be of norm $2$ and $3$, but I can't figure out how to find them except by trial and error. Any help would be really appreciated. Thanks in advance.

• Seems that some people do not like British English... (by editing to replace "factorise" with "factorize") – Jérémy Blanc Mar 29 '14 at 13:06

1 Answer

If $\frak p$ is an ideal of norm $p$ prime in $\Bbb Z[\sqrt{10}]$, one knows that ${\frak p}\subset(p)=p\Bbb Z[\sqrt{10}]$, i.e. $\frak p$ appears in the decomposition of the ideal $(p)$.

The preliminary listing of primes ramifying, splitting or remaining inert helps. For instance if you know that $p$ is inert, then there are no ideals of norm $p$. The behaviour of the prime $p$ is easily linked to the behaviour of the polynomial $X^2-10\bmod p$.

All of this remains unchanged after replacing $\Bbb Z[\sqrt{10}]$ with the ring of integers of any quadratic extension, and in fact works--in principle--also for other algebraic extensions.