The ideal $(9, 2 + 2\sqrt{10})$ of $\mathbb{Z}[\sqrt{10}]$ is a principal ideal; it is generated by $1+\sqrt{10}$.
This is easy enough to check once it's been found, but can anyone tell me some way to arrive at this (or some other) generator? That could be done with pencil & paper?
Ordinarily I would use the Euclidean algorithm, but we don't have that here…
Best Answer
We have the norm (even though it is not always positive) obtained by multiplying $a+b\sqrt {10}$ with its conjugate $a-b\sqrt {10}$, that is $N(a+b\sqrt{10})=a^2-10b^2\in\mathbb Z$. So we have $N(9)=81$ and $N(2+2\sqrt{10})=-36$. We can restrict our search to elements of norm dividing both $81$ and $36$, that is we must have $N(a+b\sqrt{10})\in\{\pm1,\pm3,\pm9\}$. Looking for small solutions of this you will stumble upon $1+\sqrt {10}$, which is both a divisor of $9$ and $2+2\sqrt{10}$ and a linear combination of these, namely $(1+\sqrt {10})\cdot 9-4\cdot (2+2\sqrt{10})$.