Find generator of principal ideal 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...
 A: 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 $(1+\sqrt {10})\cdot 9-4\cdot (2+2\sqrt{10})$.
A: You can do something similar to the Euclidean algorithm, here.
Define the function $f:\Bbb Z\left[\sqrt{10}\right]\to\Bbb Z_{\ge0}$ by $$f(a+b\sqrt{10})=|a^2-10b^2|.$$ You should be able to show that it is a multiplicative function--that is, $f(xy)=f(x)f(y)$--and that $u\in\Bbb Z\left[\sqrt{10}\right]$ is a unit if and only if $f(u)=1.$
You need to find the greatest common factor $x+y\sqrt{10}$ of $9,2+2\sqrt{10}$. Note then that $f(x+y\sqrt{10})$ must be the greatest common factor of $f(9)=81$ and $f(2+2\sqrt{10})=36.$ (Why?) That is, we need $|x^2-10y^2|=9$. In particular, though, note that we have $$36=f(2+2\sqrt{10})=f(2)f(1+\sqrt{10})=4f(1+\sqrt{10}),$$ so $1+\sqrt{10}$ is a candidate, which we can quickly see is in fact a factor of $9$.
A: Hint: let $\ w = 1\!+\!\sqrt{10}.\ $ Then $\ ww' = -\color{#c00}9.\ $ Suppose $\ (9,2w) = (a).\ $
Then $\ (aa') = (a)(a') = (9,2w)(9,2w') = 9(9,2w,2w',-4) = (\color{#c00}9),\ \  {\rm by}\ \ \ (9,4) = (1).$
