$17\mathbb{Z}[\sqrt {10}]$ is prime ideal in $\mathbb{Z}[\sqrt {10}]$

Question

This seems tedious since I would have to show $(a+b\sqrt {10})(c+d\sqrt {10})=17k+17j\sqrt {10}$ implies that one of the factors belongs to $17\mathbb{Z}[\sqrt {10}]$.

It's easier to show something is not a prime ideal in this ring; e.g., $13\mathbb{Z}[\sqrt {10}]$ is not prime since $(6+\sqrt{10})(6-\sqrt{10})=26=2*13 \in 13\mathbb{Z}[\sqrt {10}]$. On the other hand, I don't want to exhaust all $17^4$ possibilities to show it is prime.

Another approach: $(a+b\sqrt {10})(c+d\sqrt {10})=17k+17j\sqrt {10} \iff\\ac+(ad+bc)\sqrt{10}=7k+7j\sqrt{10}\mod 10$.

But I am not sure what to do from here.

What's a nice way to show this ideal is prime?

Showing the corresponding quotient is an integral domain seems just as difficult.

Answer 1

We can use the fact that $ \mathbb{Z}[\sqrt{10}] \cong \mathbb{Z}[x]/(x^2 - 10) $. Quotienting by $ (17) $ gives $ \mathbb{F}_{17}[x]/(x^2 - 10) $, and we have

$$ \left( \frac{10}{17} \right) = \left( \frac{2}{17} \right)\left( \frac{5}{17} \right) = \left( \frac{1}{2} \right) \left( \frac{2}{5} \right) = -1 $$

so that $ 10 $ is not a perfect square in $ \mathbb{F}_{17} $. This means that $ x^2 - 10 $ is irreducible, so the ideal $ (x^2 - 10) $ is maximal and $ \mathbb{F}_{17}[x]/(x^2 - 10) $ is a field. Hence, $ (17) $ is a prime (and maximal) ideal of $ \mathbb{Z}[\sqrt{10}] $.