Factorising the ideal $(14)$

Question

I wish to find the prime factors of the ideal $(14)$ in $\mathbb{Q}(\sqrt{-10})$.

My working so far has been by noticing that $$14=(2+\sqrt{-10})(2-\sqrt{-10})=2\times7$$

So we have the candidates $(2+\sqrt{-10}). (2-\sqrt{-10}),(2) $ and $(7)$

Now, it would appear that none of these are prime - but now I do not know how to proceed. Any help would be appreciated!

Answer 1

Of course we have the usual factorization $14=2\times7$ and hence also $(14)=(2)(7)$. To further factor the ideals $(2)$ and $(7)$ we can use the Kummer-Dedekind theorem. The minimal polynomial of $\sqrt{-10}$ over $\Bbb{Z}$ is $X^2+10$, which factors mod $2$ and mod $7$ as $$X^2+10\equiv X^2\pmod{2}\qquad\text{ and }\qquad X^2+10\equiv(X-2)(X+2)\pmod{7}.$$ This shows that the ideals $(2)$ and $(7)$ factor into prime ideals as $$(2)=(2,\sqrt{-10})^2\qquad\text{ and }\qquad (7)=(7,2+\sqrt{-10})(7,2-\sqrt{-10}).$$

Answer 2

How exactly do you know that none of $\langle 2 \pm \sqrt{-10} \rangle$, $\langle 2 \rangle$, $\langle 7 \rangle$ are prime? They're not prime because those are principal ideals that are generated by numbers that are irreducible but not prime.

If 2 was prime in this domain, it would divide either of $2 \pm \sqrt{-10}$, but it doesn't, and neither does 7. Try it in the opposite direction and you will also come up empty.

So what you need here are ideals that are prime but not principal. Given numbers $a, b, x, y$ in the relevant domain, the ideal $\langle a, b \rangle$ consists of all numbers of the form $ax + by$.

The candidate ideals are then $\langle 2, \sqrt{-10} \rangle$, $\langle 7, 2 \pm \sqrt{-10} \rangle$. To check if they are prime you need to check if they are maximal: are they contained in a "larger" ideal that is not the whole ring?