Find Inverse of ideal $I= \langle 3, 1+2\sqrt{-5} \rangle$

Question

Question- Find Inverse of ideal $I = \langle 3, 1 + 2\sqrt{-5} \rangle$ of $O_K$ (ring of integers of algebraic number field $K$), where $K = \mathbb{Q}(\sqrt{-5})$.

First of all can $I$ be simplified more than $I = \langle 3, 1 + 2\sqrt{-5} \rangle = \langle 3, 2 - 2\sqrt{-5} \rangle$ which is not much help.

Now if because $I$ is a prime ideal then, $I^{-1} = I' = {\alpha \in K :\ \alpha I \subseteq O_K}$.

While attempting to find $I'$, I started by letting $x + y{\sqrt-5} \in K$ and then when proceeded, what I got is $3x \in \mathbb{Z}$ and $3y, 11y \in \mathbb{Z}$ but this says that no rational $y$ can exist. So obviously I did something wrong. I guess my answer should be in terms of $I$, can somebody help here?

Answer 1

Hints:

• $I=\langle 3,1-\sqrt{-5}\rangle$.
• If $J=\langle 3,1+\sqrt{-5}\rangle$, then prove that $IJ=\langle 3\rangle$.
• Deduce that $I^{-1}$ is $J$ times a principal fractional ideal.