Prove that $I^2$ is principal.

Question

Consider the ideal $I=(2,\sqrt{-10})$ of $\mathbb{Z}[\sqrt{-10}]$. Prove that $I^2$ is principal.

My Try:

$I^2=(4,-10,2\sqrt{-10})$. I tried to prove that $I^2=(\sqrt{-10})$. But failed. Is my claim correct? Can anybody please help me how to show this?

Answer 1

No, $I^{\cdot2}=(2)$. Note that $I^{\cdot2}$ contains $(4,-10)=(2)$. On the other hand, every element of $I^{\cdot2}$ is clearly a multiple of $2$.