TITLE: Checking whether $\sqrt{2}$ is contained in the ideal $(2,\sqrt{-5}+1)$
QUESTION [5 upvotes]: Let $L=\mathbb{Q}(\sqrt{-5},\sqrt{2})$ and $I=(2,\sqrt{-5}+1)$ the ideal in $\mathcal{O}_L$, the ring of algebraic integers in $L$, which is generated by $2$ and $\sqrt{-5}+1$
I want to show that $I=(\sqrt{2})$. Once I can show that $\sqrt{2}\in I$, this is easy.
But I don't know how to show that $\sqrt{2}$ is contained in $I$.
Every element in $I$ must be of the form $2a+(\sqrt{-5}+1)b$ for some $a,b\in \mathcal{O}_L$. I cannot find such $a$ and $b$.

REPLY [4 votes]: The distributive property states $(a,b)(c,d)=(ac,ad,bc,bd)$. So we compute
$$\begin{array}{ll} (2,1+\sqrt{-5})^2 & =(4,2+2\sqrt{-5},-4+2\sqrt{-5}) \\ & =(2)(2,1+\sqrt{-5},-2+\sqrt{-5}) \\ & = (2)(2,3,-2+\sqrt{-5}) \\ & = (2)(1)=(2). \end{array}$$
by $3=(1+\sqrt{-5})-(-2+\sqrt{-5})$, valid since $(a,b)=(a-b,b)$. Thus $I^2=(\sqrt{2})^2$. 
By unique factorization, $I^2=J^2\Leftrightarrow\, I=J$ in any Dedekind domain, e.g. number rings.