Ideal $(2, 1 + \sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$ 
Consider the ideal $J = (2, 1 + \sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$. Prove that $J \neq (1)$.

I am a bit stuck. If $J = (1)$, then $J = \{1 \cdot r \mid r \in \mathbb{Z}[\sqrt{-5}\}$, so $J = \mathbb{Z}[\sqrt{-5}]$, so I would only need to find one element of $\mathbb{Z}[-5]$. I feel this isn't the correct strategy, and instead I should in some way use the norm function and the fact that the norm function is multiplicative.
Any help would be appreciated.
 A: Suppose $$(1)=(2,1+\sqrt{-5})$$
then $$1 \in (2,1+\sqrt{-5})$$
thus $\exists$ $a+b\sqrt{-5}, c+d\sqrt{-5} \in \mathbb{Z}[\sqrt{-5}]$ such that
\begin{align*}
   1&=2(a+b\sqrt{-5})+(1+\sqrt{-5})(c+d\sqrt{-5})\\
 &=(2a+c-5d)+(2b+c+d)\sqrt{-5}
 \end{align*}
Thus $2a+c-5d=1$ and $2b+c+d=0$. if so then $$(2a+c-5d)+(2b+c+d)=2a+2b+c-4d=1+0=1$$
reducing $\operatorname{modulo}$ $2$ gives you $0=1$ which is a contradiction.
A: You can show that if $a+b\sqrt{-5}\in(2,1+\sqrt{-5})$ then $a+b$ is even. Thus, $1=1+0\sqrt{-5}$ is not in the ideal.
More generally:

Theorem: Given a square-free integer $D\neq 1$ and $a+b\sqrt{D}\in \mathbb Z[\sqrt D]$ and positive integer $d\mid a^2-Db^2$ then the ideal $I=(d,a+b\sqrt D)$ has the property that if $u+v\sqrt D\in I$ then $d\mid u^2-Dv^2.$


In particular, then if $d\neq\pm 1,$ you have $I\neq (1).$

I’ll leave the reader to prove this.
When $D$ is odd, and $a=b=1,$ then we can choose $d=2\mid 1-D,$ and $u^2-Dv^2$ is even iff $u+v$ is even.
Your case above is $D=-5,\,a=b=1,\,d=2.$
A: An alternative strategy is to work out the quotient ring ${\mathbb Z}[\sqrt{-5}]/(2,1+\sqrt{-5})$ and see that it is not the trivial ring.
It is probably easiest to do see this ring as ${\mathbb Z}[x]/(x^2+5)/(2,1+x)$. After a short and straightforward calculation, this turns out to be ${\mathbb F}_2$.
