TITLE: Norm and square of the ideal $(2,1+\sqrt{-5})$ in the ring of integers
QUESTION [7 upvotes]: Let $I=(2,1+\sqrt{-5})$ be an ideal of the ring of integers of $\mathbb Q(\sqrt{-5})$. What is its norm $N(I)$? And is $I^2$ principal?

My notes say:

$1$, $\sqrt{-5}$ is a $\mathbb Z$-basis for $\mathcal O_K=\mathbb Z[\sqrt{-5}]$, and especially $1$, $1+\sqrt{-5}$, hence N$(I)=2$. Also, $I^2=(4,2(1+\sqrt{-5}),2(-2+\sqrt{-5}))\subseteq(2)$ and $N(I^2)=N(I)^2=4=N((2))$. So $I^2=(2)$ is principal.

Well, now I am trying to figure out why the ideal $I$ has norm $2$. We know that $N(2)=4$ and $N(1+\sqrt{-5})=6$. So $N(I)$ divides $4$ and $6$, i.e. it is $1$ or $2$. It cannot be $1$, because $I$ is a proper ideal. So it is $2$. But how is this related to the $\mathbb Z$-basis of $\mathcal O_K$?
And I am really clueless as to how I can compute $I^2$. I understand the norm argument, but I just don't know how to compute $I^2$.

REPLY [2 votes]: About the second point:
In fact, the class number of $\mathbb Q(\sqrt{-5})$ is $2$, so that $I^2$ is principal, whatever the ideal $I$ is. Moreover, by Kummer theory, or Dedekind(?), we know that prime $p$ splits in $\mathbb Q(\theta)$ according to how $f(x)$ factors modulo $p$, where $f$ is the minimal polynomial of $\theta$ over $\mathbb Q$. Now, the minimal polynomial of $\sqrt{-5}$ is $x^2+5\equiv (x+1)^2\pmod2$, so $(2,1+\sqrt{-5})^2=(2)$ by the above-mentioned theory.
Please tell me if some errors occur. Thanks.