Why I should show that no element of $R$ divides both $2$ and $1 + \sqrt{-5}$?

Question

Here is the question I want to answer:

Let $R = \mathbb Z[\sqrt{-5}]$ and $ I = (2, 1 + \sqrt{-5})$ be an ideal of $R$.

Show that $I$ is not a free $R-$module.

But here in this question How to show that an $\mathbb{Z}[\sqrt{-5}]$-module is not free? in its answer the author said that: to prove that it is free of rank 1, we should try to prove that no element of $R$ divides both $2$ and $1 + \sqrt{-5}.$ could anyone explain to me why this will prove that it is not free of rank $1$?

Here is the answer I am speaking about:

It is generated by two elements so if it were free, it would be free of rank $1$ or $2$.

If it were free of rank $1$, it would be generated by a single element: you can then try to prove that no element of $R$ divides both $2$ and $1+\sqrt{-5}$.

If it were free of rank $2$, then $2$ and $1+\sqrt{-5}$ would have to be free generators (can you see why ?). You can then try to prove that this is not the case, that is, that $2$ and $1+\sqrt{-5}$ are not linearly independent.

EDIT:

I know that we are using the contrapositive of this statement for the proof:

$I$ is free $R$ module $\implies$ I is a principal ideal.

And I know that we are trying to show that $I$ is not principal.

Answer 1

If $I$ were free of rank $1$ there would be an element $b \in I$ such that the $R$-linear map $\varphi: R \to I, x \mapsto xb$ would be an isomorphism. This would mean that there exist elements $r_1,r_2 \in R$ s.t. $r_1b=2$ and $r_2b=1+\sqrt{-5}$. If you are able to show that no $b$ in $R$ divides both elements this is a contradiction.

Edit. Let $b=\alpha+\beta \sqrt{-5}$, $\alpha, \beta \in \mathbb{Z}$. It holds that $N(2)=4, N(b)=\alpha^2+5\beta^2$. Since the norm is multiplicative $N(b)$ has to divide $N(2)=4$ in $\mathbb{Z}$ and thus $N(b) \in \{1,2,4\}$. Note that in all of these cases $\beta$ has to be zero and that the norm can never be two. Assume that the norm is one, then $\alpha=1$ or $\alpha=-1$ and $\beta=0$. On the other hand $N(b)$ has to divide (in $\mathbb{Z}$) $N(1+\sqrt{-5})=6$. This means that $b=1$ or $b=-1$ are in fact the only options. This however implies that $I=R$ which is a contradiction.