$(2,1+\sqrt{-5}) \not \cong \mathbb{Z}[\sqrt{-5}]$ as $\mathbb{Z}[\sqrt{-5}]$-module

Question

Let $R=\mathbb{Z}[\sqrt{-5}]$, $I= (2,1+\sqrt{-5})$. Show now that $I\not \cong R$ as $R$-module.

Any hints?

My progress

Assume $I\cong R$ then there is a $\theta: I\to R$ a $R$-module isomorphism.

I want to find a contradiction, finding two different elements which map to the same value, or a non-trivial kernel, or ... I haven't really found a method yet, I was thinking of something like

Let $\theta(2) = a\in R, \qquad \theta(1+\sqrt{-5})=b\in R$ then $3\theta(2) = 3a$ and $(1-\sqrt{-5})\theta(1+\sqrt{-5}) = (1-\sqrt{-5})b$ which implies $3a= (1-\sqrt{-5})b$. Replacing $a=\eta + \sqrt{-5}\rho$ and $b=\tilde \eta+\sqrt{-5}\tilde \rho$ didn't result in anything useful.

Any other ideas? (I prefer hints instead of solutions)

Answer 1

Hint $R$ is a cyclic $R$-module (i.e. generated by one element). Prove that your ideal cannot be generated by one element.