Let $M=\mathbb{(Z[\sqrt{-5} ])^2/ \begin{bmatrix} 2 \\ 1+\sqrt{-5} \end{bmatrix} Z[\sqrt{-5}]}$, a quotient $\mathbb{Z}[\sqrt{-5}]$-module. Can we find a basis $B$ of $M$ such that $|B|=1$?
I believe that the answer is "no," but I am having trouble proving this.
If I start by assuming $B=\{\begin{bmatrix} a \\ b \end{bmatrix}+\begin{bmatrix} 2 \\ 1+\sqrt{-5} \end{bmatrix} \mathbb{Z}[\sqrt{-5}]\}$ is a basis for $M$, it is not clear to me how I may obtain a contradiction from this. Please help.
Let $$t=\sqrt{-5}$$ to simplify further typing. Let $R=\mathbb Z[t]$. Assume as in the post that $$\begin{bmatrix}a\\b\end{bmatrix}$$ freely generates the quotient module $$M=R^2\Big/R \begin{bmatrix}2\\1+t\end{bmatrix}\ .$$ Then the canoical vectors of $R^2$, mapped to $M$, admit (unique) representations \begin{align} \begin{bmatrix}1\\0\end{bmatrix} &= p\begin{bmatrix}a\\b\end{bmatrix} + q\begin{bmatrix}2\\1+t\end{bmatrix} = \begin{bmatrix}a & 2\\b & 1+t\end{bmatrix} \begin{bmatrix}p\\q\end{bmatrix} \\\\ \begin{bmatrix}0\\1\end{bmatrix} &= r\begin{bmatrix}a\\b\end{bmatrix} + s\begin{bmatrix}2\\1+t\end{bmatrix} = \begin{bmatrix}a & 2\\b & 1+t\end{bmatrix} \begin{bmatrix}r\\s\end{bmatrix} \\\\ &\text{ or in one block} \\\\ \begin{bmatrix}1&0\\0&1\end{bmatrix} &= \begin{bmatrix}a & 2\\b & 1+t\end{bmatrix} \begin{bmatrix}p&r\\q&s\end{bmatrix} \end{align} Now we pass to determinants. Since all entries live in $R$, on the R.H.S. we have a product of two units. This shows that $2,1+t$ generate a unit in $R$, i.e. they can be used to generate $1$, i.e. $(2,1+t)=R\cong\mathbb Z[T]/(T^2+5)$. Taking this isomorphism modulo $2$, we get that $(1+T)$ is a unit (i.e. it generates the one) in $\mathbb F_2[T]/(T^2 +5)=\mathbb F_2[T]/(T+1)^2$. Contradiction.