Prove that $V$ is not a free module. I am attempting to solve this:


  Let $R=\mathbb{Z}[\sqrt{-5}]$, and let $V$ be the R-module presented by the matrix $\begin{bmatrix} 2 \\ 1+δ \end{bmatrix}$ where $δ=\sqrt{-5}$. Prove that $V$ is not a free module.


Note on definitions

I want to note that my professor defines ($V$ is a free R-module) $\iff \exists k\in\{1,2,...\}[V\cong R^k]$. He also considers the empty set to NOT be a valid basis. So there is no basis for the trivial vector space. I know that these definitions are controversial, but bear with me for now.

My attempted solution

Let $T=\begin{bmatrix} 2 \\ 1+δ \end{bmatrix}R$. We know already $V\cong R^2/T$.

Assume $V$ is a free module. We want to obtain a contradiction.

Now, I see that the rank of $\begin{bmatrix} 2+P \\ 1+δ+P \end{bmatrix}$ (when $P$ is a prime ideal of $R$) can be either $0$ or $1$, depending on $P$. My professor says that the rank being non-constant contradicts the fact that $V$ is free, but alas, I do not see the contradiction.
For any maximal ideal $P\subset R$, consider $V/PV$, which is an $R/P$-vector space.  If $V$ were isomorphic to $R^k$, then $V/PV$ would be isomorphic to $R^k/PR^k\cong (R/P)^k$, so it would have dimension $k$ as an $R/P$-vector space.

But now note that $V/PV$ is presented as an $R/P$-vector space by the matrix $\begin{bmatrix} 2+P \\ 1+\delta+P \end{bmatrix}$.  So if $V/PV$ has dimension $k$, that matrix must have rank $2-k$.  In particular, if $V$ is free, the rank of $\begin{bmatrix} 2+P \\ 1+\delta+P \end{bmatrix}$ would have to be the same for all $P$.  Since this is not true, $V$ cannot be free.

To prove that $V/PV$ is presented as an $R/P$-vector space by the matrix $\begin{bmatrix} 2+P \\ 1+\delta+P \end{bmatrix}$, first consider the following general situation.  We have a module $M$ with submodules $N,K,$ and $L$ with $K,L\subseteq N$.  Note then that $$(M/K)/(N/K)\cong M/N\cong (M/L)/(N/L).$$  To apply this here, let $M=R^2$, $K=PM$, $L=\begin{bmatrix} 2 \\ 1+\delta \end{bmatrix}R$, and $N=K+L$.  Then $(M/K)/(N/K)$ is the quotient of $(R/P)^2$ by the subspace generated by $\begin{bmatrix} 2 + P \\ 1+\delta + P \end{bmatrix}$; that is, it is the $R/P$-vector space presented by the matrix $\begin{bmatrix} 2 + P \\ 1+\delta + P \end{bmatrix}$.  On the other hand, $(M/L)/(N/L)$ is $V/PV$.
Hint: There is a standard example of failure of unique factorization in $\mathbb{Z}[\sqrt{-5}]$.  Maybe you can adapt that?

Blunter Hint: 


   What's $(1+ \delta)(1- \delta)$.  Is that also divisible by the other element of the presentation matrix?


Bazooka hint:


    The quotient is generated by the images of the standard basis elements in $R^2$ under the projection onto the quotient.  Free things don't have relations and $\begin{pmatrix}1 \\ 0\end{pmatrix} \cdot 2 + \begin{pmatrix}0 \\ 1 \end{pmatrix} \cdot (1 + \delta) =0$.  So this is not a free module on two generators.  Consequently, if it is a free module on one generator, it is generated by one of these.  Call these $e_1$ and $e_2$.  Keep adding copies of $e_1$ to the equation until you realize there's a multiple of $e_1$ that is a multiple of $e_2$ that is a multiple of $e_1$, i.e., $e_1$ satisfies a relation...  (Prior hints might shorten this search a bit.)