Let $V$ be the module generated by the column matrix $A= (2, 1+ \sqrt{-5})^T$. Prove that the residue of $A$ in $\mathbb{Z}[\sqrt{-5}]/ \mathfrak{P}$ has rank $1$ for every prime ideal $\mathfrak{P}$ of $\mathbb{Z}[\sqrt{-5}]$, but $V$ is not a free module.

I know that $\mathbb{Z}[\sqrt{-5}]$ is not a PID, hence I don't have many theorem in hands. please help.

  • $\begingroup$ It is not clear to me exactly what is meant by $V$ being the module generated by a column matrix. do you just mean $V\subset \mathbb{Z}[\sqrt{-5}]$ is generated by $2$ and $1+\sqrt{-5}$? $\endgroup$ – WSL May 2 '15 at 11:14

$V$ is not free as $2, 1+\sqrt{-5}\in V$ and $(1-\sqrt{-5})(1+\sqrt{-5}) = -4$, giving that

$2\cdot 2 +(1-\sqrt{-5})(1+\sqrt{-5}) = 0$.

If I am interpreting your other statement correctly, then the statement isn't exactly true as stated, because if $\mathfrak{P} = (2, 1+\sqrt{-5})$, then $A\equiv (0,0)^T$ over $\mathbb{Z}[\sqrt{-5}]/ \mathfrak{P}$.

This makes me feel as though I am not understanding the question, so if you edit to clarify, I will edit my answer.


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