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.

  • $\begingroup$ What is $\delta$? $\endgroup$ – user14972 Apr 12 '18 at 3:13
  • $\begingroup$ $δ=\sqrt{-5}$. I have now added that info to the question. $\endgroup$ – Pascal's Wager Apr 12 '18 at 3:14
  • $\begingroup$ @Pascal'sWager : Well, yes. I imagined it to be a formal variable, possibly lingering from some external context. $\endgroup$ – Eric Towers Apr 12 '18 at 3:17
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    $\begingroup$ Hm, how does one read "an $R$ module presented by a matrix"? I can't say I've heard the expression. From the context it looks like $R^2/(2,1+\sqrt{-5})$? $\endgroup$ – rschwieb Apr 12 '18 at 10:54
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    $\begingroup$ @Pascal'sWager OK, that's a good start for explaining to me. The issue I have is that "$AR^n$ (the product of a matrix with direct product of copies of $R$) doesn't have any meaning to me. I think you mean to say it's the submodule of $R^m$ generated by the $n$ columns of $A$. Right? $\endgroup$ – rschwieb Apr 12 '18 at 14:18

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 $V/PV$.

  • $\begingroup$ This is a very good solution. However, it is not obvious to me that $V/PV \cong (R/P)^2/ \begin{bmatrix} 2+P \\ 1+δ+P \end{bmatrix}(R/P)$. I tried defining a map $φ:V/PV \to (R/P)^2/ \begin{bmatrix} 2+P \\ 1+δ+P \end{bmatrix}(R/P)$ by $φ((\begin{bmatrix} a \\ b \end{bmatrix}+T)+PV)=\begin{bmatrix} a+P \\ b+P \end{bmatrix}$. But, alas, it seems like a very ugly function to work with and I have not been able to prove that it is well-defined, even though I conjecture it is. $\endgroup$ – Pascal's Wager Apr 20 '18 at 3:53

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.)

  • $\begingroup$ I figured that this would play a part in the solution, but I still don't see the contradiction. I have a feeling that I need to multiply the one-and-only element of $B$ by a nonzero element of $R$ to obtain zero, but I'm still stuck. If you don't mind, I could use an even blunter hint :-) $\endgroup$ – Pascal's Wager Apr 12 '18 at 3:29
  • $\begingroup$ I understand that $\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 2 \\ 1+δ \end{bmatrix}R, \begin{bmatrix} 0 \\ 1 \end{bmatrix} + \begin{bmatrix} 2 \\ 1+δ \end{bmatrix}R \}$ is not a valid basis for the quotient. However, I do not see how this implies that there exists no basis at all of size 2 for the quotient. $\endgroup$ – Pascal's Wager Apr 12 '18 at 14:32
  • $\begingroup$ Actually, I think I now understand how to solve the problem. Basically the key is 6 times any basis vector is zero. $\endgroup$ – Pascal's Wager Apr 12 '18 at 15:03
  • $\begingroup$ $\{e_1, e_2\}$ generates the quotient. So some subset of these is a minimal generating set. Use relations among them to figure out which ones are redundant. Are they all redundant? When you get to the last one, you are asking if it has a relation with itself that is not forced by the ring. For instance, $\mathbb{Z}/6\mathbb{Z}$ as a module over itself is freely generated by $[1] \pmod{6}$, because its relation is inherited from the ring, but $2 \pmod{6}$ is not, because $4 \cdot [2] \cong 1 \cdot [2] \pmod{6}$ is not forced by a "$4 = 1$" relation in the ring. $\endgroup$ – Eric Towers Apr 12 '18 at 15:03
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    $\begingroup$ But there is no $r\in R-\{0\}$ such that $r$ times the coset I wrote is zero .... or is there? $\endgroup$ – Pascal's Wager Apr 12 '18 at 20:16

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