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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.

  • What is $\delta$? –  Apr 12 '18 at 03:13
  • $δ=\sqrt{-5}$. I have now added that info to the question. – Pascal's Wager Apr 12 '18 at 03:14
  • @Pascal'sWager : Well, yes. I imagined it to be a formal variable, possibly lingering from some external context. – Eric Towers Apr 12 '18 at 03:17
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    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 $R^2/(2,1+\sqrt{-5})$? – rschwieb Apr 12 '18 at 10:54
  • @rschwieb Let $A$ be an $m\times n$ matrix with coefficients in $R$. Let $V$ be an R-module. We say "$V$ is presented by $A$" or "$A$ is a presentation matrix for $V$" to mean $V\cong R^m/AR^n$. – Pascal's Wager Apr 12 '18 at 13:53
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    @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? – rschwieb Apr 12 '18 at 14:18
  • I think you are correct. $AR^n={x\in R^m : x=Ay$ for some column vector $y\in R^n }$ – Pascal's Wager Apr 12 '18 at 14:38

2 Answers2

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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 $V/PV$.

Eric Wofsey
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  • 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. – Pascal's Wager Apr 20 '18 at 03:53
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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.)

Eric Towers
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  • 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 :-) – Pascal's Wager Apr 12 '18 at 03:29
  • 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. – Pascal's Wager Apr 12 '18 at 14:32
  • Actually, I think I now understand how to solve the problem. Basically the key is 6 times any basis vector is zero. – Pascal's Wager Apr 12 '18 at 15:03
  • ${e_1, e_2}$ 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. – Eric Towers Apr 12 '18 at 15:03
  • I must confess that I am relatively new to module theory, and that I am not entirely which rules for vector spaces carry over. I know that, in both vector space theory and module theory, a set is a basis if it is both 1. independent and 2. a spanning set. Now, I remember that, if a subset of the vector space has the same number of elements as the vector space's dimension, you only need to check one of the two properties to show that it is a basis. But I'm not sure if this is true for modules also. – Pascal's Wager Apr 12 '18 at 19:50
  • Some of the discussion here might be of use. There may be no independent, spanning set for a module. If there is an independent, spanning set, the module is free. What this problem is getting at is that there is a spanning set, ${e_1, e_2}$ that is not independent -- not even ${e_1}$ is independent, since it has a relation not forced by the ring. – Eric Towers Apr 12 '18 at 19:55
  • Just to make sure we're on the same page, $e_1$ is shorthand for $\begin{bmatrix} 1 \ 0 \end{bmatrix}+\begin{bmatrix} 2 \ 1+δ \end{bmatrix}R$, right? It seems to me then that ${e_1}$ would be independent. – Pascal's Wager Apr 12 '18 at 20:05
  • There's a little abuse of notation. $e_1$ is $\begin{bmatrix} 1 \ 0 \end{bmatrix}$. But we usually also write $e_1$ for its image along the projection onto the quotient. That image is explicitly the coset you wrote. An easier example might help: In $\mathbb{Z}/6\mathbb{Z}$, treated as a module over itself (so, having a free presentation), $e_1 = [1]$ and $e_1 + e_1 + e_1 + e_1 + e_1 + e_1 = 0$. This is a relation on $e_1$, but in this toy example, it is forced by the ring. $\mathbb{Z}[\sqrt{-5}]$ doesn't have torsion, so a relation on $e_1$ will be a little different... – Eric Towers Apr 12 '18 at 20:12
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    But there is no $r\in R-{0}$ such that $r$ times the coset I wrote is zero .... or is there? – Pascal's Wager Apr 12 '18 at 20:16
  • @Pascal'sWager : If you believe that, then you believe $V$ is freely generated by $e_1$. – Eric Towers Apr 12 '18 at 22:53
  • In the bazooka hint, you say "one of these". What does "these" refer to? By $e_1$ and $e_2$, do you mean the standard basis elements $(1, 0)$ and $(0, 1)$? Or what? – Rob Arthan Apr 13 '18 at 19:57
  • $e_1$, $e_2$ are "these", are the standard basis elements in $R^2$ and, by typical abuse, are their images in $V$. – Eric Towers Apr 13 '18 at 20:05
  • You can't be sure that some subset of ${e_1,e_2}$ forms as basis. You might instead need some linear combination of them. (For instance, consider the generators $2$ and $3$ of $\mathbb{Z}$ as a $\mathbb{Z}$-module.) – Eric Wofsey Apr 20 '18 at 02:10
  • @EricWofsey : I have not (and would not) claim that some subset of ${e_1, e_2}$ forms a basis. I claimed some subset of those is a minimal generating set. A generating set need not be linearly independent. – Eric Towers Apr 20 '18 at 22:22
  • OK, but the fact that a minimal generating set is not linearly independent does not imply that there doesn't exist some different minimal generating set that is. – Eric Wofsey Apr 20 '18 at 22:25
  • @EricWofsey : Again, I'm not seeing a claim by me anywhere in your objections. My claim: there is a two element generating set. There is a relation between the two generators, so there is a one element generating set. There is a relation on this one generator, so the resulting object cannot be free. – Eric Towers Apr 20 '18 at 22:28
  • What I said in my last comment is exactly what you are implicitly asserting. You have found one (minimal) generating set which has a relation, and are claiming therefore that the module is not free. This does not follow at all. – Eric Wofsey Apr 20 '18 at 22:47
  • In the case of the module $V$ in this problem, neither of the two generators $e_1$ and $e_2$ is redundant. And, neither one has a relation on its own ($e_1$ generates a free submodule of $V$, as does $e_2$!). So I don't know what argument you were trying to hint at in your earlier comments. – Eric Wofsey Apr 20 '18 at 22:50
  • @EricWofsey : The "Bazooka Hint" explicitly presents the relation between $e_1$, and $e_2$ in $V$, so your chain of objections is incorrect. Having shown that $V$ is cyclic, any (non-ring) relation on the surviving generator forces $V$ non-free. – Eric Towers Apr 21 '18 at 20:22
  • $V$ is NOT cyclic, though! I don't know where you got the idea that it was. And in any case, as I said, there is NO relation on $e_1$ (or $e_2$) alone. – Eric Wofsey Apr 21 '18 at 20:26
  • It is true that because there is a relation between $e_1$ and $e_2$, if you assume that $V$ is free, then you can conclude that $V$ is cyclic. (So, you could use its non-cyclicity to prove it is not free.) But you definitely can't conclude that it is cyclic generated by either $e_1$ or $e_2$: the generator might be something else. – Eric Wofsey Apr 21 '18 at 20:30
  • Again, the example of $\mathbb{Z}$ being generated as a $\mathbb{Z}$-module by ${2,3}$ is instructive. Observe that if $e_1=2$ and $e_2=3$, then they satisfy the relation $3e_1-2e_2=0$. If your argument worked for $V$, why wouldn't it work in this case to show that $\mathbb{Z}$ is not a free module? – Eric Wofsey Apr 21 '18 at 20:54