I am trying to solve the following exercise on basic module theory and I am stuck. Any help would be more than welcome!

So let $M\subseteq \mathbb{Z}^3$ the solutions to the following problem:




What are the invariant factors of $M$ and $\mathbb{Z}^3/M$? Or, put in other words, what is the structure of $M$ and $\mathbb{Z}^3/M$?

What I have so far:

$M$ must be free since it is a submodule of a free module over a PID. But I think in order to answer anything else, I must be able to solve the above system. If I knew a basis for $M$ then I could calculate the map $f$ injecting $M$ in $\mathbb{Z}^3$ and then I would reduce the matrix of $f$ using Smith's algorithm to something simple. Is there any way to avoid solving the system? Also, since the "injection" matrix will be part of the identity matrix, in the end both $M$ and $\mathbb{Z^3}/M$ will be free of rank depending on the rank of the space of solutions.

NOTE: There is some similar problem which is basically to understand an abelian group given a representation, e.g., $<a,b|2a+4b,3a-8b>$. Are the solutions to these two problems in any way related?

  • $\begingroup$ When you say you are stuck, do you mean that you don't know how to solve the system of equations or that you don't want to solve it? $\endgroup$ – Rob Arthan Feb 7 '19 at 21:12
  • $\begingroup$ @RobArthan I don't know how to solve it, since I assume I can't just copy the vector space method of solving linear equations. More importantly though, I want to know if it nescessary to solve the system or not. Because my teacher provided us with a very handwavy solution which I am not satisfied with. $\endgroup$ – Nick A. Feb 7 '19 at 21:20
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    $\begingroup$ Solving the equations isn't very different from the vector space method except you have to avoid division by multiplying by a suitable factor before subtracting. The Smith normal form method is a way of doing this that yields a very convenient normal form. We can't really comment on your teacher's solution if you can't provide the details and I can't see of any way of using the data in your problem that is easier than solving the equations. $\endgroup$ – Rob Arthan Feb 7 '19 at 21:27
  • $\begingroup$ @RobArthan Thank you. The reason I didn't post my teacher's answer is simply because I have to translate it into english and I neither understand it's maths, nor I can put it here verbatim. But it goes something like this: let $f:Z^3 \rightarrow Z^3$ be a map such that $ker(f)=M$ . Then the matrix of $f$ has the same coefficients as the system. By reducing the matrix with smiths algorithm he concludes that $Z^3/M=Z_3\times Z$. I find this completely wrong for many resons. The most important being that $Z^3/kerf=imf$ and (in his answer) $im(f)$ has torsion. $\endgroup$ – Nick A. Feb 7 '19 at 21:37
  • $\begingroup$ @DavidHill: thanks for the correction: i've deleted my erroneous comment that was based on a misreading of the OP's outline of the teacher's argument. The key point is that the you need to find the Smith normal form. $\endgroup$ – Rob Arthan Feb 7 '19 at 22:54

Well, $M$ is the kernel of the map $$ f:\mathbb{Z}^3\to\mathbb{Z}^3,\;\;\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto\begin{pmatrix}-3x+5y+3z\\-6x+20y+9z\\-3x+25y+9z\end{pmatrix} $$ Using the standard basis $\{e_1,e_2,e_3\}$ for $\mathbb{Z}^3$, we can represent $f$ above as a matrix $$ A=\begin{pmatrix}-3&5&3\\-6&20&9\\-3&25&9\end{pmatrix}. $$

If $U,V\in M_3(\mathbb{Z})$ are $3\times 3$-matrices with determinant $\pm1$, then $\{Ue_1,Ue_2,Ue_3\}$ is again a basis for the domain and $\{V^{-1}e_1,V^{-1}e_2,V^{-1}e_3\}$ is a basis for the codomain (both of which are $\mathbb{Z}^3$). The matrix for $f$ in these bases is $UAV$.

If $U$ and $V$ are as above, then they are a product of elementary matrices and these elementary matrices act by elementary row/column operations (since we are working over $\mathbb{Z}$, we can only rescale rows/columns by $\pm1$). Smith's algorithm describes how to find a basis for the domain ($\mathbb{Z}^3$) and the codomain ($\mathbb{Z}^3$) so that the matrix for $f$ is diagonal. My computation shows that we can find $U$ and $V$ such that $$ UAV=\begin{pmatrix}1&0&0\\0&3&0\\0&0&0\end{pmatrix} $$ I read from this that $M$ is 1-dimensional, $\mathrm{im}(f)\cong\mathbb{Z}^2$, and $\mathbb{Z}^3/\mathrm{im}(f)\cong\mathbb{Z}_3\times\mathbb{Z}$.

  • $\begingroup$ Thank you. Nevertheless, what I asked is to compute $\mathbb{Z}^3/M$ and that is what gave me trouble. I like the thorough answer and I have upvoted it, but it doesn't adress the main issue on my question. Maybe, from your answer, one could argue that $Ue_3$ is the kernel and therefore compute $\mathbb{Z^3}/ <Ue_3>$ ? $\endgroup$ – Nick A. Feb 8 '19 at 6:22
  • $\begingroup$ As I said in the post $M\cong\mathbb{Z}$. I think your teacher is confused and meant $M$ to be $im(f)$. $\endgroup$ – David Hill Feb 8 '19 at 15:05
  • $\begingroup$ Yes, $M \cong Z$ but that doesn't mean that $Z/M=Z^2$. For example, $Z \cong 2Z$ but $Z/Z={0}$ whereas $Z/2Z=Z_2$. I mean that we must know excatly the generator of $M$,not only that it has one. Maybe I am undestanding something wrong ? $\endgroup$ – Nick A. Feb 8 '19 at 16:18
  • $\begingroup$ I'm not sure I can state this more clearly: Smith's algorithm determines a basis $\{u_1,u_2,u_3\}$ for the domain and a basis $\{v_1,v_2,v_3\}$ for the codomain such that $f(u_1)=v_1$, $f(u_2)=3v_2$ and $f(u_3)=0$, so $M=\mathbb{Z}u_3$. This yields $\mathbb{Z}^3/M=\mathbb{Z}u_1+\mathbb{Z}u_2=\mathbb{Z}^2$. $\endgroup$ – David Hill Feb 8 '19 at 16:27
  • $\begingroup$ Ok, now I understand. Thank you very much for your thorough answer! $\endgroup$ – Nick A. Feb 8 '19 at 16:33

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