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Lately, I've been working on direct limits. In particular, given

$$\mathbb{Z}^n \xrightarrow{M} \mathbb{Z}^n \xrightarrow{M} \mathbb{Z}^n \xrightarrow{M} \cdots$$

where $M$ is an $n \times n$ matrix over $\mathbb{Z}$, then

(1) if the eigenvalues of $M$ are integers, the direct limit is isomorphic to the sum of $\mathbb{Z}[\frac{1}{e_i}]$'s, where $e_i$ is the $i$th eigenvalue. If $0$ is an eigenvalue, then just reduce the degree $n$ by the number of $0$ eigenvalues.

Is there a good reference to draw this conclusion from? Any book or online articles would be of great help!

(2) on the event that the eigenvalues are irrational, then the determinant is considered. For instance, the direct limit becomes $\mathbb{Z}[\frac{1}{d}]^n$, where $d$ is the determinant.

I am struggling a bit on this, and any help would be appreciated. Thanks!

Eric
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2 Answers2

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Let $M$ be an $R$-module and $A\in R$ not a zero divisor. Suppose we wish to compute the direct limit of the system $M\to M\to M\to\cdots$ given by repeated application of $A$. There is a nice trick that can be used to reinterpret this system as taking place entirely within the "localization" $(A)^{-1}M$:

$$\large\begin{array}{cccccccc} M & \xrightarrow{~A~} & M & \xrightarrow{~A~} & M & \xrightarrow{~A~} & M & \xrightarrow{~A~} & \cdots \\[4pt] \uparrow & & \uparrow & & \uparrow & & \uparrow \\[4pt] M & \hookrightarrow & A^{-1}M & \hookrightarrow & A^{-2}M & \hookrightarrow & A^{-3}M & \hookrightarrow & \cdots \end{array} $$

The up arrows represent multiplication by $A$. Check that all squares in the diagram commute. Thus the upper and lower direct systems are equivalent and their direct limits are equal.

By $(A)$ here I mean the set of all powers of $A$. Another way to think about it as the "extension of scalars" given by the tensor product $(A)^{-1}M\cong (A)^{-1}R\otimes_RM$. Note $(A)^{-1}R$ (as a ring) is the same as $R[A^{-1}]$. The direct limit, then, is $\bigcup A^{-n}M=R[A^{-1}]M=(A)^{-1}M$. An easier example (than integer matrices) is the integers by itself, two examples are covered in this question.

The rest of the exercise I assume wants Smith normal form to be used. Not sure about the ${\Bbb Z}[d^{-1}]$.

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anon
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  • Thank you for this. To be honest, I am not so comfortable with tensor products, and so I will need time to understand every detail of your response. I appreciate it. Thanks! – Eric Mar 14 '13 at 14:06
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    @Eric: The tensor product is only a way to formalize the intuition going on (behind extension of scalars). For your purposes, you can view ${\bf Z}^n$ as sitting inside ${\bf Q}^n$, and $E={\rm End}({\bf Z}^n)$ as sitting inside ${\rm End}({\bf Q}^n)$. Then the direct limit of ${\bf Z}^n\to{\bf Z}^n\to\cdots$ under multiplication by a matrix $A\in E$ (which is invertible in ${\rm End}({\bf Q}^n)$ but not necessarily in $E$) is isomorphic to $E[A^{-1}]{\bf Z}^n\subseteq{\bf Q}^n$; to explicitly characterize this submodule it may help to recoordinatize ${\bf Z}^n$ so that eigenvalues kick in. – anon Mar 14 '13 at 19:47
  • Thanks pal! Just a clarification. If $A$ is not a zero divisor, does it imply that $A$ doesn't have any zero eigenvalues? I am just thinking on the premise that $R = \mathbb{Z}$ and $M$ is an $n\times n$ matrix over $\mathbb{Z}$. – Eric Oct 08 '13 at 09:53
  • @Eric A nonzero matrix $A$ is a zero divisor iff it has a zero eigenvalue, so yes. (You are actually thinking about $M=\Bbb Z^n$ as a module over $R=M_n(\Bbb Z)$ with the element $A\in R$ an integer matrix.) – anon Oct 08 '13 at 10:04
  • Alright, last clarification. I already understand your explanation above. However, what's a possible explanation to simply consider a non-zero divisor $A$ rather than a more general matrix, say $A'$, where $A$ and $A'$ have the same set of non-zero eigenvalues?

    I mean, computationally, the direct limit of the system $(\mathbb{Z}^n, A)$ is isomorphic to $(\mathbb{Z}^n, A')$, but is there a general way to reduce $A'$ to $A$?

    Again, I cannot express how grateful I am for your help!

     – Eric Oct 08 '13 at 14:04
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    @Eric One can of course consider a more general matrix $A$. The only reason I do not in this answer is because this trick only works directly if the element $A\in R$ has an inverse in some bigger ring (which requires it to not be a zero divisor in $R$). Some kind of coordinate change could probably handle when the more general situation, but I haven't thought about it deeply enough. – anon Oct 08 '13 at 21:41
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Let more generally $F$ be a finitely generated free module over a PID $A$ and $M : F \to F$ be an endomorphism. Then the smith normal form tells us that there is a basis of $F$ such that $M$ becomes diagonal, say $M = \mathrm{diag}(d_1,\dotsc,d_n)$ with $d_1 | \dotsc | d_n$ in $A \setminus \{0\}$. In other words, $M$ is isomorphic to the direct sum of the the endomorphisms $d_i : A \to A$. The colimit of $A \xrightarrow{d_i} A \xrightarrow{d_i} \dotsc$ is isomorphic to the localization $A_{d_i}$. Hence, the colimit of $F \xrightarrow{M} F \xrightarrow{M} $ is isomorphic to $\oplus_i A_{d_i}$.

Edit. This is not correct, since the smith normal form uses row and column operations, which means that we have to choose two bases which are different a priori. The isomorphisms above only hold when these bases are equal.

Martin Brandenburg
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