# Is ${\varprojlim}^1 (\mathbb{Z}, \cdot n_k)$ torsion(free)?

Let $$\mathbb{Z} \xleftarrow{\cdot n_1} \mathbb{Z} \xleftarrow{\cdot n_2} \mathbb{Z} \xleftarrow{\cdot n_3} \cdots$$ be an inverse system with $n_i$ a natural number bigger 1. Actually I'm trying to find out if the ${\varprojlim}^1$ term is not a torsion group, where ${\varprojlim}^1 (\mathbb{Z}, \cdot n_k)$ is defined as the cokernel of the map $$\mu : \prod_{k \in \mathbb{N}} \mathbb{Z} \longrightarrow \prod_{k \in \mathbb{N}} \mathbb{Z} \text{ , } (z_k)_k \longmapsto (z_k - n_k \cdot z_{k+1})_k \text{ .}$$ In the literature I found similiar examples saying it follows by an easy argument that ${\varprojlim}^1 (\mathbb{Z}, \cdot n_k)$ is torsionfree, but I don't seem to see it even after I worked on it for a while.

My first idea was to take this inverse system into an exact sequence $$0 \rightarrow \left(\mathbb{Z}, \cdot n_k \right) \rightarrow \left(\mathbb{Z}, \text{id} \right) \rightarrow \left( \mathbb{Z} \bigg/ \prod_{i=1} ^k n_i, \text{projections} \right) \rightarrow 0$$ so by the long exact lim-lim$^1$ sequence the problem reduces to take a look at $\varprojlim \left(\mathbb{Z} \big/ \prod_{i=1} ^k n_i \right) \bigg/ \mathbb{Z}$ where $\mathbb{Z}$ is canonically embedded in $\varprojlim \left(\mathbb{Z} \big/ \prod_{i=1} ^k n_i \right)$ by taking a number $z$ to the sequence $(z)_k$. But also at this point I got stuck, so I tried working with the definition of ${\varprojlim}^1$ even so without any result.

Has anybody a hint how to solve this problem?

I've decoded the problem a little, so someone might have an idea for the important step (comments welcome). As I mentioned above we have ${\varprojlim}^1 (\mathbb{Z}, \cdot n_k) \cong \varprojlim \mathbb{Z} \big/ \prod_{i=1}^{k} n_i \bigg/ \mathbb{Z}$. Every element in $\mathbb{Z} \big/ \prod_{i=1}^{k} n_i$ is uniquely represented by an $0 \leq a_k < \prod_{i=1}^{k} n_i$, which we can uniquely write as $a_k = x_0 + x_1 n_1 + ... + x_{k-1} \prod_{i=1}^{k-1} n_i$, where $0 \leq x_j < \prod_{i=1}^{j} n_i$. Since $\varprojlim \mathbb{Z} \big/ \prod_{i=1}^{k} n_i$ is the kernel of the above map $\mu$ with $\prod_{k \in \mathbb{N}}\mathbb{Z} \big/ \prod_{i=1}^{k} n_i$ instead of $\prod_{k \in \mathbb{N}} \mathbb{Z}$, we have $a_{k+1} \equiv a_k \text{ mod } \prod_{i=1}^{k} n_i$ for all $k \in \mathbb{N}$, so $a_{k+1}$ has a (unique) representation $x_0 + x_1 n_1 + ... + x_{k-1} \prod_{i=1}^{k-1} n_i + x_k \prod_{i=1}^{k} n_i$. Then an element in $\varprojlim \mathbb{Z} \big/ \prod_{i=1}^{k} n_i$ is not an element of $\mathbb{Z}$ if $x_k \neq 0$ for infinitely many $k$ in the above representation. So let $a_k = x_0 + x_1 n_1 + ... + x_k \prod_{i=1}^{k} n_i$ for every $k \in \mathbb{N}$ and $z \in \mathbb{Z} \backslash \{0\}$. We like to show $([za_k])_k \notin \mathbb{Z}$. For every $k$ we have

$$za_{k+1} = z (x_0 + ... + x_k \prod_{i=1}^{k} n_i) \equiv y_0 + ... + y_k \prod_{i=1}^k n_i \text{ mod } \prod_{i=1}^{k+1} n_i$$

for some $0 \leq y_j < \prod_{i=1}^j n_i$.

(More precisely for each $x_m$ we can write $zx_m \equiv b_0 + ... + b_{m-1} \prod_{i=1}^{m-1} n_i \text{ mod } \prod_{i=1}^{m} n_i$ for $0 \leq b_j < \prod_{i=1}^j n_i$.)

Now the question is if we can choose infintely many $x_j \neq 0$ such that there are infinitely many of those $y_j$ which are nonzero.

• Are all the $n_k$'s distinct? – leibnewtz Jun 29 '17 at 14:36
• No, just any natural numbers greater 1. – Lukas Jun 29 '17 at 14:43
• Your question seems to be quite interesting in fact and probably can be generalized, though I don't understand why you define $\varprojlim^1$ to be the cokernel. Don't you mean by $\varprojlim^1$ the 1-higher inverse limit? – user321268 Jun 29 '17 at 20:13
• Perhaps my previous comment seems a bit vague how do you prove that this is the cokernel of $\mu$? – user321268 Jun 29 '17 at 20:16
• I've never heard of 'higher inverse limits'. Actually I'm working on a topological problem along Hatcher's book and he defines lim1 exactly in this cokernel way. But nevertheless it sounds interesting, so I might take a look at this. – Lukas Jun 30 '17 at 5:54

Here is a counterexample. We set $n_{k} = 3$ for all $k \in \mathbb{N}$. Take $\mathbf{y} := (1,1,1,\dotsc) \in \prod_{k \in \mathbb{N}} \mathbb{Z}$. Then the image of $\mathbf{y}$ in $\mathrm{coker}(\mu)$ is $2$-torsion since $2\mathbf{y} = \mu(-\mathbf{y})$. We show that $\mathbf{y}$ is not in the image of $\mu$. Suppose $\mathbf{y} = \mu(\mathbf{z})$ where $\mathbf{z} = (z_{1},z_{2},z_{3},\dotsc) \in \prod_{k \in \mathbb{N}}$. We have $z_{k+1} = \frac{z_{k}-1}{3}$ for all $k \in \mathbb{N}$. Note that all the $z_{k}$ are $1 \pmod{3}$, in particular are all nonzero. If $x$ is a real number, then either $x < -\frac{1}{2}$ or $\frac{1}{4} < x$ implies that $|x| > |\frac{x-1}{3}|$. This implies that the sequence $|z_{1}|,|z_{2}|,|z_{3}|,\dotsc$ is a strictly decreasing sequence of positive integers, which is a contradiction.