Showing that $\hat{\mathbb{Z}}/m\hat{\mathbb{Z}} \cong \mathbb{Z}/m\mathbb{Z}$. In particular showing that $ker(\phi) \subseteq m\hat{\mathbb{Z}}.$

Question

As the title says I have to show that $\hat{\mathbb{Z}}/m\hat{\mathbb{Z}} \cong \mathbb{Z}/m\mathbb{Z}$. I know this question has been asked before at: How to show $\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}\cong \mathbb{Z}/n\mathbb{Z}$. But I'm having trouble with showing that $ker(\phi) \subseteq m\hat{\mathbb{Z}}$ in particular. Here $\phi$ is defined as the projection from $\hat{\mathbb{Z}}$ to the $m$th coordinate. I've already proven that $\phi$ is surjective and that $m\hat{\mathbb{Z}} \subseteq ker(\phi)$. So if I prove that $ker(\phi) \subseteq m\hat{\mathbb{Z}}$ then I will have shown by the isomorphism theorem that $\hat{\mathbb{Z}}/m\hat{\mathbb{Z}} \cong \mathbb{Z}/m\mathbb{Z}$.

Note that I don't know what inverse limits are or what topoligical groups are. In the exercise the profinite completion of $\mathbb{Z}$ is just defined as the subring $\{(a_0,a_1,...)\in \prod_{n\geq 1}\mathbb{Z}/n\mathbb{Z}: \text{ for all }n\in \mathbb{N} \text{ and } d|n, a_n = a_d \text{ mod } d\}$.

So given an $(a_0,a_1,...)\in ker(\phi)$ I want to show that there is a $(b_0,b_1,...)\in \hat{\mathbb{Z}}$ such that $(a_0,a_1,...) = m(b_0,b_1,...)$.

I was able to show that for an $(a_0,a_1,...)\in ker(\phi)$ there is a $(b_0,b_1,...)\in \prod_{n\geq 1}\mathbb{Z}/n\mathbb{Z}$ with $m(b_0,b_1,...) = (a_0,a_1,...)$. But this element was not unique nor was it always in $\hat{\mathbb{Z}}$. I do see that for an $n\in \mathbb{N}$ with $gcd(n,m) = 1$ the $b_n$ is uniquely defined and I was also able to show that this element does satisfy for all $d|n, a_n = a_d \text{ mod } d$. That is to say that because $gcd(n,m) = 1$ we know that there is a unique $m^{-1}\in \mathbb{Z}/n\mathbb{Z}$. And so we must choose $b_n = m^{-1}a_n \text{ mod } n$. Now say $d|n$ then $gcd(d,m) = 1$ and so $b_d = m^{-1}a_d \text{ mod } d$. Note that if $mm^{-1} = 1 \text{ mod } n$ then also $mm^{-1} = 1\text{ mod } d$. By definition we also have that $a_n = a_d \text{ mod } d$. And so $b_n = m^{-1}a_n = m^{-1}a_d = b_d \text{ mod } d$.

But if $gcd(n,m) = r > 1$ then I don't know what to do. I know that $a_n = a_r \text{ mod } r$ and $a_r = a_m = 0 \text{ mod } r$. And so $a_n = xr \text{ mod } n$ for some $x\in \mathbb{Z}/n\mathbb{Z}$. Then you could find integers $a,b$ such that $an+bm = r$ and so $a_n = xbm \text{ mod } n$. But taking $b_n = xb \text{ mod }n$ doesn't generally give an element in $\hat{\mathbb{Z}}$.

I've worked out an example to try and find a pattern. It sort of looks like the $(b_0,b_1,...)$ is uniquely defined, not that this is necessary of course. It looks like the restriction that $a_n = xr \text{ mod } n$ and that $a_k = a_n \text{ mod } n$ for some factor $k$ of $n$ are enough to determine $a_n \text{ mod } n$. But I stuck trying to do this.

A hint or a new approach to prove this would be appreciated.

Answer 1

Your definition of $\widehat{\Bbb{Z}}$ is exactly the projective limit one $\varprojlim \Bbb{Z}/n\Bbb{Z}$.

$a \in \ker(\phi)$ iff $a_m = 0\bmod m$.

Dividing by primes repeatedly to get the division by $m$ we can reduce to the case where $m=p$ is prime.

Let $k$ be the least integer such that $a_{p^j}\not \equiv 0 \bmod p^j$. If it doesn't exist then let $j=1$.

• For the $d=p^j k$ let $c_d = a_d/p$ where $a_d$ is viewed as an integer $\in 0\ldots d-1$ (which must be divisible by $p$) and $c_d$ is viewed as an element of $\Bbb{Z}/d\Bbb{Z}$.

• For the remaining $d$ not divisible by $p^j$ let $c_d \equiv c_{p^jd}\bmod d$.

Then $c\in \widehat{\Bbb{Z}}$ and $a=\underbrace{c+\ldots +c}_p$.

Answer 2

I have found an answer to my own question.

We define $b_n = a_{mn}/m \mod n$.

The reason why this is possible is because $a_{mn} = a_m = 0 \mod m$ and so $a_{mn} = mx \mod mn$ for some $x \in \mathbb{Z}/mn\mathbb{Z}$. At first this seems to give multiple answer depending on which $x$ was chosen because there could be multiple $x$ which are a solution to $a_{mn} = mx \mod mn$. But we see that if $x$ is a solution then $x+n,x+2n,...$ are the other solutions because $$a_{mn} = m(x+k) \mod mn$$ $$0 = mk \mod mn$$ $$0 = k \mod n$$ This means that when we reduce the solutions of $a_{mn} = mx \mod mn$ modulo $n$ we will always get the same outcome. So $b_n$ is not dependent on which solution was found.

Now we just have to verify that $a_n = mb_n \mod n$ and that for $d|n$ we have $b_n = b_d \mod d$. Once you do this you will have shown that $(b_1,b_2,...)\in \hat{\mathbb{Z}}$ and that $(a_1,a_2,...)\in m\hat{\mathbb{Z}}$. And so $ker(\phi) \subset m\hat{\mathbb{Z}}$.

In the spirit of giving a hint I'll leave the last two verifications in a spoiler block so that you can try it out yourself.

To show that $a_n = mb_n \mod n$ we do the following: $$mb_n = a_{mn} = a_n \mod n$$ And to prove that for $d|n$ we have $b_n = b_d \mod d$ we just see that: $$a_{mn} = a_{md} \mod mn$$ And so $$a_{mn}/m = a_{md}/m \mod n$$ Therefore $$b_n = b_d \mod n$$