Problem. (a) Write down explicitly the rules for addition and multiplication in $\mathbb{Z}_{p}$.

(b) Show that $(\mathbb{Z}_{p^{i}},\varphi_{ij})$ where $\varphi:\mathbb{Z}_{p^{j}} \to \mathbb{Z}_{p^{i}}$ is given by $\varphi_{ij}(n + p^{j}\mathbb{Z}) = n + p^{i}\mathbb{Z}$ is a inverse system of finite rings and groups.

(c) Show that $$\mathbb{Z}_{p} = \varprojlim_{i}\mathbb{Z}_{p^{i}}.$$

(d) Show that the multiplicative group of units in $\mathbb{Z}_{p}$ is isomorphic to the direct product of $\mathbb{Z}_{p}$ with a cyclic group of order $\max\{p-1,2\}$.

I have no problems with (a), (b) and (c).

For item (d), I dont know how to approach this problem. I take an arbitrary cylic group of order $\max\{p-1,2\}$ and I tried to construct a isomorphism between the groups, but it doesn't works (at least, I cannot see how to do that). I would like some hints and approaches to follow.

Also, in the item (c), I used the inverse system given in (b) and I built a bijection between $\mathbb{Z}_{p}$ and $\varprojlim \mathbb{Z}_{p^{i}}$, considering $\varprojlim \mathbb{Z}_{p^{i}}$ as a subgroup of $\prod_{i}\mathbb{Z}_{p^{i}}$. Generally, this is the standard approach to problems like that. But I would like to know any other alternative approach.

Thanks for the advance.

Edit. I read the references in comment below, but it uses some things that the book doesn't comment on (like exact sequences, Hensel's lemma, for example). Basically, the book defines $\mathbb{Z}_{p}$ as the set of formal infinite sums and proves only the necessary results to find the completion of $\mathbb{Z}$, I mean: the book doesn't develop until now $p$-adic theory. So, I would like to know if there is another proof.

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    $\begingroup$ I think this is answered here together with here, (although there may be a bit more work to do when $p=2$). $\endgroup$ – Derek Holt Apr 26 at 12:02
  • $\begingroup$ Thank you, @DerekHolt! I will check the links! $\endgroup$ – Lucas Corrêa Apr 29 at 23:07
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    $\begingroup$ With the restrictions imposed on what methods are allowed, the natural thing to do is to compute the unit groups of the finite rings $\Bbb Z/p^j \Bbb Z$ which make up the inverse limit (and for which, by the way, I find "$\Bbb Z_{p^j}$" a horrible notation -- what book is that from?). $\endgroup$ – Torsten Schoeneberg May 7 at 18:12
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    $\begingroup$ @TorstenSchoeneberg let me see if I got your idea. I should to show that the unit group of $\mathbb{Z}/p^j \mathbb{Z}$ is isomorphic to something and when I calculate the inverse limit of something, the answer will be $\mathbb{Z}_{p} \times C_{n}$? $\endgroup$ – Lucas Corrêa May 9 at 14:37
  • $\begingroup$ @TorstenSchoeneberg, this question was rewritten for a friend, but its from Wilson's book "Profinite Groups", but Wilson uses $\mathbb{Z}/p^j \mathbb{Z}$. $\endgroup$ – Lucas Corrêa May 9 at 14:40

I'll follow up my comment with a sketch of how I assume the exercise is meant to be solved. People with knowledge in $p$-adics will see I try to recover the standard (for $p \neq 2$) $\Bbb Z_p^\times \simeq \mu(\Bbb Q_p) \times (1+p\Bbb Z_p)$ -- and Teichmüller representatives/roots of unity on the first factor, and the second factor being $\simeq (1+p)^{\Bbb Z_p} \simeq (\Bbb Z_p, +)$ -- with arithmetic congruences modulo $p$-powers.

Let's look at the rings $\Bbb Z/p^j\Bbb Z$. Everyone since Euler knows that the unit group $(\Bbb Z/p^j\Bbb Z)^\times$ has order $(p-1)\cdot p^{j-1}$, which suggests it's of the form $C_{p-1} \times C_{p^{j-1}}$, where $C_n$ denotes the cyclic group of order $n$. (That's actually not the case for $p=2$ ($j\ge 3$), see below, but indeed for all other primes.) Well great: If that is the case, and furthermore if we can establish these isomorphisms in a compatible way when we let $j$ vary, then we have $$\Bbb Z_p^\times \simeq\varprojlim (\Bbb Z/p^j\Bbb Z)^\times \simeq \varprojlim (C_{p-1} \times C_{p^j}) \simeq \varprojlim C_{p-1} \times \varprojlim C_{p^j} \simeq C_{p-1} \times \Bbb Z_p$$ (there's a few things to check here maybe, but this chain of isomorphisms should be straightforward).

For $p \neq 2$, such compatible isomorphisms $(\Bbb Z/p^j\Bbb Z)^\times \simeq C_{p-1} \times C_{p^{j-1}}$can be found like this:

  • First, the $C_{p^{j-1}}$-part. I claim that the element $1+p$ (formally, its various residues mod $p^j$) generates such a cyclic subgroup. Namely, Let $y_j=$ the residue of $1+p$ in $\Bbb Z/p^j\Bbb Z$.

Check that the order of $y_j$ in $(\Bbb Z/p^j\Bbb Z)^\times$ is $p^{j-1}$, i.e. that $(1+p)^{p^k} \not \equiv 1$ mod $p^{j}$ for $k <j-1$, but $(1+p)^{p^{j-1}} \equiv 1$ mod $p^{j}$. For this, you need some playing with $p$-divisibility of binomial coefficients $\binom{p^k}{n}$, and at some point you really need $p \neq 2$.

Compatibility is clear, as the projection $(\Bbb Z/p^j\Bbb Z) \rightarrow (\Bbb Z/p^i\Bbb Z)$ sends $y_j$ to $y_i$.

  • The $C_{p-1}$ part needs different considerations. A crucial fact here, proved by induction, is $$a \equiv b \text{ mod } p \Rightarrow a^{p^{j-1}} \equiv b^{p^{j-1}} \text{ mod } p^j \quad (*)$$ With this, for $a \in\lbrace 1, ..., p-1\rbrace$, check that the residues of $a^{p^{j-1}}$ in $\Bbb Z/p^j\Bbb Z$ are distinct from each other and form a subgroup isomorphic to $(\Bbb Z/p\Bbb Z)^\times$, which is cyclic of order $p-1$.

Choose and fix $2 \le a \le p-1$ such that its residue mod $p$ is a generator of $(\Bbb Z/p\Bbb Z)^\times$. Then $x_j :=$ the residue of $a^{p^{j-1}}$ in $(\Bbb Z/p^j\Bbb Z)^\times$ is a generator of the $C_{p-1}$ we want, and by the fact $(*)$ and iterations of Fermat's little theorem, we also have that the projection $(\Bbb Z/p^j\Bbb Z) \rightarrow (\Bbb Z/p^i\Bbb Z)$ sends $x_j$ to $x_i$.

Finally, for $p=2$, the unit group turns out to be not $C_{2^{j-1}}$, but $C_2 \times C_{2^{j-2}}$ (for $j \ge 3$). To adapt the above, for $y_j$ instead of $1+p$ take $1+p^2$ (i.e. $5$) mod $2^j$, and show its order in the unit group is $2^{j-2}$. Funnily, here the constant $C_2$ part is easy, namely it's generated by $x_j :=$ the residue of $-1$ in $(\Bbb Z/2^j\Bbb Z)$.

  • $\begingroup$ Thank you for this great answer! I will check the details with attention, but it seems very clear to me! $\endgroup$ – Lucas Corrêa May 13 at 19:57

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