Structure of the group $\{1+p\mathbb Z_p \}$

Question

In preparation for algebraic number theory I am reading Serre : A course in Arithmetic.

I stuck in understanding a proof (p.17):

Notation: $U_n=1+p^n\mathbb Z_p$

Actually there are many things which I don't understand...

• Why it holds that $(\alpha_n)^{p^{n-2}}\neq 1$ and $(\alpha_n)^{p^{n-1}}= 1$?

• Why $\phi_{n,\alpha}$ is an isomorphism? I see that is is an homomorphism, but proving that its also bijective is not that easy for me, because its hard for me to imagine the group $U_1$/$U_n$

• I also dont understand the last tree lines

I know that I did understand only few things. In my opinion this book leaves out many arguments, so i hope that someone can make this clear for me.

Thanks in advance!

Answer 1

1) You have $\alpha^{p^i}\in U_{i+1}-U_{i+2}$:

a) For i=n-1 you get that $\alpha^{p^{n-1}}\in U_{n}$, therefore the image of $(\alpha_n)^{p^{n-1}} $ of $(\alpha)^{p^{n-1}} $ in the quotient $U_1/U_n$ is 1.

b) Taking i=n-2, you get that $\alpha^{p^{n-2}}\notin U_n$ hence $\alpha_n^{p^{n-2}}\neq 1$. ($\alpha^{p^{n-2}}$ is not in the kernel $U_n$ of the quotient map $U_1\to U_1/U_n$)

2) The fact that $\alpha_n^{p^{n-2}}\neq 1$ and $\alpha^{p^{n-1}}=1$ shows that the order of $\alpha_n$ is $p^{n-1}$ wich is exactely the order of $U_1/U_n$ so the map is byjective. (the cardinal of a subgroup generated by an element of finite order is the element's order).

3)This is a general fact about inverse limits : Denote $\pi_i: \mathbb Z/p^i \mathbb Z \to\mathbb Z/p^{i-1} \mathbb Z $ and $f_i:U_1/U_{i}\to U_1/U_{i-1}$ the maps corresponding to the vertical arrows of the diagram.

Recall that $\underset{\leftarrow}{\mathrm{lim}} \mathbb Z/p^i \mathbb Z$ is the set of tuples $(a_1,a_2,a_3,a_4\cdots )$ ($a_i\in \mathbb Z/p^i \mathbb Z$) of the infinite product $\mathbb Z/p \mathbb Z \times \mathbb Z/p^2 \mathbb Z\times \cdots$,such that $a_{i-1}=\pi_i(a_{i+1})$ for all $i$ ;

and $\underset{\leftarrow}{\mathrm{lim}} U_1/U_{i}$ is the set of tuples $(b_1,b_2,b_3,\cdots )$ ($b_i\in \mathbb U_1/U_{i+1}$) of the infinite product $ U_1/U_{2} \times U_1/U_{3}\times U_1/U_{4}\cdots$,such that $b_{i-1}=f_i(b_{i+1})$ dor all $i$.

Know the collection of maps $\theta_{n,\alpha}$ gives an isomorphisme betwenn the two infinite products $\mathbb Z/p \mathbb Z \times \mathbb Z/p^2 \mathbb Z\times \cdots$ and $U_1/U_{2} \times U_1/U_{3}\times U_1/U_{4}\cdots$.

The commutativity of the diagrams shows that if $(a_1,a_2,\cdots ) \in \underset{\leftarrow}{\mathrm{lim}} \mathbb Z/p^i \mathbb Z$ then $(\theta_{1,\alpha}(a_1),\theta_{2,\alpha}(a_2),\cdots)\in \underset{\leftarrow}{\mathrm{lim}} U_1/U_{i}$.

This gives an injective map $\underset{\leftarrow}{\mathrm{lim}} \mathbb Z/p^i \mathbb Z \to \underset{\leftarrow}{\mathrm{lim}} U_1/U_{i}$ i let u finish the proof ....

Answer 2

$\alpha\in U_1\setminus U_2$ is a $p$-adic integer of the form $1+p\beta$, where $v_p(\beta)=0$. Think of $p$-adic integers more or less as power series in $p$. The n $\alpha\in U_1\setminus U_2$ does have a non-zero term in $p$. An element $a$ lies in $U_n\setminus U_{n+1}$ if its first term with positive valuation has valuation $n$, i.e. if it can be written as $1+p^nu$, and $u$ has valuation $0$. In other words, $v_p(a-1)=n$.

A simple induction shows that $v_p(\alpha^{p^k}-1)=k$. For instance $\alpha^p=(1+p\beta)^p=1+p^2\beta+\frac{p(p-1)}2p^2\beta^2+\dots\,$ has the form $1+p^2\gamma$, where $\gamma$ is a unit.

Elements of $\mathbf Z/p^n\mathbf Z $ can be written in base $p$ with no more than $n$ digits, i.e. as a sum $\displaystyle\sum\limits_{k=0}^{n-1} c_kp^k$, where $c_k\in \{0 .. p-1\}$. That is exactly the way $\beta$ can be represented.

Now $\theta_{n,\alpha}$ is injective: indeed, if it were not, there would be a $z<p^n$ such that $\alpha^z=1+ p^{n+1}\gamma$, for some $\gamma$, i. e. $v_p(\alpha^z-1)\ge n+1$. But is is easy to check $v_p(\alpha^z-1)=v_p(\alpha)+v_p(z)=1+v_p(z)$. So $v_p(z)\ge n$, which is impossible by the above description.

This map is a bijection because it is injective and the sets have the same number of elements.

The last three line only mean that the inverse limit of an inverse system of isomorphisms is an isomorphism.

Answer 3

I guess there is a mistake: It should be $a_{i-1}=\pi_i(a_{i})$ similarly for $f_i$. A question: I see that if the diagram is commutative then: If we take for example $a_2\in \mathbb Z/p^2\mathbb Z$, then $\theta_{2,\alpha}(\pi_2(a_2))=f_2(\theta_{3,\alpha}(a_2))$, hence an element of $\lim_{\leftarrow}\mathbb Z/p^{n-1}\mathbb Z$ maps to an element of $\lim_{\leftarrow}U_1/U_n$. Is this correct? Now we must show that $\theta$ is an isomorphism, but this is actually clear, right? Because the $\theta_{n,\alpha}$ are. There is also another thing: Why this diagram commutes? Is this a clear fact? Also thank you very much! Your answer helped a lot!