6
$\begingroup$

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!

$\endgroup$
5
  • $\begingroup$ I remember finding that argument hard to follow the first time I saw it. I recommend that you find another book on $p$-adic analysis and learn about the $p$-adic exponential and logarithm. For $p \not= 2$, the $p$-adic exponential function defines an isomorphism $p\mathbf Z_p \rightarrow 1 + p\mathbf Z_p$ and its inverse is the $p$-adic logarithm. Since $p\mathbf Z_p \cong \mathbf Z_p$, we get $1 + p\mathbf Z_p \cong \mathbf Z_p$. The $2$-adic exponential function defines an isomorphism $4\mathbf Z_2 \rightarrow 1 + 4\mathbf Z_2$. Thus $1 + 4\mathbf Z_2 \cong \mathbf Z_2$. $\endgroup$ – KCd May 24 '15 at 22:50
  • $\begingroup$ The $p$-adic exponential is an isometry where it converges: $|e^x - e^y|_p = |x-y|_p$, so it identifies $p^n\mathbf Z_p$ with $1 + p^n\mathbf Z_p$ for $n \geq 1$ when $p \not= 2$ and for $n \geq 2$ when $p = 2$. $\endgroup$ – KCd May 24 '15 at 22:51
  • $\begingroup$ Our teacher told us that he will use this book. So i preferably would like to understand this above, because I think I will see similar things in the lecture and exercises. $\endgroup$ – Epsilondelta May 24 '15 at 23:06
  • $\begingroup$ And I too took a course with that book. I stand by what I wrote before that trying to make sense of the isomorphism using the $p$-adic exponential and logarithm is more intuitive. I didn't understand that proof when I first saw it and I was able to handle the rest of the book. $\endgroup$ – KCd May 24 '15 at 23:16
  • 1
    $\begingroup$ What Serre is essentially saying is that $1 + p$ (for $p \not= 2$) or $5 = 1 + 4$ (for $p = 2$) are topological generators of $U_1$ (for $p \not= 2$) or $U_2$ (for $p = 2$). That is, the homomorphism $f \colon \mathbf Z \rightarrow U_1$ where $f(n) = (1+p)^n$ for $p \not= 2$ is $p$-adically continuous and it extends by (uniform) continuity to an isomorphism $\mathbf Z_p \rightarrow U_1$ (similar result for $f(n) = 5^n$ when $p = 2$). If you understand inverse limits then the commutative diagram says the projective system on the right can be identified with the projective system on the left. $\endgroup$ – KCd May 24 '15 at 23:19
2
$\begingroup$

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 ....

$\endgroup$
3
  • $\begingroup$ Yes sorry for the typing errors (copy paste) also fot the $\theta$'s i think u have noticed. Yes the commutativity gives that an element of $lim←Z/p^{n}Z$ maps to an element of $lim←U1/Un$, and the map is injective since the $\theta$'s are injective. But you should give an argument to shcow that is surjective..... u can use the maps $\theta_{n,\alpha}^{-1}$ to get an inverse map or to show surjectivity. $\endgroup$ – user225222 May 25 '15 at 23:38
  • $\begingroup$ The commutativity comes from the fact that the maps are "natural", i.e the map $Z/p^i Z\to U_1/U_i $ comes from a morphism $Z\to U_1/U_i$ given by $z\mapsto \alpha_n^z$. Anyway u can check that : $f_n\theta_{n+1,\alpha}(z)=f_n(\alpha_{n+1}^z)=f_n(\alpha_{n+1})^z=\alpha_{n}^z= \theta_{n,\alpha}\pi_n(z)$ $\endgroup$ – user225222 May 25 '15 at 23:46
  • $\begingroup$ Hello. I got an open question: How can I see that $\underset{\leftarrow}{\mathrm{lim}} U_1/U_{i}=U_1$? $\endgroup$ – Epsilondelta May 26 '15 at 23:50
1
$\begingroup$

$\alpha\in U_1\setminus U_2$ 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.

$\endgroup$
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.