There are a few ways to define the $p$-adic numbers.

If one defines the ring of $p$-adic integers $\mathbb Z_p$ as the inverse limit of the sequence $(A_n, \phi_n)$ with $A_n:=\mathbb Z/p^n \mathbb Z$ and $\phi_n: A_n \to A_{n-1}$ (like in Serre's book), how to prove that $\mathbb Z_p$ is the same as $$\mathbb Z_p=\left \{ \sum_{i=n}^\infty a_i p^i \ | \ n \in \mathbb Z, \ a_i \in\left \{ 0,1,...,p-1 \right \} \right \} \ \ ?$$

I found a proof here but it's very long and technical. So maybe there are other ways to prove it? I'm looking for a shorter proof.

Best regards.

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3 Answers3


Your displayed equation is wrong. The power series should start with non-negative powers of $p$, i.e. the sum should start at 0 instead of an arbitrary integer $n$.

Intuitively, the two definitions are the same because both say that a $p$-adic integer is a compatible choice of residues modulo higher and higher powers of $p$. First, there are $p$ choices for a residue modulo $p$. Once you have made your choice (which is given by $a_0$ in your second definition), you have $p$ possible choices for the residue modulo $p^2$. There are of course $p^2$ choices for this residue altogether, but only $p$ of those will be compatible with the first choice you have made, i.e. will reduce to $a_0$ modulo $p$. This second choice is given by $a_0+pa_1$ in your second definition. You keep going in this way, fixing a residue modulo $p^3$, $p^4$, etc. Every time, there are only $p$ possible new choices that will be compatible with the ones you have already made. If this was the ordinary integers, your sequence $a_i$ would stabilise at 0 at some point, but in the $p$-adics you are allowed to keep going forever.

To sum it up, the isomorphism between the two rings takes an infinite sequence $\sum_{i=0}^\infty a_ip^i$ and sends it to the inverse system $$\left(\sum_{i=0}^n a_ip^i\in \mathbb{Z}/p^n\mathbb{Z}\right)_n,$$ with $\phi_n$ given by reduction-mod-$p^n$ maps. It is completely trivial (and you should do it, not look it up!) to check that this is indeed a ring isomorphism.

Alex B.
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You could try to prove that $\left\{ \sum_{i=0}^\infty a_i p^i \mid a_i \in\{ 0,1,...,p-1 \} \right\}$, together with all the natural maps $\pi_n$ to each of the $\mathbb Z/p^n \mathbb Z$, satisfies the universal property of an inverse limit.

That is, suppose you have any commutative ring $B$ with ring homomorphisms $q_n \colon B \to \mathbb Z/p^n \mathbb Z$ satisfying $\phi_n \circ q_n = q_{n-1}$, show there exists a unique ring homomorphism $\psi \colon B \to \left\{ \sum_{i=0}^\infty a_i p^i \mid a_i \in\{ 0,1,...,p-1 \} \right\}$ with $q_n = \pi_n \circ \psi$.

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Let $ p$ be a prime number. Consider ring of formal power series in $ p$ with coefficients from $ \mathbb{Z}/p\mathbb{Z}$ i.e., the set $$F=\{a_0+a_1p+a_2p^2+\cdots : a_i\in \mathbb{Z}/p\mathbb{Z}\}.$$ Addition and multiplication is defined as addition and multiplication of formal power series.

For the same $p$ as above, we define what is called completion of $\mathbb{Z}$ with respect to $p$. What ever it may be, as a set it is $$C=\{(x_1,x_2,\cdots): x_n\in \mathbb{Z}/p^n\mathbb{Z}, x_{n+1}-x_n\in p^n\mathbb{Z}\}.$$ Addition and multiplication is defined componentwise.

We want to see that these two gives isomorphic rings. For that, we at least need to have some bijective correspondence between $F$ and $G$.

Let $(x_1,x_2,\cdots)\in C$. Then, we want to get an element of the form $a_0+a_1p+\cdots$ with $a_i\in \mathbb{Z}/p\mathbb{Z}$.

We want $a_0\in \mathbb{Z}/p\mathbb{Z}$. We have $x_1\in \mathbb{Z}/p\mathbb{Z}$. Set $a_0=x_1$.

We want $a_1\in \mathbb{Z}/p\mathbb{Z}$. We see that $x_2\in \mathbb{Z}/p^2\mathbb{Z}$ such that $x_2-x_1\in p\mathbb{Z}$ i.e., $p$ divides $x_2-x_1$. Let $x_2-x_1=pt$ then $t<p$ as $x_2,x_1<p^2$.

Set $a_1=t$. See that $x_2=a_0+a_1p$.

With this we can guess what should the other $a_n$ have to be. We define $a_n=\dfrac{x_{n+1}-x_{n}}{p^n}\in \mathbb{Z}/p\mathbb{Z}$.

We also have $x_n=a_0+a_1p+a_2p^2+\cdots+a_{n-1}p^{n-1}$.

So, given $(x_1,x_2,\cdots)$ we have an element $a_0+a_1p+a_2p^2+\cdots$.

Let $a_0+a_1p+a_2p^2+\cdots$. We have to assign $(x_1,x_2,\cdots)$ with properties as above. By above observation, it is natural to assign $x_n=a_0+a_1p+a_2p^2+\cdots$

It is clearly bijective and it is upto you to check the isomorphism as rings.

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    those two rings are not isomorphic! for example, in the ring of formal power series, $1+1+1..+1=0$ if we sum $p$ times however in the $p$-adic numbers this is $p \neq 0$! – M. Van Feb 06 '20 at 19:26
  • -1 indeed. There is a a bijection between $C (=\mathbb Z_p)$ and the infinite tuples $(a_0, a_1, ...) \in \mathbb F_p^{\mathbb N}$ as you construct, but this bijection is **NOT** compatible with addition nor multiplication if addition and multiplication on those tuples are defined via viewing them as formal power series. Also not if multiplication is defined "componentwise". Rather, following through this approach the "right" addition and multiplication on those tuples is given by the Witt polynomials. Cf. https://en.wikipedia.org/wiki/Witt_vector, https://mathoverflow.net/a/113943/27465 – Torsten Schoeneberg Nov 16 '21 at 17:48