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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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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.

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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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