The $p$-adics and Related Concepts

Here we give a practical representation of $p$-adic numbers in terms of a power series in powers of $p.$ This perspective will connect our understanding of the $p$_adics as a Cauchy completion and as an inverse limit. We first do this for the $p$-adic integers and then explain how to extend it to all of $\mathbb Q_p$.

Recall that a $p$-adic integer $x$ can be given as an equivalence class of Cauchy sequences. We may take a representative to be a sequence of rational integers $(m_n)$ which is Cauchy with respect to $| \cdot |_p$. Every subsequence of $(m_N)$ is equivalent to $(m_N)$ and we choose a special subsequence as follows. The Cauchy condition implies that given $\epsilon = p^{-\ell}$ there exists $N_\ell$ that if $n > N_{\ell}$ then $| m_n – m_{N_\ell} | \leq p^{-\ell-1}$. We may do this in such a way that $N_{\ell+1} > N_{\ell}$, and we define the subsequence $(M_\ell)$ of $(m_n)$ by setting $M_{\ell} = m_{N_{\ell}}.$ Our construction implies that $$|M_{\ell} – M_{\ell + 1}| \leq p^{-\ell-1}.$$

Now we can expand any integer $M$ base $p$, $$M = \sum_{n=0}^{N(M)} a_n(M) p^n \quad \mbox{where} \quad a_n(M) \in \{0, 1, \ldots, p-1\}.$$ Here $a_n(M)$ is the $n$th digit of $M$ and $N(M)$ is the index of the last non-zero digit. Given $M_{\ell}$ we define $N_{\ell}$ to be the truncation of the base $p$ expansion of $M_{\ell}$ given by $$N_{\ell} = \sum_{n=0}^{\ell} a_n(M_{\ell}) p^n.$$

**Exercise:**

- Show that $(N_{\ell})$ is a Cauchy sequence equivalent to $(M_n)$.
- Show that $|N_{\ell + 1} – N_{\ell}|_p \leq p^{-\ell-1}$ implies $$N_{\ell + 1} = N_{\ell} + a_{\ell + 1}(M_{\ell + 1}).$$

This exercise implies that $N_{\ell}$ is the $\ell$th partial sum of the infinite base $p$ expansion $$\sum_{n=0}^{\infty} a_n(M_n) p^n.$$

**Definition:** The *canonical series* representation for $x \in \mathbb Z_p$ is given by the sequence of partial sums of $$\sum_{n=0}^{\infty} a_n(M_n) p^n.$$

The canonical series representation for elements in $\mathbb Z_p$ is also connected to the inverse limit definition of the $p$-adic integers. To see this, suppose $$x = \sum_{n=0}^N a_n x^n.$$ Then, the $(\ell+1)$th truncation of the canonical representation of $x$ is $$\sum_{n=0}^{\ell} a_n p^{\ell} \in \{0, 1, \ldots, p^{\ell + 1}-1 \},$$ which we may use to produce the projections $\pi_{\ell+1} : \mathbb Z \rightarrow \mathbb Z/p^{\ell + 1} \mathbb Z$. The sequence of projections, $(\pi_{\ell}(x))$ is then in the inverse limit $\varprojlim \mathbb Z/p^{\ell} \mathbb Z$. The homomorphisms $\varphi_{i,j} : \mathbb Z/p^j \mathbb Z \rightarrow \mathbb Z/p^i \mathbb Z$ are given by truncation, $$\varphi_{i,j}\left( \sum_{n=0}^{j-1} a_n p^n \right) = \sum_{n=0}^{i-1} a_n p^n,$$ which is equivalent to reduction modulo $p^{i}$.

**Exercise:** Verify that the $\varphi_{i,j}: \mathbb Z/p^j \mathbb Z \rightarrow \mathbb Z^i/p^i$ are homomorphisms such that for $i < j < k$, $\varphi_{i,k} = \varphi_{j,k} \circ \varphi_{i,j}$.

#### Series, Cosets & Balls

How do we represent balls in $\mathbb Z_p$ using series expansions? We get a clue by observing that under our correspondence between canonical series representatives and $\mathbb Z_p$ so that $$p \mathbb Z_p = \left\{ \sum_{n=1} a_n p^n : a_n \in \{0, 1, \ldots, p-1\} \right\}.$$ That is $p \mathbb Z_p$ consists of all series with constant term equal to 0. The cosets of $p \mathbb Z_p$, that is the other balls of radius $1/p$ consist of series whose constant coefficient is equal to one of the other representatives in $\{1, \ldots, p-1\}$

This correspondence continues. Two series $x$ and $y$ are in the same ball of radius $p^{\ell}$ if and only if their $\ell$th projections are equal: $\pi_{\ell}(x) = \pi_{\ell}(y)$. Put another way, $x$ and $y$ are in the same ball of radius $p^{\ell}$ if their series representations agree up to (at least) the $p^{\ell}$th term. In this situation, $r = \pi_{\ell}(x)$ is an integer in $\{0, 1, \ldots, p^{\ell} – 1 \}$ and $B(x, p^{-n}) = r + p^n \mathbb Z_p$.

#### Canonical Series on $\mathbb Q_p$

Recall that to compute the $p$-adic absolute value for a rational number $x$ we factor $x = p^{n} \frac{a}{b}$ (in lowest terms), and take $|x|_p = p^{-n}$. Put another way, if $x$ is a rational number that is not an integer, then there exists a negative power of $p$ such that $x = p^{-n} y$ for some $p$-adic integer $y$.

This is more generally true, if $x$ is in $\mathbb Q_p \setminus \mathbb Z_p$ then there exists a negative power of $p$ and $y \in \mathbb Z_p$ such that $p^{-n} x = y$. We can then get a series expansion for $x$ by taking the canonical series for $y$ and multiplying it by $p^{-n}$.

**Theorem:** If $x \in \mathbb Q_p$ then there exists negative integer $-n$ such that $$x = \sum_{m=-n}^{\infty} a_m p^m \quad \mbox{where} \quad a_m \in \{0, 1, \ldots, p-1\}.$$ That is, if $p$-adic integers correspond to power series, then $p$-adic rational numbers correspond to Laurent series.

#### Arithmetic On Series

Laurent series have their own arithmetic. That is we can perform addition, subtraction, multiplication and division of series using ordinary rules of algebra. Because $\mathbb Q_p$ has its own arithmetic, we may ask how the arithmetic of the $p$-adics interacts with the arithmetic when viewing them as Laurent series. Thankfully (perhaps?) the two different arithmetics produce the same results. That is, the canonical series representative for, say $x + y$ is the same as the sum of the series representatives of $x$ and $y$ (so long as we ‘carry the $p$’ appropriately). The same is true of the other operations (in the case of division we have to make sure we are not dividing by 0, whose canonical series has all zero coefficients), and we need not leave the world of canonical series in order to do arithmetic.