The $p$-adic Number System and the Artin-Hasse Exponential

June 2023

Introduction #

In this paper, we explore the $p$-adic system, by defining it in multiple ways: as an extension of the $p$-adic integers, as well as an extension of the rationals. We then proceed to perform analysis in the $p$-adics, by defining convergence, continuity and discs. We then describe exponentiation and logarithmic functions over the $p$-adics as functions derived from power series. We explore the radius of convergence and other properties of these functions. We then explore the Artin-Hasse exponential, which, though seemingly random, turns out to be an integral power series.

Formal Power Series #

The Ring $R[[x]]$ #

Prove that $R[[x]]$ is a ring under the natural operations of addition and multiplication.

Addition : Let $f,g \in R[[x]]$ such that $f(x)= \sum_{i=0}^\infty a_i x^i$ and $g(x)= \sum_{j=0}^\infty b^j x^j$. Then $f(x) + g(x) = \sum_{i=0}^\infty a_i x^i + \sum_{j=0}^\infty b^j x^j = \sum_{i,j=0}^\infty (a_i + b_j) x^i$. Since ${a_i + b_j }$ is closed under addition as $a_i$and $b_j \in R$ we can say, $R[[x]]$ is a ring under addition.

Multiplication : For proving the multiplication, we need to prove the commutative property in a formal power series. Let $f,g \in R[[x]]$ such that $f(x)= \sum_{i=0}^\infty a_i x^i$ and $g(x)= \sum_{j=0}^\infty b^j x^j$. Then $$f(x)g(x) = \sum_{i=0}^\infty a_i x^i \times \sum_{j=0}^\infty b^j x^j = \sum_{k=0}^\infty\left(\sum_{i = 0}^ka_ib_{k-i}\right)x^k$$

Since the sum $\sum_{i=0}^ka_ib_{k-i}$ is a sum of products of elements of $R$, the sum itself is in $R$ since $R$ is closed under addition and multiplication. Therefore, the above sum is a power series in $R$, that is the products $fg \in R[[x]]$.
 
Now, since the ring is commutative over both $+$ and $\cdot$, it follows that $R[[x]]$ is also commutative. Moreover, it is clearly associative over $+$ since the ring $R$ is itself associative over $+$. Next, the ring has a zero element, i.e. $0 \in R[[x]]$ since for any $f\in R[[x]]$, we have $f + 0 = f$. Similarly, it also has an identity element, i.e. $1$. Also, the additive inverse of $f$ is the power series with the coefficients each being the additive inverse of the coefficients of $f$. Thus, every element of $R[[x]]$ has an inverse element. Distributivity can also be easily seen from the fact that $R$ is itself distributive.
 
We will now prove associativity over multiplication. Let $$a = \sum_{i=0}^\infty a_ix^i, b = \sum_{i=0}^\infty b_ix^i, c = \sum_{i=0}^\infty c_ix_i$$ Now, $$ab = \sum_{k=0}^\infty\left(\sum_{i=0}^ka_ib_{k-i}\right)x^k = \sum_{i=0}^\infty p_ix^i$$ So, $$(ab)c = \sum_{k=0}^\infty \left(\sum_{i=0}^k p_ic_{k-i}\right)x^k = \sum_{k=0}^\infty\left(\sum_{i=0}^k \left(\sum_{j=0}^ia_jb_{i-j}\right)c_{k-i}\right)x^k$$ The above can be expanded as $$\sum_{k=0}^\infty (c_0a_0b_0 + c_1(a_0b_1 + a_1b_0) + \cdots c_k(a_0b_k + \cdots a_kb_0)x^k$$ Now, $a(bc) = (bc)a$ by commutativity. Therefore, we may use a similar approach as above to show that $$a(bc) = \sum_{k=0}^\infty (a_0b_0c_0 + a_1(b_0c_1 + b_1c_0) + \cdots + a_k(b_0c_k + \cdots b_kc_0))x^k$$ By rearranging the terms above, we get $$a(bc) = \sum_{k=0}^\infty (c_0a_0b_0 + c_0(a_0b_1 + a_1b_0) + \cdots + c_k(a_0b_k + \cdots a_kb_0))x^k = (ab)c$$ This proves associativity over multiplication.

Units in $R[[x]]$ #

Definition 1. $f(x) \in R[[x]]$ then $f(0) = a_0$

Definition 2. Units in $R[[x]]$ are all $f$ and $g$ in $R[[x]]$ such that $[f \cdot g](x) = 1$.

Examples of these units are:

1. $f(x) = -(x-1)$

2. $f(x) = \sum_{i=0}^{\infty} x^i$

3. $f(x) = 1 + \sum_{i=1}^n a_ix^i$ for $a_i \in R$

4. $f(x) = (\pm (x-1) \cdot \sum_{i=0}^{\infty} x^i)^n$ for $n \in \mathbb{N}$

Proposition 1. If $f(x) \in R[[x]]$ and $f(0) = 1$, then $f(x)$ is a unit in $R[[x]]$.

Proof. Let $f(x), g(x)$ be in $R[[x]]$ such that $f \cdot g = 1$. We need to prove that such a power series, $g$, exists if and only if we have $f(0) = 1$. Let $$f(x) = a_0 + a_1x + \cdots$$ and $$g(x) = b_0 + b_1x + \cdots.$$ Thus, proving the existence of $g$ is equivalent to proving that there exists a sequence $\lbrace b_i \rbrace$ such that $$(a_0 + a_1x + \cdots)(b_0 + b_1x + \cdots) = 1.$$ Expanding this product, and combining the terms of equal degree, we get: $$a_0b_0 + (a_0b_1 + a_1b_0)x + \cdots + (a_0b_n + \cdots + a_nb_0)x^n + \cdots = 1$$ Since the RHS is simply $1$, we need all terms $(a_0b_n + \cdots a_nb_0) = 0$ for all $n\geq1$ and $a_0b_0 = 1$.
 
Now, if $f(0) = a_0$ is not a unit, then we cannot find a $b_0$ in $R$ such that $a_0b_0 = 1$. This shows that we cannot find the desired sequence $\lbrace b_i \rbrace$. Thus, if $f(x)$ is a unit in $R[x]$, it follows that $f(0)$ is a unit.
 
We now prove the converse. That is, if $f(0)$ is a unit in $R$, then there exists some sequence $\lbrace b_i\rbrace$ so that $$g(x)=\sum_{i\geq0}b_ix^i$$ and $f(x)g(x) = 1$. We use induction in order to do so. We will first prove that there exists some sequence $\lbrace b_i\rbrace_{i=0}$ (i.e. a single term $b_0$) so that $a_0b_0 = 1$. This finishes our base case. Now, assume that there exists some sequence $\lbrace b_i\rbrace_{i=0}^n$ so that $a_0b_0 = 1$ and for all $0 < k \leq n$ we have $$(a_0b_k + \cdots a_kb_0) = 0.$$ We prove that there exists a sequence $\lbrace b_i\rbrace_{i=0}^{n+1}$ so that the same holds for $k = n+1$ as well. We have $$a_{n+1}b_0 + \cdots a_0b_{n+1} = 0.$$ Therefore, we have $$b_{n+1} = -\frac{a_{n+1}b_0 + \cdots + a_1b_n}{a_0}$$ Since we know $b_i$ exist for all $i$ less than or equal to $n$, we have that $b_{n+1}$ also exists, completing our inductive step. Thus, if $f(0)$ is a unit, then $f(x)$ has an inverse. ◻

Compositions of formal power series #

We will now generalize as to when $f(g(x))$ where $g(x) \in R[[x]]$ is an element of $R[[x]]$ itself.

Proposition 2. Let $f, g \in R[[x]]$. Then, $f(g(x)) \in R[[x]]$ if and only if $g(0) = 0$.

Proof. Suppose $g(0) = 0$. Then, we can write $g(x) = x^k + h(x)$ where $k$ is the smallest power of $x$ in $g(x)$, and $k \neq 0$ since $g(0)=0$. Moreover, the smallest power of $x$ in $h(x)$ is more than $k$. Also, let $$f(x) = \sum_{i=0}^\infty a_ix^i$$ Therefore, $$f(g(x)) = a_0 + a_1g(x) + a_2g(x)^2 + \cdots = a_0 + a_1(x^k + h(x)) + a_2(x^k + h(x))^2 + \cdots$$ Also let $$f(g(x)) = c_0 + c_1x + c_2x^2 + \cdots$$ In order to prove that $f(g(x)) \in R[[x]]$ it suffices to prove that $c_n \in R$ for all $n$. So, we will work on finding the coefficient of the $x^n$ term, i.e. $c_n$.
 
Consider the term $a_{n+1}(x^k + h(x))^{n+1}$ in the expansion for $f(g(x))$. The term with the lowest power in this expansion is $x^{(n+1)k}$. Since $k \neq 0$, it follows that $(n+1)k > n$. Therefore, the coefficient $c_n$ of $x^n$ is independent of $a_{n+1}$. By a similar argument, $c_n$ is independent of $a_j$ for all $j > n$. In other words, it follows that $$c_n = \sum_{i = 0}^n a_it_i$$ where $t_i \in R$. Note that there will be no powers of $a_i$ in the expansion since the coefficients $a_i$ are not raised to a power in the expansion of $f(g(x))$. Since $a_i, t_i \in R$, it follows that $a_it_i \in R$. Thus, $\sum_{i=0}^na_it_i \in R$. Therefore, for all $n$, we have $c_n \in R$, where $$f(g(x)) = c_0 + c_1x + \cdots$$ Therefore, by definition of a formal power series, $f(g(x)) \in R[[x]]$.
 
Now, suppose $g(0) = c \neq 0$. Let $g(x) = c + h(x)$. Then, $$f(g(x)) = a_0 + a_1(c + h(x)) + a_2(c+h(x))^2 + \cdots = c_0 + c_1x + \cdots$$ In the $n$th term of the above expansion, we have a constant $c^n$. Since there are infinitely many terms in the expansion, $c_0$ is an infinite sum, which is not defined in $R$. Therefore, $c_0 \notin R$. It follows that $f(g(x)) \notin R[[x]].$ ◻

Multivariate Power Series #

We can define the system $R[[x,y]]$ to $(R[[x]])[[y]]$. Inductively, we may define $$R[[x_1, \dots, x_k]] = (R[[x_1, \dots, x_{k-1}]])[[x_k]]$$ Since $R$ is a ring implies $R[[x]]$ is a ring (base case) Then, we assume that $R[[x_1,x_2, x_3, … x_k]]$ is a ring Now, $R[[x_1, x_2, X_3, … x_k+1]]$ is $(R[[x_1,x_2, x_3, … x_k]])[[x_k+1]]$ which is a ring (inductive case). Thus, by induction we can say that $R[[x_1, x_2, X_3, … x_n+1]]$ is a ring.

We will now generalize. For which power series $g \in R[[x,y]]$ can we define $f(g(x,y))$ for any $f \in R[[t]]$?

Proposition 3. For $f, g \in \mathbb{R}[[x, y]]$, we have that $f(g(x, y))$ is defined in $R[[x, y]]$ if and only if $g(0, 0) \neq 0$.

Proof. Suppose $g(0, 0) = 0$. Thus, the smallest power of $x$ and $y$ can be represented as $b_{mn}x^my^n$ where both $m$ and $n$ are not $0$ at the same time. So, we let $g(x, y) = b_{mn}x^my^n + h(x, y)$. Thus, $$\begin{aligned} f(g(x, y)) &= f(b_{mn}x^my^n + h(x, y)) \\ &= a_0 + a_1(b_{mn}x^my^n + h(x, y))+a_2(b_{mn}x^my^n + h(x, y))^2+\cdots \\ &= c_{00} + c_{10}x + c_{01}y + \cdots \end{aligned}$$ Now, consider the coefficient, $c_{pq}$. We know that the term $$a_{p+q+1}(b_{mn}x^m+y^n + h(x, y)^{p+q+1})$$ has smallest power $x^{m(p+q+1)}y^{n(p+q+1)}$ which is clearly greater than the power of $x^py^q$. Hence, $c_{pq}$ is independent of $a_k$ for all $k> p+q$. Therefore, we have that $$c_{pq} = \sum_{i=0}^{p+q}a_id_i$$ which is clearly an element of $R$. Therefore, $f(g(x, y))$ makes sense in $R[[x, y]]$.
 
Now, if $g(0, 0) = c$ which is a constant, then let $g(x, y) = c + h(x, y)$. Therefore, we have $$f(g(x, y)) = a_0 + a_1(c+h(x, y)) + a_2(c+h(x, y))^2 + \cdots$$ The constant term of the above expansion will be $a_0 + a_1c + a_2c^2 + \cdots$ which is an infinite sum. This does not make sense in $R$ since we can only compute finite sums in $R$. Therefore, if $c\neq 0$, then there is no way to make sense of $f(g(x, y))$ as an element of $R[[x, y]]$. ◻

Polynomial Fields and Rational Functions #

For some field $k$, we write $k[x]$ to denote the set of polynomials with coefficients in $k$. We write $k(x)$ to denote the set $$\left\lbrace \left .\frac{g(x)}{h(x)} \right \rvert g(x), h(x) \in R[[x]]\right \rbrace$$ We will now explore the relation between $k(x), k[x]$ and $k[[x]]$. Firstly, note that $k[x] \in k(x)$ by letting $g(x) = 1$. Also, clearly $k[x] \in k[[x]]$. This is because a polynomial is a formal power series with the coefficients equal to $0$ for all $x^n$ where $n$ is greater than the degree of the polynomial. We now check when an element of $k(x)$ can be written as a formal power series. Consider an example of the power series expansion of a rational function $g(x)/h(x)$, $t(x)$. Let

$g(x) = 2x+1$ and $h(x) = x^1 + 1$. Then, $$\frac{g(x)}{h(x)} = \frac{2x+1}{x^2 + 1} = 1 + x + 2x -x^3 -2x^4 + x^5 + 2x^6 - \cdots$$ Notice how the terms of the power series are recursive. The $n$th term of $t(x)$ can be found in terms of the previous terms of $t(x)$. In fact, this holds in general too!

Proposition 4. We claim that a rational function $g(x)/h(x)$, i.e. an element of $k(x)$ with $h(0) \neq 0$ can be written as a power series $t(x)$.

Proof. If $h(0) \neq 0$, then we have already proved that $h(x)$ is a unit. Thus, $1/h(x) \in R[[x]]$. So, since formal power series are closed under multiplication, $g(x). (1/h(x)) \in R[[x]]$. Thus, $g(x)/h(x)$ can be written as a power series $t(x)$. If $h(0)=0$, then $h(x)$ is not a unit which implies $1/h(x) \notin R[[x]]$ and thus, $g(x)/h(x)$ cant be expressed as a formal power series. ◻

Proposition 5. Let $t(x) = g(x)/h(x)$ be the power series expansion of a rational function. Then, the terms $a_n$ of $t(x)$ satisfy a linear recursion. Formally, this means that there exists a number $m\geq 1$ and constants $c_1, \dots, c_m \in k$ such that for all $n$ that is sufficiently large, $$a_n = \sum_{i=1}^mc_ia_{n-i}$$

Proof. Firstly, let the degree of $h(x)$ be $m$ and the degree of $g(x)$ be $k$. Let $h(x) = b_0 + b_1x + \cdots b_mx^m$ and $t(x) = a_0 + a_1x + \cdots$. Now, since $t(x) = g(x)/h(x)$, we may write $g(x) = t(x)h(x)$.
 
Since the degree of $g(x)$ is $k$, it follows that the coefficients of $x^n$ in the expansion of the product on the RHS is $0$ for all $n$ that is greater than $k$.
 
Now, consider some $n > \max(m, k)$. Clearly, $n>k$. Therefore, the coefficient of $x^n$ in the expansion will be $0$. Let us find the coefficient using the fact that $t(x) = a_0 + a_1x + \cdots$ and $h(x) = b_0 + b_1x + \cdots + b_mx^m$. This will be $$a_nb_0 + a_{n-1}b_1 + \cdots + a_{n-m}b_m$$ Since $n>k$, the above must be $0$. Rearranging the terms, we get $$a_nb_0 = -(a_{n-1}b_1 + \cdots a_{n-m}b_m)$$ Since $k$ is a field and $b_0 = h(0) \neq 0$, we have $$a_n = -b_0^{-1}\left(a_{n-1}b_1 + \cdots a_{n-m}b_m\right) = \sum_{i=1}^mc_ia_{n-i}$$ where $c_i = -b_0^{-1}b_i$. This proves that the coefficients of the power series $t(x)$ satisfy a linear recursion. ◻

Proposition 6. The converse of the previous lemma also holds true. That is, if $t(x) = \sum_{n\geq}a_nx^n\in k[[x]]$ is such that the coefficients $a_n$ satisfy a linear recursion, then there exists some rational function $g(x)/h(x)$ which has a power series expansion of $t(x)$.

Proof. We know that for $t(x)$, we have $$a_n = c_1a_{n-1} + c_2a_{n-2} + \cdots c_ma_{n-m}$$ for $n$ that is sufficiently large enough, since we are given that the coefficients of $t$ satisfy a linear recursion. Therefore, we have $$a_n - c_1a_{n-1} - \cdots - c_ma_{n-m} = 0$$ Let $b_0 = 1$ and $b_i = -c_i$ when $i\geq1$. Therefore, we may write $$a_nb_0 + a_{n-1}b_1 + \cdots a_{n-m}b_m = 0$$ Now, consider $h(x) = b_0 + b_1x + \cdots b_mx^m$. Consider the product $t(x)h(x)$. The coefficient of $x^n$ in this product will be $$b_0a_n + b_1a_{n-1} + \cdots b_ma_{n-m}$$ which is $0$ by our assumption. Therefore, all terms of $t(x)h(x)$ have a coefficient of $0$ for $n$ sufficiently large. Suppose for all $n>k$ we have that the coefficient of $x^k$ is $0$ (and not $0$ for $n$ not more than $k$). Therefore, $t(x)h(x) = g(x)$ is a polynomial of degree $k$. Therefore, $g(x)/h(x)$ is a rational function that is represented by $t(x)$. Therefore, we have found some rational function that is represented by $t(x)$ given that the coefficients of $t(x)$ satisfy a linear recursion. This completes our proof. ◻

$p$-adic numbers #

The $p$-adic numbers are defined as follows:

Definition 3. The set of $p$-adic numbers $\mathbb{Z_p}$ is defined as:

$$\mathbb{Z_p} = \lbrace (a_1, a_2, \dots) \mid a_i \in \mathbb{Z}/p^i \mathbb{Z} \text{ and } a_{i+1} \equiv a_i \pmod{p^i}$$

One can prove that the set of $p$-adic numbers, $\mathbb{Z_p}$ forms a ring under term-wise addition and multiplication.

Proposition 7. The ring $\mathbb{Z_p}$ is an integral domain, i.e., if for some $a, b \in \mathbb{Z_p}$ we have $ab = 0$, then either $a=0$ or $b=0$.

Proof. In $\mathbb{Z_p}$, we have $0 = (0, 0, \dots)$. Thus, we need to prove that either $a = (0, 0, \dots)$ or $b = (0, 0, \dots)$ given that $ab = (0, 0, \dots)$. First, let $a = (a_1, a_2, \dots)$ and $b = (b_1, b_2, \dots)$. Then, $$ab = (a_1b_1, a_2b_2, \dots) = (0, 0, \dots)$$ This implies that we must have $a_ib_i \equiv 0 \pmod{p^i}$ for all $i$. Clearly, we must also have $a_1b_1 \equiv 0 \pmod{p}$. Since $\mathbb{Z_p}$ is an integral domain, we must have $a_1 \equiv 0 \pmod p$ or $b_1 \equiv 0 \pmod{p}$. Assume without loss of generality that $a_1 \equiv 0 \pmod{p}$. We will prove that either $a_i = 0$ for all $i$ or $b_i = 0$ for all $i$.
 
We will prove this by contradiction. Suppose that $a_i \neq 0$ and $b_i \neq 0$. Then, there exists some $a_k$ such that $a_k \equiv 0 \pmod {p^k}$. Also assume without loss of generality that $b_i \neq 0$ for some $i \leq k$ (if the smallest such $i$ is greater than $k$, then we just exchange $a$ and $b$). If $a_k \equiv 0 \pmod{p^k}$, then by the definition of a $p$-adic number, we must have $a_{k+1} \equiv a_k \not \equiv 0 \pmod{p^k}$. Thus, $a_{k+1} \neq 0$. It follows that for any $i \geq k$, we must have $a_i \neq 0$ as well as $b_i \neq 0$.
 
Now, since $ab = 0$, we must have $a_ib_i \equiv 0 \pmod{p^i}$ for all $i \geq k$. Now, we have $a_{k-1} \equiv 0 \pmod{p^{k-1}}$ but $a_k \equiv 0 \pmod{p^k}$. Moreover, by definition, we have $a_k \equiv a_{k-1} \pmod{p^{k-1}}$. Therefore, it follows that $$a_k \equiv p^{k-1}m_1 \pmod{p^k}$$ where $0 < m_1 < p$. Next, we have $$a_{k+1} \equiv p^km_0 + p^{k-1}m_1 \pmod{p^{k+1}}$$ where $0 < m_0 < p$. In general, we therefore have $$a_{k+i} \equiv p^{k+i - 1}m_{i-1} + \cdots + p^km_0 \pmod{p^{k+i}}$$ Let us say the smallest value of $i$ for which $b_i \neq 0$ is $i = r$. We assumed earlier (WLOG) that $r < k$. Now, just as before, we may write: $$b_{r+j} \equiv p^{r+j-1}n_{j-1} + \cdots + p^rn_0 \pmod{p^{r+j}}$$ Since $k > r$, let $k = r + l$ where $l$ is some natural number. Also, let $i = r - k + j$. Therefore, $$b_{r+j} = b_{k+i} \equiv p^{k+i - 1}n_{j-1} + \cdots + p^rn_0 \pmod{p^{k+i}}$$ Now, $v_p(a_{k+i}b_{k+i}) = v_p(a_{k+i}) + v_p(b_{k+i})$. Since the smallest power of $p$ in the expansion of $a_{k+i}$ is $k$. Therefore, we have $v_p(a_{k+i}) = k$. Since the smallest power of $p$ in the expansion of $b_{k+i}$ is $r$, we have that $v_p(b_{k+i}) = r < k$. Thus, $v_p(a_{k+i}b_{k+i}) = kr \leq k^2$. Now, consider the case when $i = k^2 - k + 1$. We therefore have $v_p(a_{k^+1}b_{k^2+1}) \leq k^2$. This implies that $$a_{k^2+1}b_{k^2 +1} \not \equiv 0 \pmod{p^{k^2+1}}$$ However, since $ab = 0$, we must have $a_{k^2+1}b_{k^2+1} \equiv 0 \pmod{p^{k^2+1}}$. A contradiction. Thus, we cannot have both $a \neq 0$ as well as $b \neq 0$. ◻

Let us explore what the units in $\mathbb{Z_p}$ are. We first take an example. Consider $\mathbb{Z_3}$. Consider $a = (1, 4, 13, 40, \dots)\in \mathbb{Z_p}$. Also, let $b = (1, 7, 25, 79, \dots)$. Clearly, $b$ also is an element of $\mathbb{Z_p}$. Moreover, $$ab = (1\cdot1, 4\cdot7, 13\cdot25, 40\cdot 79, \dots) = (1, 1, 1, 1, \dots)$$ Thus, $b$ is the inverse of $a$ in $\mathbb{Z_p}$. Since $a$ has an inverse in $\mathbb{Z_p}$, it follows that $a$ is a unit in $\mathbb{Z_p}$. Note that however, $a = (0, 3, 12, \dots)$ is not a unit in $\mathbb{Z_p}$. It seems that in general, $a\in \mathbb{Z_p}$ is a unit if and only if the first term of $a$ is non-zero. In other words, we claim that $a \in \mathbb{Z_p}$ is a unit if and only if $a \equiv 0 \pmod{p}$.

Proposition 8. In $\mathbb{Z_p}$, a $p$-adic number $u$ is a unit if and only if $u_1 \not \equiv 0 \pmod{p}$

Proof. Let $u = (u_1, u_2, \dots)$. Suppose $u_1 = 0$. Then, there is clearly no inverse for $u$ since there is no $b_1 \in \mathbb{Z}/p\mathbb{Z}$ such that $u_1b_1 \equiv 1 \pmod{p}$ ($0$ is not a unit modulo $p$). Hence, if $u$ is a unit, then it follows that $u_1 \not \equiv 0$ or equivalently, $u \equiv 0 \pmod{p}$.
 
We now prove the other direction. Suppose $u \not \equiv 0 \pmod{p}$. Then, we must prove that $u$ is a unit in $\mathbb{Z_p}$. Since $p$ is prime, if $u \neq 0 \pmod{p}$, it means that $u_1 \not \equiv 0 \pmod{p}$. Since $p$ is a prime, it follows that $(u_1, p) = 1$. Thus, $u_1$ is a unit in $\mathbb{Z}/p\mathbb{Z}$.
 
Now, suppose that $(u_k, p) = 1$. Now, by definition of a $p$-adic number, we have $u_{k+1} \equiv u_k \pmod{p^k}$. This in turn implies that $u_{k+1} \equiv u_k \pmod{p}$. Thus, $(u_{k+1}, p) = 1$. Hence, $(u_{k+1}, p^{k+1}) = 1$. Thus, $u_{k+1}$ is a unit in $\mathbb{Z}/p^{k+1}\mathbb{Z}$.
 
By induction, this implies that for all $k$, we have $u_k$ is a unit in $\mathbb{Z}/p^k\mathbb{Z}$. Since the multiplication operation is termwise in $\mathbb{Z_p}$, it follows that $u$ is a unit in $\mathbb{Z_p}$. ◻