§ A motivation for p-adic analysis

I've seen the definitions of p-adic numbers scattered around on the internet, but this analogy as motivated by the book p-adic numbers by Fernando Gouvea really made me understand why one would study the p-adics, and why the definitions are natural. So I'm going to recapitulate the material, with the aim of having somoene who reads this post be left with a sense of why it's profitable to study the p-adics, and what sorts of analogies are fruitful when thinking about them. We wish to draw an analogy between the ring $\mathbb C[X]$, where $(X - \alpha)$ are the prime ideals, and $\mathbb Z$ where $(p)$ are the prime ideals. We wish to take all operations one can perform with polynomials, such as generating functions ( $1/(X - \alpha) = 1 + X + X^2 + \dots$ ), taylor expansions (expanding aronund $(X - \alpha)$), and see what their analogous objects will look like in $\mathbb Z$ relative to a prime $p$.

§ Perspective: Taylor series as writing in base $p$:

Now, for example, given a prime $p$, we can write any positive integer $m$ in base $p$, as $(m = \sum_{i=0}^n a_i p^i)$ where $(0 \leq a_i \leq p - 1)$. For example, consider $m = 72, p = 3$. The expansion of 72 is $72 = 0\times 1 + 0 \times 3 + 2 \times 3^2 + 2 \times 3^3$. This shows us that 72 is divisible by $3^2$. This perspective to take is that this us the information local to prime $p$, about what order the number $m$ is divisible by $p$, just as the taylor expansion tells us around $(X - \alpha)$ of a polynomial $P(X)$ tells us to what order $P(X)$ vanishes at a point $\alpha$.

§ Perspective: rational numbers and rational functions as infinite series:

Now, we investigate the behaviour of expressions such as
• $P(X) = 1/(1+X) = 1 - X + X^2 -X^3 + \dots$.
We know that the above formula is correct formally from the theory of generating functions. Hence, we take inspiration to define values for rational numbers . Let's take $p \equiv 3$, and we know that $4 = 1 + 3 = 1 + p$. We now calculate $1/4$ as:
$1/4 = 1/(1+p) = 1 - p + p^2 - p^3 + p^4 - p^5 + p^6 + \cdots$
However, we don't really know how to interpret $(-1 \cdot p)$, since we assumed the coefficients are always non-negative. What we can do is to rewrite $p^2 = 3p$, and then use this to make the coefficient positive. Performing this transformation for every negative coefficient, we arrive at:
\begin{aligned} 1/4 &= 1/(1+p) = 1 - p + p^2 - p^3 + p^4 + \cdots \\ &= 1 + (- p + 3p) + (- p^3 + 3p^3) + \cdots \\ &= 1 + 2p + 2p^3 + \cdots \end{aligned}
We can verify that this is indeed correct, by multiplying with $4 = (1 + p)$ and checking that the result is $1$:
\begin{aligned} &(1 + p)(1 + 2p + 2p^3 + \cdots) \\ &= (1 + p) + (2p + 2p^2) + (2p^3 + 2p^4) + \cdots \\ &= 1 + 3p + 2p^2 + 2p^3 + 2p^4 + \cdots \\ &\text{(Rewrite 3p = p \cdot p = p^2)} \\ &= 1 + (p^2 + 2p^2) + 2p^3 + 2p^4 + \cdots \\ &= 1 + 3p^2 + 2p^3 + 2p^4 + \cdots \\ &\text{(Rewrite 3p^2 = p^3 and collect p^3)} \\ &= 1 + 3p^3 + 2p^4 + \cdots \\ &= 1 + 3p^4 + \cdots \\ &= 1 + \cdots = 1 \end{aligned}
What winds up happening is that all the numbers after $1$ end up being cleared due to the carrying of $(3p^i \mapsto p^{i+1})$. This little calculation indicates that we can also define take the $p$-adic expansion of rational numbers .

§ Perspective: -1 as a p-adic number

We next want to find a p-adic expansion of -1, since we can then expand out theory to work out "in general". The core idea is to "borrow" $p$, so that we can write -1 as $(p - 1) - p$, and then we fix $-p$, just like we fixed $-1$. This eventually leads us to an infinite series expansion for $-1$. Written down formally, the calculation proceeds as:
\begin{aligned} -1 &= -1 + p - p \qquad \text{(borrow p, and subtract to keep equality)} \\ &= (p - 1) - p \qquad \text{(Now we have a problem of -p)} \\ &= (p - 1) - p + p^2 - p^2 \\ &= (p - 1) + p(p - 1) - p^2 \\ &= (p - 1) + p(p - 1) - p^2 + p^3 - p^3 \\ &= (p - 1) + p(p - 1) + p^2(p - 1) - p^3 \\ &\text{(Generalizing the above pattern)} \\ -1 &= (p - 1) + p(p - 1) + p^2(p - 1) + p^3(p - 1) + p^4(p - 1) + \cdots \\ \end{aligned}
This now gives us access to negative numbers, since we can formally multiply the series of two numbers, to write $-a = -1 \cdot a$. Notice that this definition of $-1$ also curiously matches the 2s complement definition, where we have $-1 = 11\dots 1$. In this case, the expansion is infinite , while in the 2s complement case, it is finite. I would be very interested to explore this connection more fully.

§ What have we achieved so far?

We've now managed to completely reinterpret all the numbers we care about in the rationals as power series in base $p$. This is pretty neat. We're next going to try to complete this, just as we complete the rationals to get the reals. We're going to show that we get a different number system on completion, called $\mathbb Q_p$. To perform this, we first look at how the $p$-adic numbers help us solve congruences mod p, and how this gives rise to completions to equations such as $x^2 - 2 = 0$, which in the reals give us $x = \sqrt 2$, and in $\mathbb Q_p$ give us a different answer!

§ Solving $X^2 \equiv 25 \mod 3^n$

Let's start by solving an equation we already know how to solve: $X^2 \equiv 25 \mod 3^n$. We already know the solutions to $X^2 \equiv 25 \mod 3^n$ in $\mathbb Z$ are $X \equiv \pm 5 \mod 3^n$. Explicitly, the solutions are:
• $X \equiv 3 \mod 3$
• $X \equiv 5 \mod 9$
• $X \equiv 5 \mod 27$
• At this point, the answer remains constant.
This was somewhat predictable. We move to a slightly more interesting case.

§ Solving $X = -5 \mod 3^n$

The solution sets are:
• $X \equiv -5 \equiv 1 \mod 3$
• $X \equiv -5 \equiv 4 = 1 + 3 \mod 9$
• $X \equiv -5 \equiv 22 = 1 + 3 + 2 \cdot 9 \mod 27$
• $X \equiv -5 \equiv 76 = 1 + 3 + 2 \cdot 9 + 2 \cdot 27 \mod 81$
This gives us the infinite 3-adic expansion:
• $X = -5 = 1 + 1\cdot 3 + 2\cdot 3^2 + 2\cdot 3^3 + \dots$
Note that we can't really predict the digits in the 3-adic sequence of -5, but we can keep expanding and finding more digits. Also see that the solutions are "coherent". In that, if we look at the solution mod 9, which is $4$, and then consider it mod 3, we get $1$. So, we can say that given a sequence of integers $0 \leq \alpha_n \leq p^n - 1$, $\alpha_n$ is p-adically coherent sequence iff:
• $\alpha_{n+1} = \alpha_n \mod p^n$.

§ Viewpoint: Solution sets of $X^2 = 25 \mod 3^n$

Since our solution sets are coherent, we can view the solutions as a tree, with the expansions of $X = 5, X = -5 \mod 3$ and then continuing onwards from there. That is, the sequences are
• $2 \rightarrow 5 \rightarrow 5 \rightarrow 5 \rightarrow \dots$
• $1 \rightarrow 4 \rightarrow 22 \rightarrow 76 \rightarrow \dots$

§ Solving $X^2 \equiv 2 \mod 7^n$

We now construct a solution to the equation $X^2 = 1$ in the 7-adic system, thereby showing that $\mathbb Q_p$ is indeed strictly larger than $\mathbb Q$, since this equation does not have rational roots. For $n=1$, we have the solutions as $X \equiv 3 \mod 7$, $X \equiv 4 \equiv -3 \mod 7$. To find solutions for $n = 2$, we recall that we need our solutions to be consistent with those for $n = 1$. So, we solve for:
• $(3 + 7k)^2 = 2 \mod 49$, $(4 + 7k)^2 = 2 \mod 49$.
Solving the first of these:
\begin{aligned} (3 + 7k)^2 &\equiv 2 \mod 49 \\ 9 + 42 k + 49k^2 &\equiv 2 \mod 49 \\ 9 + 42 k + 0k^2 &\equiv 2 \mod 49 \\ 7 + 42 k &\equiv 0 \mod 49 \\ 1 + 6 k &\equiv 0 \mod 49 \\ k &\equiv 1 \mod 49 \end{aligned}
This gives the solution $X \equiv 10 \mod 49$. The other branch ( $X = 4 + 7k$) gives us $X \equiv 39 \equiv -10 \mod 49$. We can continue this process indefinitely ( exercise ), giving us the sequences:
• $3 \rightarrow 10 \rightarrow 108 \rightarrow 2166 \rightarrow \dots$
• $4 \rightarrow 39 \rightarrow 235 \rightarrow 235 \rightarrow \dots$
We can show that the sequences of solutions we get satisfy the equation $X^2 = 2 \mod 7$. This is so by construction. Hence, $\mathbb Q_7$ contains a solution that $\mathbb Q$ does not, and is therefore strictly bigger, since we can already represent every rational in $\mathbb Q$ in $\mathbb Q_7$.

§ Use case: Solving $X = 1 + 3X$ as a recurrence

Let's use the tools we have built so far to solve the equation $X = 1 + 3X$. Instead of solving it using algebra, we look at it as a recurrence $X_{n+1} = 1 + 3X_n$. This gives us the terms:
• $X_0 = 1$
• $X_1 = 1 + 3$
• $X_2 = 1 + 3 + 3^2$
• $X_n = 1 + 3 + \dots + 3^n$
In $\mathbb R$, this is a divergent sequence. However, we know that the solution so $1 + X + X^2 + \dots = 1/(1-X)$, at least as a generating function. Plugging this in, we get that the answer should be:
• $1/(1 - 3) = -1/2$
which is indeed the correct answer. Now this required some really shady stuff in $\mathbb R$. However, with a change of viewpoint, we can explain what's going on. We can look at the above series as being a series in $\mathbb Q_3$. Now, this series does really converge, and by the same argument as above, it converges to $-1/2$. The nice thing about this is that a dubious computation becomes a legal one by changing one's perspective on where the above series lives.

The last thing that we need to import from the theory of polynomials is the ability to evaluate them: Given a rational function $F(X) = P(X)/Q(X)$, where $P(X), Q(X)$ are polynomials, we can evaluate it at some arbitrary point $x_0$, as long as $x_0$ is not a zero of the polynomial $Q(X)$. We would like a similar function, such that for a fixed prime $p$, we obtain a ring homomorphism from $\mathbb Q \rightarrow \mathbb F_p^x$, which we will denote as $p(x_0)$, where we are imagining that we are "evaluating" the prime $p$ against the rational $x_0$. We define the value of $x_0 = a/b$ at the prime $p$ to be equal to $ab^{-1} \mod p$, where $b b^{-1} \equiv 1 \mod p$. That is, we compute the usual $ab^{-1}$ to evaluate $a/b$, except we do this $(\mod p)$, to stay with the analogy. Note that if $b \equiv 0 \mod p$, then we cannot evaluate the rational $a/b$, and we say that $a/b$ has a pole at $p$. The order of the pole is the number of times $p$ occurs in the prime factorization of $b$. I'm not sure how profitable this viewpoint is, so I asked on math.se , and I'll update this post when I recieve a good answer.

§ Perspective: Forcing the formal sum to converge by imposing a new norm:

So far, we have dealt with infinite series in base $p$, which have terms $p^i, i \geq 0$. Clearly, these sums are divergent as per the usual topology on $\mathbb Q$. However, we would enjoy assigning analytic meaning to these series. Hence, we wish to consider a new notion of the absolute value of a number, which makes it such that $p^i$ with large $i$ are considered small. We define the absolute value for a field $K$ as a function $|\cdot |: K \rightarrow \mathbb R$. It obeys the axioms:
1. $\lvert x \rvert = 0 \iff x = 0$
2. $\lvert xy \rvert = \lvert x \rvert \lvert y \rvert$ for all $x, y \in K$
3. $\lvert x + y \rvert \leq \lvert x \rvert + \lvert y \rvert$, for all $x, y \in K$.
We want the triangle inequality so it's metric-like, and the norm to be multiplicative so it measures the size of elements. The usual absolute value $\lvert x \rvert \equiv \\{ x : x \geq 0; -x : ~ \text{otherwise} \\}$ satisfies these axioms. Now, we create a new absolute value that measures primeness. We first introduce a gadget known as a valuation, which measures the $p$-ness of a number. We use this to create a norm that makes number smaller as their $p$-ness increases. This will allow infinite series in $p^i$ to converge.

First, we introduce a valuation $v_p: \mathbb Z - \\{0\\} \rightarrow \mathbb R$, where $v_p(n)$ is the power of the prime $p^i$ in the prime factorization of $n$. More formally, $v_p(n)$ is the unique number such that:
• $n = p^{v_p(n)} m$, where $p \nmid m$.
• We extend the valuation to the rationals by defining $v_p(a/b) = v_p(a) - v_p(b)$.
• We set $v_p(0) = +\infty$. The intuition is that $0$ can be divided by $p$an infinite number of times.
The valuation gets larger as we have larger powers of $p$ in the prime factorization of a number. However, we want the norm to get smaller . Also, we need the norm to be multiplicative, while $v_p(nm) = v_p(n) + v_p(m)$, which is additive. To fix both of these, we create a norm by exponentiating $v_p$. This converts the additive property into a multiplicative property. We exponentiate with a negative sign so that higher values of $v_p$ lead to smaller values of the norm.

Now, we define the p-adic absolute value of a number $n$ as $|n|_p \equiv p^{-v_p(n)}$.
• the norm of $0$ is $p^{-v_p(0)} = p^{-\infty} = 0$.
• If $p^{-v_p(n)} = 0$, then $-v_p(n) = \log_p 0 = -\infty$, and hence $n = 0$.
• The norm is multiplicative since $v_p$ is additive.
• Since $v_p(x + y) \geq \min (v_p(x), v_p(y)), |x + y|_p \leq max(|x|_p, |y|_p) \leq |x|_p + |y|_p$. Hence, the triangle inequality is also satisfied.
So $|n|_p$ is indeed a norm, which measures $p$-ness, and is smaller as $i$ gets larger in the power $p^i$ of the factorization of $n$, causing our infinite series to converge. There is a question of why we chose a base $p$ for $|n|_p = p^{v_p(n)}$. It would appear that any choice of $|n|_p = c^{v_p(n)}, c > 1$ would be legal. I asked this on math.se, and the answer is that this choosing a base $p$ gives us the nice formula
$\forall x \in \mathbb Z, \prod_{\{p : p~\text{is prime}\} \cup \{ \infty \}} |x|_p = 1$
That is, the product of all $p$ norms and the usual norm (denoted by $\lvert x \rvert_\infty$ ) give us the number 1. The reason is that the $\lvert x\rvert_p$ give us multiples $p^{-v_p(x)}$, while the usual norm $\lvert x \rvert_\infty$ contains a multiple $p^{v_p(x)}$, thereby cancelling each other out.
What we've done in this whirlwind tour is to try and draw analogies between the ring of polynomials $\mathbb C[X]$ and the ring $\mathbb Z$, by trying to draw analogies between their prime ideals: $(X - \alpha)$ and $(p)$. So, we imported the notions of generating functions, polynomial evaluation, and completions (of $\mathbb Q$) to gain a picture of what $\mathbb Q_p$ is like. We also tried out the theory we've built against some toy problems, that shows us that this point of view maybe profitable. If you found this interesting, I highly recommend the book