## § 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.

#### § Viewpoint: 'Evaluation' for p-adics

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.

#### § p-adic abosolute value: Definition

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.

#### § Conclusion

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