p-adic Numbers — Topology, Algebra, and Hensel's Lemma

💡 The Core Insight

The p-adic numbers complete $\mathbb{Q}$ with respect to a non-Archimedean absolute value where divisibility by high powers of $p$ measures closeness. This topology makes $\mathbb{Z}_p$ compact and allows polynomial equations to be solved by successive approximation via Hensel's Lemma.

Definition

p-adic Numbers — Topology, Algebra, and Hensel's Lemma Fix a prime $p$ . The p-adic absolute value on $\mathbb{Q}$ is defined by:

|x|_p = p^{-v_p(x)} \text{ for } x \neq 0, \quad |0|_p = 0

where $v_p(x)$ is the p-adic valuation: if $x = p^n \frac{a}{b}$ with $\gcd(a,p) = \gcd(b,p) = 1$ , then $v_p(x) = n$ . The p-adic metric is $d_p(x,y) = |x-y|_p$ . The completion of $\mathbb{Q}$ with respect to this metric is the field $\mathbb{Q}_p$ of p-adic numbers. The ring $\mathbb{Z}_p = \{x \in \mathbb{Q}_p : |x|_p \leq 1\}$ is the ring of p-adic integers.

Hensel's Lemma (Classical Form): Let $f(X) \in \mathbb{Z}_p[X]$ be a polynomial. If there exists $a \in \mathbb{Z}_p$ such that $|f(a)|_p < |f'(a)|_p^2$ , then there exists a unique $\alpha \in \mathbb{Z}_p$ with $f(\alpha) = 0$ and $|\alpha - a|_p < |f'(a)|_p$ .

TheoremMain Results

Theorem [1] (Ostrowski): Every non-trivial absolute value on $\mathbb{Q}$ is equivalent either to the standard absolute value $|\cdot|_\infty$ or to a p-adic absolute value $|\cdot|_p$ for some prime $p$ .

Theorem [2] (Ultrametric Inequality): For all $x, y \in \mathbb{Q}_p$ , |x + y|_p \leq \max\{|x|_p, |y|_p\} with equality if $|x|_p \neq |y|_p$ .

Theorem [3] (Compactness of $\mathbb{Z}_p$ ): The ring $\mathbb{Z}_p$ is a compact topological ring. Every element of $\mathbb{Z}_p$ has a unique representation x = \sum_i=0^\infty a_i p^i where $0 \leq a_i < p$ , and this series converges in the p-adic topology.

Theorem [4] (Hensel's Lemma — Strong Form): Let $f(X) \in \mathbb{Z}_p[X]$ . If $a \in \mathbb{Z}_p$ satisfies $v_p(f(a)) > 2v_p(f'(a))$ , then there exists a unique $\alpha \in \mathbb{Z}_p$ with $f(\alpha) = 0$ , $v_p(\alpha - a) > v_p(f'(a))$ , and $f'(\alpha) = f'(a)$ .

Proof

Proof of Theorem [4] (Hensel's Lemma): We construct $\alpha$ as a limit of a sequence using Newton iteration. Let $a_0 = a$ and define recursively:

a_{n+1} = a_n - \frac{f(a_n)}{f'(a_n)}

Step 1 (Sequence is well-defined): We have $f'(a_0) = f'(a) \neq 0$ in $\mathbb{Z}_p$ since $v_p(f'(a))$ is finite. By induction, we'll show $f'(a_n) = f'(a)$ for all $n$ , so division is always possible.

Step 2 (Improvement at each step): We claim that

v_p(f(a_{n+1})) > 2v_p(f(a_n)) - 2v_p(f'(a))

By Taylor expansion in $\mathbb{Z}_p$ :

f(a_{n+1}) = f(a_n - \frac{f(a_n)}{f'(a_n)}) = f(a_n) - f'(a_n) \cdot \frac{f(a_n)}{f'(a_n)} + \frac{f''(\xi)}{2}\left(\frac{f(a_n)}{f'(a_n)}\right)^2

for some $\xi$ between $a_n$ and $a_{n+1}$ . This gives:

f(a_{n+1}) = \frac{f''(\xi)}{2f'(a_n)^2} \cdot f(a_n)^2

Taking valuations:

v_p(f(a_{n+1})) \geq 2v_p(f(a_n)) - 2v_p(f'(a_n))

Step 3 (Derivative is constant): For $n = 0$ , we have $v_p(a_1 - a_0) = v_p(f(a_0)/f'(a_0)) = v_p(f(a_0)) - v_p(f'(a_0)) > v_p(f'(a_0))$ by hypothesis. By Taylor:

f'(a_1) = f'(a_0) + f''(\eta)(a_1 - a_0)

Thus $v_p(f'(a_1) - f'(a_0)) \geq v_p(a_1 - a_0) > v_p(f'(a_0))$ , which by the ultrametric inequality implies $v_p(f'(a_1)) = v_p(f'(a_0))$ . By induction, $f'(a_n) = f'(a)$ for all $n$ .

Step 4 (Cauchy sequence): We have

v_p(a_{n+1} - a_n) = v_p(f(a_n)) - v_p(f'(a_n)) > v_p(f(a_{n-1})) - v_p(f'(a)) > \cdots

The sequence of valuations $v_p(a_{n+1} - a_n)$ is strictly increasing, so $|a_{n+1} - a_n|_p \to 0$ . Thus $(a_n)$ is Cauchy in $\mathbb{Z}_p$ , which is complete, so $a_n \to \alpha \in \mathbb{Z}_p$ .

Step 5 (Limit is a root): Since $f$ is continuous in the p-adic topology and $v_p(f(a_n)) \to \infty$ , we have $f(\alpha) = \lim f(a_n) = 0$ .

Step 6 (Uniqueness): Suppose $\beta \in \mathbb{Z}_p$ also satisfies the conditions. Then:

0 = f(\alpha) - f(\beta) = f'(\alpha)(\alpha - \beta) + O((\alpha-\beta)^2)

If $\alpha \neq \beta$ , then $v_p(f'(\alpha)(\alpha - \beta)) < v_p((\alpha-\beta)^2)$ since $v_p(\alpha - a), v_p(\beta - a) > v_p(f'(a))$ implies both have derivative $f'(a)$ . This contradicts the equation above. □

∎

⚠️ Common mistake

Common Misconception: The p-adic absolute value is not Archimedean. In particular, all integers have norm at most 1: $|n|_p \leq 1$ for $n \in \mathbb{Z}$ . This means "large" integers (in the usual sense) can be "small" p-adically if they're divisible by high powers of $p$ . The triangle inequality becomes stronger (ultrametric), which implies all triangles are isosceles: if $|x|_p \neq |y|_p$ , then $|x+y|_p = \max\{|x|_p, |y|_p\}$ . This violates geometric intuition from Euclidean geometry.

mathematician

Kurt Hensel (1897) — Introduced p-adic numbers as a tool for applying analytic methods to number theory, developing the theory in his book Theorie der algebraischen Zahlen and subsequent works on p-adic analysis.

visualization

Example1

Computing in $\mathbb{Q}_7$ : Find the 7-adic expansion of $\frac{1}{3}$ . We seek $x = \sum_{i=0}^\infty a_i 7^i$ with $0 \leq a_i < 7$ such that $3x = 1$ . Start with $3a_0 \equiv 1 \pmod{7}$ . Since $3 \cdot 5 = 15 = 2 \cdot 7 + 1$ , we have $a_0 = 5$ . Now $3 \cdot 5 = 15 = 1 + 2 \cdot 7$ , so we carry 2. For the next digit: $3a_1 + 2 \equiv 0 \pmod{7}$ , giving $3a_1 \equiv -2 \equiv 5 \pmod{7}$ , so $a_1 = 4$ (since $3 \cdot 4 = 12 = 5 + 7$ ). Continuing: $3 \cdot 4 + 2 = 14 = 2 \cdot 7$ , carry 2. $3a_2 + 2 \equiv 0 \pmod{7}$ gives $a_2 = 4$ again. This pattern repeats: $\frac{1}{3} = 5 + 4 \cdot 7 + 4 \cdot 7^2 + 4 \cdot 7^3 + \cdots$ in $\mathbb{Z}_7$ . Verification: $3(5 + 4 \cdot 7 + 4 \cdot 7^2 + \cdots) = 15 + 12 \cdot 7 + 12 \cdot 7^2 + \cdots = (1 + 2 \cdot 7) + (1 + 2 \cdot 7) \cdot 7 + \cdots = 1 + 2 \cdot 7 + 2 \cdot 7^2 + \cdots$ . Actually, more carefully: the geometric series gives $\frac{1}{3} = 5 + 4 \cdot 7 \cdot \frac{1}{1-7} = 5 - \frac{28}{6}$ ... Let me recalculate. Actually: $\frac{1}{3} = \frac{1}{3} \cdot \frac{1}{1-0} = \sum_{k=0}^\infty 5 \cdot 7^k$ where we solve $3x \equiv 1 \pmod{7^k}$ iteratively. The answer is $\frac{1}{3} = 5 + 2 \cdot 7 + 2 \cdot 7^2 + \cdots = 5(1 + \frac{2 \cdot 7}{5} + \cdots)$ ... Direct: $\frac{1}{3} = \overline{524}$ in base 7 (periodic).

Example2

Applying Hensel's Lemma: Solve $x^2 \equiv 2 \pmod{7^n}$ for all $n$ . First, check if $2$ is a quadratic residue mod $7$ : we need $x^2 \equiv 2 \pmod{7}$ . Testing: $1^2=1, 2^2=4, 3^2=2 \pmod 7$ . So $x_0 = 3$ works. Apply Hensel: $f(x) = x^2 - 2$ , $f'(x) = 2x$ . At $x_0 = 3$ : $f(3) = 7$ , $f'(3) = 6$ . Check condition: $|f(3)|_7 = |7|_7 = 7^{-1}$ and $|f'(3)|_7^2 = |6|_7^2 = 1$ . Since $7^{-1} < 1$ , Hensel applies. Newton iteration: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 3 - \frac{7}{6}$ . In $\mathbb{Z}_7$ , we need $6^{-1}$ . Since $6 \cdot 6 = 36 = 1 + 5 \cdot 7$ , we have $6^{-1} \equiv 6 \pmod{7}$ . So $x_1 = 3 - 7 \cdot 6 = 3 - 42 = -39 \equiv 3 + 7 \cdot 4 = 31 \pmod{49}$ . Check: $31^2 = 961 = 2 + 959 = 2 + 7 \cdot 137 = 2 + 7^2 \cdot 19 + 7 \cdot 4$ . So $31^2 \equiv 2 + 7 \cdot 4 \pmod{49}$ ... Actually, $31 = 3 + 4 \cdot 7$ , and $(3 + 4 \cdot 7)^2 = 9 + 24 \cdot 7 + 16 \cdot 49 = 2 + 7 + 24 \cdot 7 \pmod{49} = 2 + 25 \cdot 7 \equiv 2 + 4 \cdot 7 \pmod{49}$ . Hmm, let me use the formula: $x_1 = 3 - \frac{7}{6} \pmod{49}$ . We have $\frac{7}{6} = 7 \cdot 6^{-1} = 7 \cdot 6 = 42 \equiv -7 \pmod{49}$ , so $x_1 = 3 + 7 = 10 \pmod{49}$ . Check: $10^2 = 100 = 2 + 98 = 2 + 2 \cdot 49 \equiv 2 \pmod{49}$ . ✓ Hensel guarantees this lifts to a unique square root of $2$ in $\mathbb{Z}_7$ : $\sqrt{2} = 3 + 1 \cdot 7 + a_2 \cdot 7^2 + \cdots$

Counterexample

Failure without derivative condition: Consider $f(x) = x^2$ over $\mathbb{Q}_2$ with initial approximation $a = 0$ . We have $f(0) = 0$ and $f'(0) = 0$ . The condition $v_2(f(0)) > 2v_2(f'(0))$ becomes $\infty > \infty$ , which fails. Indeed, $0$ is a "simple" root modulo any power of $2$ , but there's no unique lift: both $x = 0$ and $x = 2^k$ (for any $k \geq 1$ ) satisfy $x^2 \equiv 0 \pmod{2^{2k}}$ . Without the derivative condition, we lose uniqueness and even existence. For instance, $f(x) = x^2 + x$ has $f(0) = 0$ , $f'(0) = 1$ , so the derivative is good. But $f(1) = 0$ and $f'(1) = 1$ as well mod $2$ . The equation $x^2 + x \equiv 0 \pmod{2^n}$ has multiple solutions for all $n$ , demonstrating that the valuation condition is essential for Hensel's conclusion.

🧠 Intuition

Mental Model: Think of $\mathbb{Z}_p$ as an infinite tree where each node has $p$ children, labeled $0, 1, \ldots, p-1$ . A p-adic integer is an infinite path from the root, specifying digits $a_0, a_1, a_2, \ldots$ . The p-adic metric measures how long two paths stay together: $d_p(x,y) = p^{-n}$ where $n$ is the first level where the paths diverge. This topology makes "nearby" elements share many initial digits. Balls are clopen (both closed and open), and any point in a ball is its center—there's no distinguished origin for a ball! This is radically different from Euclidean space. Hensel's Lemma says: if you have an approximate root modulo $p^n$ with a good derivative, you can systematically improve it, like focusing in on a specific branch of the tree. The non-vanishing derivative ensures you're not at a "branch point" where multiple paths could work. Geometrically, $\mathbb{Q}_p$ is totally disconnected (every point is a connected component) but locally compact. It's the p-adic analogue of the real line, but with a fractal, tree-like structure rather than a continuum.

connection

<span class="wiki-link-missing">Iwasawa Theory</span> studies Galois representations and L-functions using infinite field extensions of $\mathbb{Q}_p$ . The Iwasawa algebra $\mathbb{Z}_p[[\text{Gal}(\mathbb{Q}_\infty/\mathbb{Q})]]$ for the cyclotomic extension $\mathbb{Q}_\infty$ is a complete local ring where Hensel-like techniques control the arithmetic of ideal class groups. The Main Conjecture of Iwasawa theory (proved by Mazur-Wiles for cyclotomic fields) relates p-adic L-functions to Selmer groups, fundamentally relying on p-adic completeness and analytic continuation in the p-adic domain. This connection bridges classical number theory, algebraic K-theory, and modular forms, showing how p-adic topology enables deep arithmetic results.

p-adic Numbers — Topology, Algebra, and Hensel's Lemma

p-adic Numbers — Topology, Algebra, and Hensel's Lemma

Overview

Core Theory

Examples and Applications

Historical Development

Connections to Other Areas

What You Need First

What This Opens Up

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