p-adic valuation

In number theory, the p-adic valuation, denoted v_p, is a discrete valuation on the field of rational numbers \mathbb{Q} associated to a fixed prime number p, defined for a nonzero rational x = a/b in lowest terms as v_p(x) = v_p(a) - v_p(b), where v_p(n) for a nonzero integer n is the highest exponent k such that p^k divides n, and v_p(0) = \infty.^[1]^[2] This valuation extends naturally to the ring of integers and provides a measure of "divisibility by p" that captures the multiplicity of p in the prime factorization of elements in \mathbb{Q}.^[1] The p-adic valuation satisfies key properties that make it a non-Archimedean valuation: it is additive under multiplication, so v_p(xy) = v_p(x) + v_p(y) for all x, y \in \mathbb{Q}, and it obeys the ultrametric inequality v_p(x + y) \geq \min\{v_p(x), v_p(y)\} for addition, with equality holding if v_p(x) \neq v_p(y).^[1]^[2] From this, one defines the p-adic absolute value or norm |x|_p = p^{-v_p(x)} for x \neq 0 (and |0|_p = 0), which induces a metric d(x, y) = |x - y|_p on \mathbb{Q}, turning it into an ultrametric space where the distance satisfies the strong triangle inequality d(x, z) \leq \max\{d(x, y), d(y, z)\}.^[1] This metric allows the completion of \mathbb{Q} to yield the field of p-adic numbers \mathbb{Q}_p, a complete normed field that extends the rationals and plays a central role in local number theory.^[2] Beyond its foundational role in constructing \mathbb{Q}_p, the p-adic valuation is instrumental in various applications, such as determining the p-adic order of factorials via Legendre's formula v_p(n!) = \sum_{k=1}^\infty \lfloor n / p^k \rfloor, analyzing Diophantine equations through lifting the exponent lemmas, and studying arithmetic in global fields via Ostrowski's theorem, which classifies all non-trivial absolute values on \mathbb{Q} as either the usual Archimedean one or the p-adic ones for primes p.^[1]^[2] In broader contexts, it facilitates p-adic analysis, interpolation (e.g., p-adic zeta functions), and connections to algebraic geometry over p-adic fields.^[2]

Definition

For integers

The p-adic valuation, denoted v_p(n), for a fixed prime number p and a nonzero integer n, is defined as the highest non-negative integer k such that p^k divides n.^[1] Equivalently, n can be expressed as n = \pm p^k m, where m is an integer not divisible by p, and thus v_p(n) = k.^[3] For example, consider n = 12 and p = 2: since $12 = 2^2 \cdot 3 and $2

does not divide &#36;3

, it follows that v_2(12) = 2.^[3] Similarly, for p = 3, $12 = 3^0 \cdot 12 and $3

does not divide &#36;12

, so v_3(12) = 0.^[1] The case n = 0 is handled by convention as v_p(0) = +\infty, which ensures consistency in arithmetic operations involving the valuation, such as treating divisions by zero appropriately in extended contexts.^[3] This concept was introduced by Kurt Hensel in 1897 as part of his foundational work on what would later be known as p-adic numbers, motivated by the study of algebraic integers through power series expansions.^[4]

For rational numbers

The p-adic valuation on the rational numbers \mathbb{Q} extends the definition from the integers by accounting for the denominator in the fraction. For a nonzero rational number r = a/b, where a and b are nonzero integers, the p-adic valuation is defined as v_p(r) = v_p(a) - v_p(b), with v_p denoting the integer valuation.^[5] The valuation of zero is set to v_p(0) = +\infty.^[5] This definition relies on the prior establishment of v_p on \mathbb{Z}, but it applies directly to \mathbb{Q} as the field of fractions of \mathbb{Z}.^[3] For example, consider p = 2 and r = 3/4. Here, v_2(3) = 0 since 3 is odd, and v_2(4) = 2 since $4 = 2^2. Thus, v_2(3/4) = 0 - 2 = -2.^[3] This negative value reflects that the denominator introduces more factors of 2 than the numerator, highlighting how the valuation on \mathbb{Q} allows for negative exponents unlike on \mathbb{Z}. This extension yields a well-defined function on \mathbb{Q} independent of the choice of representation for r. Suppose r = a/b = (a k)/(b k) for some nonzero integer k. Then v_p(ak) - v_p(bk) = v_p(a) + v_p(k) - (v_p(b) + v_p(k)) = v_p(a) - v_p(b), so the value remains unchanged.^[5] This independence follows from the unique prime factorization in \mathbb{Z}, ensuring consistency across equivalent fractions.^[2]

Properties

Multiplicativity and additivity

The p-adic valuation v_p on the nonzero rational numbers satisfies the multiplicativity property v_p(xy) = v_p(x) + v_p(y) for all x, y \in \mathbb{Q}^\times. This follows directly from the definition of v_p on \mathbb{Q}, where any nonzero rational x can be expressed uniquely (up to units) as x = \pm p^{v_p(x)} \cdot \frac{a}{b} with a, b \in \mathbb{Z} coprime to p; multiplying such expressions for x and y yields the additive exponents for p.^[6] Consequently, v_p defines a group homomorphism from the multiplicative group \mathbb{Q}^\times to the additive group \mathbb{Z}.^[6] To see this multiplicativity explicitly for integers, note that the unique prime factorization theorem in \mathbb{Z} implies that if m, n \in \mathbb{Z} with prime factorizations involving p to powers k and \ell, respectively, then v_p(mn) = k + \ell = v_p(m) + v_p(n). For rationals, the property extends via the quotient definition: v_p(a/b) = v_p(a) - v_p(b) for a, b \in \mathbb{Z} \setminus \{0\} with b \neq 0, so multiplicativity holds by combining the integer case.^[6] The multiplicativity also implies additivity under exponentiation: for any x \in \mathbb{[Q](/page/Q)}^\times and integer n \geq [0](/page/0), v_p(x^n) = n \cdot v_p(x). This follows by induction on n, using the base case n=[0](/page/0) where v_p(1) = [0](/page/0) and the inductive step v_p(x^{n+1}) = v_p(x^n \cdot x) = n v_p(x) + v_p(x) = (n+1) v_p(x).^[6] In the context of p-adic integers \mathbb{Z}_p, the units—elements invertible within \mathbb{Z}_p—are precisely those with v_p(u) = [0](/page/0). This reflects that such units are not divisible by p, preserving the valuation under multiplication by other elements.^[2] For example, consider p=2 and the product (3/4) \cdot (5/2) = 15/8. Here, v_2(3/4) = v_2(3) - v_2(4) = 0 - 2 = -2, v_2(5/2) = v_2(5) - v_2(2) = 0 - 1 = -1, and multiplicativity gives v_2(15/8) = -2 + (-1) = -3, which matches the direct computation v_2(15) - v_2(8) = 0 - 3 = -3.^[6]

Non-Archimedean inequality

One of the defining properties of the p-adic valuation v_p on the rational numbers \mathbb{Q} is its behavior under addition, which satisfies the inequality v_p(x + y) \geq \min(v_p(x), v_p(y)) for all x, y \in \mathbb{Q}.^[1] This contrasts with the additive property of the absolute value on the reals, where |x + y| \leq |x| + |y|, and highlights the "non-Archimedean" nature of v_p, as the valuation of a sum is at least as large as the smaller of the individual valuations, preventing the accumulation of "size" in the same way.^[2] To outline the proof for integers first, suppose without loss of generality that v_p(x) \leq v_p(y), so x = p^{v_p(x)} x' and y = p^{v_p(y)} y' with p \nmid x', y'. Then x + y = p^{v_p(x)} (x' + p^{v_p(y) - v_p(x)} y'), where the term in parentheses is an integer not necessarily divisible by p (unless cancellation occurs). Thus, v_p(x + y) \geq v_p(x) = \min(v_p(x), v_p(y)). The result extends to rationals by clearing denominators and applying the integer case.^[7] Equality holds in the inequality precisely when v_p(x) \neq v_p(y), as the term with the smaller valuation dominates without cancellation. When v_p(x) = v_p(y), the valuation of the sum may be strictly larger if the leading terms cancel modulo p. For example, with p=2, v_2(1 + 2) = v_2(3) = 0 = \min(v_2(1), v_2(2)) = \min(0, 1), showing equality under unequal valuations, while

v_2([2 + 2](/page/2_+_2_=_?)) = v_2(4) = 2 > 1 = \min(v_2(2), v_2(2))

, illustrating the strict inequality possible under equal valuations.^[7] This additive property implies the strict triangle inequality (or ultrametric inequality) for the associated p-adic absolute value |x|_p = p^{-v_p(x)}, yielding |x + y|_p \leq \max(|x|_p, |y|_p), which is stronger than the usual triangle inequality and underscores the non-Archimedean character.^[1] In general, a valuation on a field is called non-Archimedean if it satisfies this minimum inequality under addition, distinguishing it from Archimedean valuations like the one inducing the real absolute value.^[2]

p-adic absolute value

Definition and basic properties

The p-adic absolute value on the rational numbers \mathbb{Q} is defined using the p-adic valuation v_p. For a prime p and x \in \mathbb{Q} with x \neq 0, write x = p^{v_p(x)} \cdot \frac{m}{n} where m, n \in \mathbb{Z} are coprime to p; then |x|_p = p^{-v_p(x)}, and by convention |0|_p = 0 (noting that p^{-\infty} = 0).^[8]^[9] This normalization, with base p, ensures that |p|_p = p^{-1} < 1, distinguishing it from other possible scalings of the valuation and facilitating consistency in the study of completions.^[8] The p-adic absolute value satisfies several foundational algebraic properties: it is multiplicative, so |xy|_p = |x|_p |y|_p for all x, y \in \mathbb{Q}; |1|_p = 1; |-x|_p = |x|_p for all x \in \mathbb{Q}; and |x|_p = 0 if and only if x = 0.^[9]^[8] For example, with p = 2, |4|_2 = 2^{-2} = \frac{1}{4} since v_2(4) = 2, while |1/2|_2 = 2^{1} = 2 since v_2(1/2) = -1.^[9]^[8] Unlike the usual absolute value on \mathbb{R}, where nonzero integers have absolute value at least 1, the p-adic absolute value is non-trivial in the sense that there exist nonzero rationals x with $0 < |x|_p < 1, such as |p|_p = p^{-1}.^[8]

Ultrametric inequality

The ultrametric inequality for the p-adic absolute value states that for any rational numbers x, y \in \mathbb{Q} and prime p,

|x + y|_p \leq \max(|x|_p, |y|_p).

This is a stronger form of the triangle inequality, characteristic of non-Archimedean norms.^[10]^[11] The inequality follows directly from the corresponding property of the p-adic valuation v_p. Specifically, v_p(x + y) \geq \min(v_p(x), v_p(y)) for x, y \neq 0, with the convention v_p(0) = \infty. Since |z|_p = p^{-v_p(z)} for z \in \mathbb{Q}, it follows that

-v_p(x + y) \leq -\min(v_p(x), v_p(y)) = \max(-v_p(x), -v_p(y)),

|x + y|_p = p^{-v_p(x+y)} \leq p^{\max(-v_p(x), -v_p(y))} = \max(p^{-v_p(x)}, p^{-v_p(y)}) = \max(|x|_p, |y|_p).

To derive the valuation inequality, express x = p^{v_p(x)} x' and y = p^{v_p(y)} y' where x', y' \in \mathbb{Q} are not divisible by p. Without loss of generality, assume v_p(x) \leq v_p(y), so x + y = p^{v_p(x)}(x' + p^{v_p(y) - v_p(x)} y'). The term in parentheses has valuation at least $0, yielding v_p(x + y) \geq v_p(x) = \min(v_p(x), v_p(y))$.^[10] Equality holds in the ultrametric inequality if |x|_p \neq |y|_p. In this case, the term with the larger absolute value (smaller valuation) dominates the sum, so |x + y|_p equals the maximum. For instance, with p=2, |1 + 2|_2 = |3|_2 = 1 = \max(|1|_2, |2|_2) = \max(1, 1/2), since v_2(1) = 0 < v_2(2) = 1. However, when |x|_p = |y|_p, strict inequality may occur, as in |2 + 2|_2 = |4|_2 = 1/4 < \max(|2|_2, |2|_2) = 1/2, where the valuations add in the sum.^[11] This inequality induces a non-Archimedean metric d(x, y) = |x - y|_p on \mathbb{Q}, which generates a totally disconnected topology.^[11]