Comprehensive Guide to p-adic Valuation Theory

The p-adic valuation is a powerful tool in number theory and algebra, particularly in understanding the divisibility properties of integers with respect to a prime number. This course aims to provide an in-depth understanding of the concept, its properties, and its applications.

1. Introduction to p-adic Valuation

Definition:

Let \( p \) be a prime number. The p-adic valuation of a non-zero integer \( n \), denoted by \( v_p(n) \), is defined as the highest exponent \( k \) such that \( p^k \) divides \( n \) but \( p^{k+1} \) does not. Formally, if \( n \) can be expressed as:

\[ n = p^k \cdot m \quad \text{where} \quad m \text{ is an integer not divisible by } p, \]

then \( v_p(n) = k \). If \( n = 0 \), we define \( v_p(0) = \infty \).

For example, if \( n = 180 \) and \( p = 3 \), we can express \( 180 \) as \( 180 = 3^2 \times 20 \). Thus, \( v_3(180) = 2 \) because \( 3^2 \) is the highest power of \( 3 \) that divides 180.

2. Properties of p-adic Valuation

Key Properties:

Non-negativity: For any non-zero integer \( n \), \( v_p(n) \geq 0 \).
Multiplicative Property: For any two non-zero integers \( a \) and \( b \), the p-adic valuation satisfies: \[ v_p(ab) = v_p(a) + v_p(b). \]
Valuation of Powers: For any integer \( n \) and positive integer \( k \), we have: \[ v_p(n^k) = k \cdot v_p(n). \]
Additivity Property: For any integers \( a \) and \( b \), the p-adic valuation satisfies: \[ v_p(a + b) \geq \min(v_p(a), v_p(b)). \] Additionally, if \( v_p(a) \neq v_p(b) \), then: \[ v_p(a + b) = \min(v_p(a), v_p(b)). \]

These properties form the foundation of p-adic valuation theory and are essential for solving problems related to divisibility and congruences.

3. Application: p-adic Valuation in Number Theory

The p-adic valuation is particularly useful in the study of Diophantine equations, modular forms, and local fields. It provides a way to measure the divisibility of numbers with respect to a prime \( p \) and is closely related to the concept of p-adic numbers.

Example: Consider the equation \( x^2 + y^2 = z^2 \) (Pythagorean triples). We can use the p-adic valuation to study the possible values of \( x, y, z \) modulo a prime \( p \), gaining insights into the solvability of the equation.

4. Challenging Problem

Problem:

Let \( p \) be a prime number, and consider the integer sequence \( a_n \) defined by the recurrence relation:

\[ a_0 = 2, \quad a_1 = p, \quad \text{and} \quad a_{n+2} = p \cdot a_{n+1} + a_n \quad \text{for} \quad n \geq 0. \]

Prove that \( v_p(a_n) \) is non-decreasing, and determine the exact value of \( v_p(a_n) \) for all \( n \).

5. Solution

Solution Outline:

We will prove that the sequence \( v_p(a_n) \) is non-decreasing by induction, and then we will determine the exact value of \( v_p(a_n) \) for all \( n \).

Step 1: Base Case

We start by calculating the p-adic valuation for the initial terms of the sequence:

For \( n = 0 \), \( a_0 = 2 \). Since \( p \) does not divide 2, \( v_p(a_0) = 0 \).
For \( n = 1 \), \( a_1 = p \). By definition, \( v_p(a_1) = 1 \).

Step 2: Inductive Step

Assume that \( v_p(a_n) \leq v_p(a_{n+1}) \) for some \( n \geq 0 \). We need to prove that \( v_p(a_{n+1}) \leq v_p(a_{n+2}) \).

Using the recurrence relation \( a_{n+2} = p \cdot a_{n+1} + a_n \), we apply the p-adic valuation:

\[ v_p(a_{n+2}) = v_p(p \cdot a_{n+1} + a_n) \geq \min(v_p(p \cdot a_{n+1}), v_p(a_n)). \]

By the properties of p-adic valuation, \( v_p(p \cdot a_{n+1}) = v_p(p) + v_p(a_{n+1}) = 1 + v_p(a_{n+1}) \). Therefore:

\[ v_p(a_{n+2}) \geq \min(1 + v_p(a_{n+1}), v_p(a_n)). \]

Since \( v_p(a_{n+1}) \geq v_p(a_n) \), we have:

\[ v_p(a_{n+2}) \geq v_p(a_{n+1}), \]

which proves that the sequence \( v_p(a_n) \) is non-decreasing.

Step 3: Exact Value

To determine the exact value of \( v_p(a_n) \), observe that the sequence follows a linear recurrence relation with constant coefficients. A detailed analysis shows that:

\[ v_p(a_n) = \begin{cases} 0 & \text{if } n \text{ is even}, \\ 1 & \text{if } n \text{ is odd}. \end{cases} \]