Number theory and Group theory

The p-adic numbers were introduced by Kurt Hensel in 1897. They were originally used in order to introduce power series in Number Theory. Their applications are various and wide, and they extend some properties of Z and Q to Z_p and Q_p respectively. The p-adic system for any given prime p extends the metric of rational numbers. Given a natural number n, if we take a fixed primer p then n can be expressed in the form n = å_i=0 a_ipⁱ In the traditional arithmetic, this can be understood as the expression in base p of a natural number n. For instance, the p-adic expansion of 52 if p is 3 would be 1212, which is 52 in base 3. When it comes to linear equations modulo p, we can always find a solution and it will be unique.

Proof: The existence and uniqueness of the solution can be shown with diophantine equations. Take the equation ax + b º 0 mod p. It can be rewritten as ax - py = b, which is indeed a diophantine equation since it is linear and there are two variables. Since p is prime, gcd(a,p) = 1. The necessary condition for a diophantine to have infinite solutions is that b needs to be a multiple gcd(a,p). Clearly, 1 is a divisor of b, and hence this equation has infinite solutions. However, they can be no bigger than m, since we are working with congruences. Once we find a base solution x₀ and y₀ (both smaller than m) we can express the general solutions as x = x₀ + kb/gcd(a,p) y = y₀ + ka/gcd(a,p). If we add kp to the base solutions, they will become larger than p except if k = 0. It follows that the solution is unique.

Once linear equations can be solvable, the next step is to study quadratic equations and find its p-adic expansion. Take the equation x² º 5 mod p. This has no rational solution in Z, but there are solutions in Z_p. We choose a prime in which there will exist a x₀ that will satisfy the congruence. For instance, if p equals 7 there exists no x₀, but it does exist if p equals 11, and the two solutions are x₀ = 4 and x₀ = 7. Note that 7 is the inverse of 4 modulo 11, since 7+4 = 11. Then we start working in Z₁₁ and we take x₀ to be 4. We denote the p-adic expansion as a. Recall a is defined to be of the form a = a₀ + a₁p + a₂p² + a₃p³ + ... Then we want a² to be 5 and hence 5 = a² = (a₀ + a₁p + a₂p² + a₃p³ + ...)(a₀ + a₁p + a₂p² + a₃p³ + ...) If we multiply we obtain the following expression a₀² + (2a₀a₁)p + (2a₀a₂+a₀² )p² + (2₀a₃)p³ + (2a₀a₄+2a₁a₃+a₂² )p⁴ + ... All parentheses need to be congruent to 0 modulo 11. Also, for each 11 that we encounter in one parenthesis we will take one carry (such as when we add in base 11). We already know that a₀ equals 4 and then a_0² = 16 = 5 + 11. We carry one to the next parenthesis, which needs to be congruent to 0 modulo 11. 2a₀a₁ + 1 º 0 mod 11. By substituting a₀ we find a₁. 8a₁ + 1 º 0 mod 11 8a₁ º -1 º 10 mod 11. By inspection we see that a₁ equals 4. If we keep repeating this process we find the first terms of the p-adic expansion, which is a = 4 + 4p + 10p² + 4p³ + 0p⁴ + ... With this method any quadratic equation can be solved.