I thought it might be nice to post a summary of my project topic up here incase people can use it for the homework :) Plus, it incorporates a lot of ideas from the last chapter, which might be a nice thing to review before the exam.

Kevin, Dennis and I are presenting **Hensel's Lemma** in class April 22nd.

**BACKGROUND INFORMATION**

- Hensel's Lemma was discovered by Kurt Hensel in the early nineteen hundreds. (He was also the inventor of p-adic analysis in 1902).
- Hensel's Lemma is a generic name for analogues for
*complete commutative rings*of the*Newton method*for solving equations. - It is an important result in number theory because it gives information on finding roots of polynomials.

**KEY VOCABULARY**

Complete Commutative Rings: In ring theory, a branch of abstract algebra, a commutative ring is a ring in which the multiplication operation is commutative. The study of commutative rings is called commutative algebra.

P-adic numbers: First described by Kurt Hensel. For each prime number p, the p-adic number system is an extension of the ordinary arithmetic of the rational numbers, achieved by *an alternative interpretation of the concept of absolute value*. The p-adic numbers were motivated primarily by an attempt to bring the ideas and techniques of power series methods into number theory. Their influence now extends far beyond this. For example, the field of p-adic analysis essentially provides an alternative form of calculus! For more information follow this link: http://en.wikipedia.org/wiki/P-adic_number#Properties.

Newton's Meathod: Newton's method is perhaps the best known method for finding successively better approximations to the zeroes (or roots) of a real-valued function.

**HENSEL'S LEMMA**

Let f(x) = x^{d} + a_{1}x^{d - 1} + … + a_{d} be a (monic) polynomial with integer coefficients a_{i}. Let n be an integer so that p|f(n), and f'(n) = dn^{d - 1} + a_{1}(d - 1)n^{d - 2} + … + a_{d - 1} is not divisible by p. Then there is a sequence of integers n_{k} for every k >= 1, so that n_{1} = n, n_{k + 1} - n_{k} is divisible by p^{k} and f (n_{k}) is divisible by p^{k}. Moreover, n_{k} is uniquely determined modulo p^{k}.

In other words:

- Suppose that
**r**is a solution of the conrguence f(r) = 0 mod p^{k-1}, where k is an integer and f(x) is a polynomial with integer coefficients.

- If f'(r) NOT = 0 mod p, then there is a unique integer t, 0<= t <= (p-1) such that f(r + tp
^{k+1}) = 0 mod p^{k}, with t defined as tf'(r) = -(f(r)/p^{k-1}) mod p

- If f'(r) = 0 mod p and f(r) = 0 mod p
^{k}, then f(r + tp^{k-1}) = 0 mod p^{k}for all integers t.

- Also, if f'(r) = 0 mod p, and f(r) NOT = 0 mod p
^{k}, then f(x) = 0 mod p^{k}has no solution for any x = r mod p^{k-1}.

**EXAMPLES**