Math 445 Number Theory

Introduction/Review of concepts from abstract algebra

An integer p is prime if whenever p = ab with a,b \Bbb Z, either a = n or b = n .

[For sanity's sake, we will take the position that primes should also be 2 .]

Fundamental Theorem of Arithmetic: Every integer n 2 can be expressed as a product of primes; n = p1··pk .

If we insist that the primes are written in increasing order, p1 pk, then this representation is unique.

The Division Algorithm: For any integers n 0 and m > 0, there are unique integers q and r with n = mq+r and 0 r m-1 .

[Note: this is also true for any integers n,m with m 0, although you need to replace ``m-1'' with ``|m-1|'' .]

The basic idea: keep repeatedly subtracting m from n until what's left is less than m.

Notation: b|a = ``b divides a'' = ``b is a divisor of a'' = ``a is a multiple of b'', means a = bk for some integer k .

If b|a and a 0, then |b| |a| .

If a|b and b|c, then a|c

If a|c and b|d, then ab|cd

If p is prime and p|ab, then either p|a or p|b

Notation: (a,b) = gcd(a,b) = greatest common divisor of a and b

Different, equivalent, formulations for d = (a,b) :

(1) d|a and d|b, and if c|a and c|b, then c d .

(2) d is the smallest positive number that can be written as d = ax+by with a,b \Bbb Z .

(3) d|a and d|b, and if c|a and c|b, then c|d .

(4) d is the only divisor of a and b that can be expressed as d = ax+by with a,b \Bbb Z .

If c|a and c|b, then c|(a,b)

If c|ab and (c,a) = 1, then c|b

If a|c and b|c, and (a,b) = 1, then ab|c

If a = bq+r, then (a,b) = (b,r)

Euclidean Algorithm: This last fact gives us a way to compute (a,b), using the division algorithm:

Starting with a > b, compute a = bq1+r1, so (a,b) = (b,r1). Then compute b = r1q2+r2, and repeat: ri-1 = riqi+1 +ri+1 . Continue until rn+1 = 0, then (a,b) = (b,r1) = (r1,r2) = = (rn,rn+1) = (rn,0) = rn .

Since b > r1 > r2 > r3 > , this process must end, by well-orderedrness.

We can reverse these calculations to recover (a,b) = ax+by, by rewriting each equation in our algorithm as ri+1 = ri-1-riqi+1, and then repeatedly substituting the higher equations into the lowest one, in turn, working up through the list of equations.

Congruence modulo n : Notation: a b (mod n) (also written a\medspace \underset n \medspaceb) means n|(b-a)

Equivalently: the division algorithm will give the same remainder for a and b when you divide by n

Congruence mod n is an equivalence relation

The congruence class of a mod n is the collection of all integers congruent mod n to a:

[a]n = {b \Bbb Z : a\medspace \underset n \medspaceb} = {b \Bbb Z : n|(b-a)}

Fermat's Little Theorem. If p is prime and (a,p) = 1, then ap-1 \medspace \underset p \medspace1

Because: (a·1)(a·2)(a·3)(a·(p-1)) \medspace \underset p \medspace 1·2·3(p-1) , and (1·2·3(p-1),p) = 1 . Same idea, looking at the a's between 1 and n-1 that are relatively prime to n (and letting f(n) be the number of them), gives

If (a,n) = 1, then af(n) \medspace \underset n \medspace1 .

If the prime factorization of n is p1a1pkak, then f(n) = [p1a1(p1-1)][pkak(pk-1)]

The integers \Bbb Z, the integers mod n \Bbb Zn, the real numbers \Bbb R, the complex numbers \Bbb C are all rings.

A homomorphism is a function f:R S from a ring R to a ring S satisfying:

for any r,r R , f(r+r) = f(r) + f(r) and f(r·r) = f(r) ·f(r) .

The basic idea is that it is a function that ``behaves well'' with respect to addition and multiplication.

An isomorphism is a homomorphism that is both one-to-one and onto. If there is an isomorphism from R to S, we say that R and S are isomorphic, and write R @ S .

Example: if (m,n) = 1, then \Bbb Zmn @ \Bbb Zm×\Bbb Zn . The isomorphism is given by

f([x]mn) = ([x]m,[x]n)

The main ingredients in the proof:

If f: R S and y: R T are ring homomorphisms, then the function w: R S×T given by w(r) = (f(r),y(r)) is also a homomorphism.

If m|n, then the function f: \Bbb Zn \Bbb Zm given by f([x]n) = [x]m is a homomorphism.

Together, these give that the function we want above is a homomorphism. The fact that (m,n) = 1 implies that f is one-to-one; then the Pigeonhole Principle implies that it is also onto!

The above isomorphism and induction imply that if n1,nk are pairwise relatively prime (i.e., if i j then (ni,nj) = 1), then

\Bbb Zn1nk @ \Bbb Zn1××\Bbb Znk . This implies:

The Chinese Remainder Theorem: If n1,nk are pairwise relatively prime, then for any a1,ak \Bbb N the system of equations

x ai (mod ni), i = 1,k

has a solution, and any two solutions are congruent modulo n1nk .

A solution can be found by (inductively) replacing a pair of equations x a (mod n) , x b (mod m), with a single equation x c (mod nm), by solving the equation a+nk = x = b+mj for k and j, using the Euclidean Algorithm.

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