An integer p is prime if whenever p = ab with a,b Î \Bbb Z, either a = ±n or b = ±n .
[For sanity's sake, we will take the position that primes should also be ³ 2 .]
Fundamental Theorem of Arithmetic: Every integer n ³ 2 can be expressed as a product of primes; n = p1·¼·pk .
If we insist that the primes are written in increasing order, p1 £ ¼ £ pk, then this representation is unique.
The Division Algorithm: For any integers n ³ 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 ¹ 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 ¹ 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 Î \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 Î \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) = ¼ = (rn,rn+1) = (rn,0) = rn .
Since b > r1 > r2 > r3 > ¼ , 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 º b (mod n) (also written a\medspace \underset n® º \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 Î \Bbb Z : a\medspace \underset n® º \medspaceb} = {b Î \Bbb Z : n|(b-a)}
Fermat's Little Theorem. If p is prime and (a,p) = 1, then ap-1 \medspace \underset p® º \medspace1
Because: (a·1)(a·2)(a·3)¼(a·(p-1)) \medspace \underset p® º \medspace 1·2·3¼(p-1) , and (1·2·3¼(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® º \medspace1 .
If the prime factorization of n is p1a1¼pkak, then f(n) = [p1a1(p1-1)]¼[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¢ Î 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 n1,¼nk are pairwise relatively prime (i.e., if i ¹ j then (ni,nj) = 1), then
\Bbb Zn1¼nk @ \Bbb Zn1×¼×\Bbb Znk . This implies:
The Chinese Remainder Theorem: If n1,¼nk are pairwise relatively prime, then for any a1,¼ak Î \Bbb N the system of equations
x º ai (mod ni), i = 1,¼k
has a solution, and any two solutions are congruent modulo n1¼nk .
A solution can be found by (inductively) replacing a pair of equations x º a (mod n) , x º b (mod m), with a single equation x º c (mod nm), by solving the equation a+nk = x = b+mj for k and j, using the Euclidean Algorithm.