Integers modulo N Geoff Smith c 1998 Divisibility : b a c ∈ Z Suppose that a; b ∈ Z We say that divides exactly when there is such that a = bc: We express the fact that b divides a in symbols by writing b | a: Observations We leave the reader to verify all of the following simple facts. x | x ∈ : (a) 0 for every Z y ∈ | y; y : (b) Suppose that Zand 0 then =0 (c) Both a | b and b | a if and only if |a| = |b|: (d) If a | b and b | c; then a | c: a | b k ∈ ; a | kb: (e) If and Z then (f) If a | b and a | c; then a | (b c): Various relations : ∼ Z a ∼ b Suppose that N ∈ N We define a relation on by writing exactly ; when N | (a − b): It is easy to check that ∼ is an equivalence relation. If x ∈ Z }: then the equivalence class [x] which contains x is {x + kN | k ∈ Z The set of equivalence classes is written ZN. An Example Suppose that N =3:There are exactly three equivalence classes of ∼.Theyare {:::−9;−6;−3;0;3;6;9:::} {:::−8;−5;−2;1;4;7;10 :::} and 1 2 {:::−7;−4;−1;2;5;8;11 :::}: We could write the first of these classes as [0]; [3]; [6]; [−3]; or as the equiv- alence class of any one of its elements. However, the square brackets can get a little annoying. We can use a bold font instead, so the first equivalence class is 0(= 3 = 6 = −3 = :::) If you are making hand written notes, a neat way to indicate bold type is to underline the symbol. Thus you can write [1] = 1 =1: Addition on ZN. N; Notice that ZN is a set of size and that its distinct elements are precisely 0; 1;:::;N−1: We want to define addition of elements of ZN. We do it like a∈x;b∈y: x y a b this. Suppose x; y ∈ ZN.Choose Define + to be [ + ]. Notice that the plus sign in [a + b] indicates addition of integers. Now, there is something rather dodgy about this recipe. To illustrate the problem, we make a diversion. Let P be the set of all prime numbers, let C be the set of composite numbers and let U = {1}.ThusthesetsP; C; U are U ∪ P ∪ C: X {U; P; C}: pairwise disjoint, and N = Let = Try to define addition on X as follows: when A; B ∈ X; choose a ∈ A; b ∈ B and let A + B to be that element of X which contains a + b: Right, it is bright and early on Monday morning. The phone rings: someone needs to know P +C urgently. You choose 7 ∈ P and 6 ∈ C: Now 7+6 = 13 ∈ P; so you answer that P + C = P: The next day, the same clown phones again, claiming to have mislaid P + C and asking for it again. You choose 3 ∈ P and 9 ∈ C.Now3+9=12∈Cso you confidently answer that P + C = C: On Wednesday the punter phones once more, having found the scrap of paper on which Monday’s answer had been written. The customer is very angry. How come P + C is P on Mondays but C on Tuesdays, even though P =6 C? The problem is that you have freedom of action; you can choose a ∈ P and b ∈ C and the set where a + b lives depends on which particular a and b you happen to select. Now, this is disturbing because we have allowed this freedom : of action when trying to define addition in ZN However, in that case there is not a problem. To see this, recall that we tried to add x; y ∈ ZN by selecting a ∈ x;b∈y;and declaring x +y to be [a + b]: Suppose we do it again (it is now Tuesday!). Choose ba ∈ x; b ∈ y: Now a ∼ ba and b ∼ bb: Thus a − ba = kN for b b b − b lN l ∈ Z: a b − ba b k lN; some k ∈ Zand = for some Thus ( + ) ( + )=( + ) and so (a + b) ∼ (ba + b): We conclude that [a + b]=[ba+b] and all is well! well-defined We say that the addition on ZN is . Multiplication on ZN. x; y ∈ Z We define an operation × on ZN using the obvious recipe. If N we select a ∈ x;b∈y;and declare x × y to be [a × b]: However, we are now worldly 3 wise, and our doubts are definitely in place. We must check that this makes sense. Choose ba ∈ x;b ∈ y: Now Now a ∼ ba and b ∼ b so a − ba = kN for some k ∈ b b − b lN l ∈ Z: Zand = for some Thus a × b =(ba+kN) × (b + lN)=ba×b+(k+l+kl)N: Therefore (a × b) ∼ (ba × b)andso[a×b]=[ba×b]: Laws of algebra of ZN The following laws can all be directly verified using the definitions of addition N and multiplication in ZN. recall that is an arbitrary, but fixed, natural number. x; y ∈ Z : (a) x + y ∈ ZN whenever N x y z x y z x; y; z ∈ : (b) ( + )+ = +( + ) whenever ZN x 0 0 x x x ∈ : (c) + = + = whenever ZN x a ∈ ; a −a 0: (d) If =[ ] ZN then [ ]+[ ]= x y y x x; y ∈ : (e) + = + whenever ZN x × y ∈ x; y ∈ Z : (f) ZN whenever N x × y × z x × y × z x; y; z ∈ : (g) ( ) = ( ) whenever ZN x × 1 1 × x x x ∈ : (h) = = whenever ZN x × y y × x x; y ∈ : (i) = whenever ZN x × y z x×y x×z x; y; z ∈ : (j) ( + )=( )+( ) whenever ZN Properties (a)–(d) ensure that ZN is a group under addition. Property (e) ensures that this group is abelian (commutative). Properties (f)–(h) ensure that ZN is a monoid under multiplication (a monoid is just like a group, except that the inverse axiom is missing). Property (i) ensures that this monoid is abelian (commutative). Property (j) is the distributive law of multiplication over addition, which is the only property we have which tells us how multiplication and addition interact. : Z Notice that the laws of algebra of ZN are very familiar. If you replace N by Zthroughout the list, every single law remains valid. However, do not be : Z deceived. Some strange mathematics can happen in ZN For example, in 4 we have 2 × 2 = 0: This seems very odd at first. The product of non-zero elements 0: of ZN can sometimes be This disturbing state of affairs disappears in the case that N is a prime number, and only in that case, as we will see in the next section. We will allow ourselves to denote multiplication by juxtaposition in future. 4 Congruence notation The notation a ∼ b to indicate that N | (a − b) suffers from two drawbacks. It suppresses the rˆole of N; and it is not the notation in common use. The standard ∼ ∼ notation is a = b mod N: Here = is pronounced “is congruent to’, and “mod” is short for modulo.ThenumberNis called the modulus of the congruence. Thus ∼ ∼ 10 ∼ 1 = 3mod2;−7=2mod3and2 = 4mod10: All the fuss about addition and multiplication being well-defined amounts a; b; c; d ∈ ∼ N ∈ N: a to the following. Suppose Z and If = b mod N and ∼ ∼ ∼ c = d mod N; then both a + c = b + d mod N and ac = bd mod N: M | N; Now suppose that M ∈ N is a natural number such that it follows ∼ ∼ that if a = b mod N; thena = b mod M: Greatest Common Divisors N The structure ZN is very special when happens to be a prime number. We now develop some machinery to understand this situation. Suppose that a; b ∈ : ;d|a; d | b}: Let ∆a;b = {d | d ∈ Z Thus ∆a;b is the set of common divisors a b; : of the integers and so ∆0;0 = Z However, this is the case of least interest, so we will assume that at least one of a; b is not 0: Let m =max{|a|; |b|}; so −m ≤ d ≤ m ∀d ∈ ∆a;b: The set ∆a;b is therefore finite, and is not empty because 1 ∈ ∆a;b: Thus ∆a;b has a greatest element called the greatest common divisor of a and b: We write this divisor as g.c.d.(a; b): Notice that ∆a;b =∆b;a : so g.c.d.(a; b)=g.c.d.(b; a): Moreover g.c.d.(a; b) ≥ 1 so g.c.d.(a; b) ∈ N Thus g.c.d.(0; 1) = 1; g.c.d.(−4; 6) = 2 and g.c.d.(−9; −12) = 3: p Recall that p ∈ N is a prime number if has exactly two natural number divisors. Thus the first few prime numbers are 2; 3; 5; 7; 11; 13; 17; 19; 23; 29; 31; 37; 41; 43; 47; 53; 59; 61; 67; 71 ::: Thus p is prime exactly when Ωp;p = {−p; −1; 1;p}has size 4. From the point of view of greatest common divisors, the important point about a prime number p a ∈ ; a; p p: a; p is that if Z then g.c.d.( )mustbe1or Moreover g.c.d.( )=1 unless p | a; in which case g.c.d.(a; p)=p: Division b 6 : Theorem[Remainder Theorem] Suppose that a; b ∈ Z and =0 It follows ≤ r<|b| a qb r: that there are uniquely determines q; r ∈ Zwith 0 such that = + + }: ∩ N ∪{ } Proof Let Γa;b = {a+µb | µ ∈ Z The set Γa;b =Γa;b ( 0 is not empty (in + fact Γa;b contains arbitrarily positive and negative integers).