Download Integers modulo N

Integers modulo N
c
Geoff Smith 1998
Divisibility
Suppose that a, b ∈ Z. We say that b divides a exactly when there is c ∈ Zsuch
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.
(a) x | 0 for every x ∈ Z.
(b) Suppose that y ∈ Zand 0 | y, then y = 0.
(c) Both a | b and b | a if and only if |a| = |b|.
(d) If a | b and b | c, then a | c.
(e) If a | b and k ∈ Z, then a | kb.
(f) If a | b and a | c, then a | (b ± c).
Various relations
Suppose that N ∈ N. We define a relation ∼ on Zby writing a ∼ b 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 ∼. They are
{. . . − 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 equivalence 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
Z.
N
Notice that ZN is a set of size N, and that its distinct elements are precisely
0, 1, . . . , N − 1. We want to define addition of elements of ZN . We do it like
this. Suppose x, y ∈ ZN . Choose a ∈ x, b ∈ y. Define x + y to be [a+b]. 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}. Thus the sets P, C, U are
pairwise disjoint, and N = U ∪ P ∪ C. Let X = {U, P, C}. 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. Now 3 + 9 = 12 ∈ C so 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 b
a ∈ x, bb ∈ y. Now a ∼ b
a and b ∼ bb. Thus a − b
a = kN for
b
a + bb) = (k + l)N,
some k ∈ Zand b − b = lN for some l ∈ Z. Thus (a + b) − (b
and so (a + b) ∼ (b
a + bb). We conclude that [a + b] = [b
a + bb] and all is well!
We say that the addition on ZN is well-defined.
Multiplication on
Z.
N
We define an operation × on ZN using the obvious recipe. If x, y ∈ ZN we
select a ∈ x, b ∈ y, and declare x × y to be [a × b]. However, we are now worldly
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wise, and our doubts are definitely in place. We must check that this makes
sense. Choose b
a ∈ x, bb ∈ y. Now Now a ∼ b
a and b ∼ b
b so a − b
a = kN for some
k ∈ Zand b − bb = lN for some l ∈ Z. Thus
a × b = (b
a + kN ) × (bb + lN ) = b
a × bb + (k + l + kl)N.
Therefore (a × b) ∼ (b
a ×b
b) and so [a × b] = [b
a ×b
b].
Laws of algebra of
Z
N
The following laws can all be directly verified using the definitions of addition
and multiplication in ZN . recall that N is an arbitrary, but fixed, natural
number.
(a) x + y ∈ ZN whenever x, y ∈ ZN .
(b) (x + y) + z = x + (y + z) whenever x, y, z ∈ ZN .
(c) x + 0 = 0 + x = x whenever x ∈ ZN .
(d) If x = [a] ∈ ZN , then [a] + [−a] = 0.
(e) x + y = y + x whenever x, y ∈ ZN .
(f) x × y ∈ ZN whenever x, y ∈ ZN .
(g) (x × y) × z = x × (y × z) whenever x, y, z ∈ ZN .
(h) x × 1 = 1 × x = x whenever x ∈ ZN .
(i) x × y = y × x whenever x, y ∈ ZN .
(j) x × (y + z) = (x × y) + (x × z) whenever x, y, z ∈ 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.
Notice that the laws of algebra of ZN . are very familiar. If you replace ZN
by Zthroughout the list, every single law remains valid. However, do not be
deceived. Some strange mathematics can happen in ZN . For example, in Z4 we
have 2 × 2 = 0. This seems very odd at first. The product of non-zero elements
of ZN can sometimes be 0. 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.
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Congruence notation
The notation a ∼ b to indicate that N | (a − b) suffers from two drawbacks. It
suppresses the rôle 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. The number N is called the modulus of the congruence. Thus
1∼
= 3 mod 2, −7 ∼
= 2 mod 3 and 210 ∼
= 4 mod 10.
All the fuss about addition and multiplication being well-defined amounts
to the following. Suppose a, b, c, d ∈ Z and N ∈ N. If a ∼
= b mod N and
c∼
= d mod N, then both a + c ∼
= b + d mod N and ac ∼
= bd mod N.
Now suppose that M ∈ N is a natural number such that M | N, it follows
that if a ∼
= b mod N, thena ∼
= b mod M.
Greatest Common Divisors
The structure ZN is very special when N happens to be a prime number. We
now develop some machinery to understand this situation. Suppose that a, b ∈ .
Let ∆a,b = {d | d ∈ Z, d | a, d | b}. Thus ∆a,b is the set of common divisors
of the integers a and b, 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.
Recall that p ∈ N is a prime number if p 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 is that if a ∈ Z, then g.c.d.(a, p) must be 1 or p. Moreover g.c.d.(a, p) = 1
unless p | a, in which case g.c.d.(a, p) = p.
Division
Theorem[Remainder Theorem] Suppose that a, b ∈ Z and b 6= 0. It follows
that there are uniquely determines q, r ∈ Zwith 0 ≤ r < |b| such that a = qb + r.
Proof Let Γa,b = {a+µb | µ ∈ Z}. The set Γ+
a,b = Γa,b ∩(N ∪{0} is not empty (in
fact Γa,b contains arbitrarily positive and negative integers). Let r = min Γ+
a,b so
r ≥ 0 and a − r = qb for some q ∈ Z. Thus a = qb + r with q, r ∈ Z, 0 ≤ r < |b|.
Now suppose qb, rb ∈ Z, 0 ≤ rb < |b| and a = qbb + rb. Subtracting we find that
0 = a − a = (q − qb)b + (r − rb).
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Thus b divides r − rb but −|b| < r − rb < |b|. We conclude that r − b
r = 0, so
r = rb. Thus qb = qbb and so (q − qb)b = 0. However, b 6= 0 so q = qb and we have
established uniqueness.
Divisors
Suppose that a, b ∈ Zare not both zero. Let
Ωa,b = {λa + µb | λ, µ ∈ Z}.
The set Ωa,b contains both positive and negative integers. Let t be the least
positive element of Ωa,b .
Proposition In this notation we have:
(i) t divides both a and b.
(ii) If d divides both a and b, then d divides t.
(iii) t = g.c.d.(a, b).
Proof 0 < t ∈ Z. Thus a = qt + r according to the Remainder Theorem, so
r ≥ 0. Now r = a − qt ∈ Ωa,b violates the minimality of t unless r = 0. Thus t
divides a. Similarly t divides b, so t is a common divisor of a and b and (i) is
established.
If d divides both a and b, then d divides all elements of Ωa,b so d divides t
and (ii) is established. Now (iii) follows from (i) and(ii).
Euclid’s algorithm
Suppose a, b ∈ Zand b 6= 0. Let a0 = a and a1 = |b|. Given that ai has been
defined for all i ≤ n, if an = 0 let d = an−1 and stop the procedure. On the other
hand, if an 6= 0 then apply the Remainder Theorem to find integers qn , an+1
such that an−1 = qn an + an+1 . Notice that an > an+1 whenever n ≥ 1, so the
integers an form a decreasing sequence of non-negative integers for n ≥ 1. This
this procedure must terminate after finitely many steps, and d = am is defined
at stage m. This procedure is called Euclid’s Algorithm.
Consider the various equations an−1 = qn an + an+1 and an−1 − qn an =
an+1 . From these it follows that (i) any common divisor of an−1 and an divides
an+1 and moreover (ii) any common divisor of an and an+1 divides an−1 . Thus
g.c.d.(an−1, an ) = g.c.d.(an , an+1 ) for every 1 ≤ n ≤ m. It follows that
g.c.d.(a, b) = g.c.d.(a0 , a1 ) = g.c.d.(am , am+1 ) = g.c.d.(t, 0) = t.
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Primes and Products
Proposition Suppose that p is a prime number, and that a, b are integers. It
follows that if p | ab, then either p | a or p | b.
Proof Suppose (for contradiction) that p 6 | a and p 6 | b. Thus g.c.d.(p, a) =
1 = g.c.d.(p, b). Thus there are λ1 , λ2 , µ1 , µ2 ∈ Zsuch that 1 = λ1 p + µ1 a and
1 = λ2 p + µ2 b. Multiply these equations so
1 = λ1 λ2 p2 + λ1 pµ2 b + λ2 pµ1 a + µ1 aµ2 b.
Tidy up by putting λ = λ1 λ2 p + λ1 µ2 b + λ2 µ1 a and µ = µ1 µ2 . We now have
1 = λp + µab. It follows that p 6 | ab. However, this is absurd, so we are done.
Corollary 1 (i) If p is prime and a, b ∈ Zp are such that ab = 0, then either
a = 0 or b = 0. (ii) If a ∈ Zp and a 6= 0, then there is λ ∈ Zp such that aλ = 1.
1.
This is because if a = [a], then there are λ, µ ∈ Zsuch that λa + µb = Q
Corollary 2 If a1 , a2 , . . . , am ∈ Zand p is a prime number with p | m
i=1 ai ,
the p | aj for some 1 ≤ j ≤ m.
It follows from part (ii) of the first corollary that that the non-zero elements
of Z∗p of Zp form a group under multiplication of order p − 1. By Lagrange’s
Theorem we have ap−1 = 1 for all a ∈ Zp. Translated into the language of
congruences we obtain that if the integer a is not divisible by the prime number
p, we have ap−1 ∼
= 1 mod p. Allowing for the case that p divides a, we have
a
mod
p
for
all integers a. Either of the last two results is sometimes
ap ∼
=
called Fermat’s Little Theorem.
Bonus: The Fundamental Theorem of Arithmetic
Theorem[Fundamental Theorem of Arithmetic] Suppose
that n ∈ N, and n >
Qt
1. It follows that n can be expressed as a product i=1 pni i of distinct prime
numbers pi and that (up to commutativity) this factorization is unique.
Proof We first show, by complete induction on n, that every n ∈ N with n > 1
is the product of prime numbers. It is not strictly necessary to begin complete
inductions, but letQ
us do it for safety! The smallest natural number bigger than
t
1 is 2, and that is i=1 pni i with t = 1, p1 = 2 and n1 = 1. Now suppose that m
is an arbitrary natural number bigger than 1. Either m is prime, in which case
it is its own prime factorization (as 2 was), or it is composite. In the latter case
m = m1 m2 with 1 < m1 , m2 < m and m1 , m2 ∈ N. Now each of m1 , m2 is a
product of prime numbers by inductive hypothesis, so m is a product of prime
numbers. By complete induction we are done.
Now for uniqueness. Again we proceed by complete induction, the base
case being unecessary or a matter of staring at 2, depending on your degree
of nervousness.
Suppose
Qsthatmn ∈ N with n > 1 has two rival factorizations
Qt
n = i=1 pni i and n = j=1 qj j . Here the pi are pairwise distinct primes, and
Qs
m
the qj are pairwise distinct primes. Now p1 | n so p1 | j=1 qj j . Thus p1 | qj
for some j by the second corollary. Thus p1 = qj and n/p1 = n/qj . This last
7
equation, together with induction, ensures that the induced factorizations of
n/p1 and n/qj co-incide (up to commutativity). We are done.