18.310 lecture notes
September 2, 2013
Modular Arithmetic and Elementary Algebra
These notes cover basic notions in algebra which will be needed for discussing several topics
of this course. In particular, we will need them to describe the RSA cryptosystem, the primality
testing algorithms, and for the material on error-correcting codes that we will be covering later in
this course.
1
Euclid’s Algorithm
Euclid’s algorithm (or the Euclidean algorithm) is a very eﬃcient and ancient algorithm to ﬁnd the
greatest common divisor gcd(a, b) of two integers a and b. It is based on the following observations.
First, gcd(a, b) = gcd(b, a), and so we can assume that a ≥ b. Secondly gcd(a, 0) = a by deﬁnition.
Thirdly and most importantly, if
a = zb + c
where z is an integer then gcd(a, b) = gcd(b, c). Indeed any divisor of a and b will divide c, and
conversely any divisor of b and c will divide a. We can compute c by taking the remainder after
dividing a by b, i.e. c is a mod b. (We will discuss the mod operation in greater details in the next
section, but at this point, we only need the deﬁnition of c as the remainder of dividing a by b.) But
c < b < a and thus we have made progress by reducing the numbers we have to compute their gcd
of. And therefore, we can proceed and express b as:
b = yc + d,
(thus d = b mod c) and thus gcd(b, c) = gcd(c, d). We continue until we express gcd(a, b) as
gcd(g, 0) = g, and at that point, we have found the gcd.
Example. Let a = 365 and b = 211.
gcd(211, 154). Continuing, we get:
Then c = 154 and we have that gcd(365, 211) =
gcd(365, 211) = gcd(211, 154)
= gcd(154, 57)
= gcd(57, 40)
= gcd(40, 17)
= gcd(17, 6)
= gcd(6, 5)
= gcd(5, 1)
= gcd(1, 0)
= 1.
Algebra-1 For example, 40 was found by taking 154 mod 57. The gcd of 365 and 211 is 1, which means that
they are relatively prime.
We now state an easy consequence of Euclid’s algorithm
Lemma 1. For any positive integers, there exist integers s and t such that gcd(a, b) = sa + tb.
Indeed, Euclid’s algorithm also allows to ﬁnd such integers s and t. This clearly proves that no
common divisor to a and b is greater than gcd(a, b) since any common divisor to a and b is also a
divisor to sa + tb. To ﬁnd s and t, we proceed bottom up. Suppose we have found u and v such
that
gcd(b, c) = ub + vc.
Then, knowing that a = zb + c allows us to replace c by a − zb and therefore get:
gcd(a, b) = gcd(b, c) = ub + v(a − zb) = va + (u − vz)b.
Thus, we have expressed the gcd as an integer combination of a and b, knowing it as an integer
combination of b and c. Thus bottom up we can ﬁnd s and t such that
gcd(a, b) = sa + tb.
This procedure is often referred to as the extended Euclidean algorithm.
Example. Consider again the example with a = 365 and b = 211. We express their gcd(365, 211) =
1 by going bottom up in the derivation above, and derive:
1 =
=
=
=
=
=
=
6−5
6 − (17 − 2 · 6)
−17 + 3 · (40 − 2 · 17)
3 · 40 − 7 · (57 − 40)
10 · (154 − 2 · 57) − 7 · 57
10 · 154 − 27 · (211 − 154)
37 · (365 − 211) − 27 · 211
=
=
=
=
=
=
−17 + 3 · 6
−7 · 17 + 3 · 40
10 · 40 − 7 · 57
10 · 154 − 27 · 57
37 · 154 − 27 · 211
37 · 365 − 64 · 211
There is another way to implement the extended Euclidean algorithm, which is a little easier to do
on a spreadsheet. If you are ﬁnding gcd(a, b) = gcd(a0 , a1 ) and, after i steps, you have reduced to
it to the calculation of gcd(ai , ai+1 ) with ai ≥ ai+1 , you can keep track of two numbers xi and yi
such that xi a + yi b equals ai . Initially, x0 = 1 and y0 = 0, as a0 = 1a + 0b. To ﬁnd xi and yi , we
know that
ai−2 = xi−2 a + yi−2 b
and
ai−1 = xi−1 a + yi−1 b.
As ai = ai−2 − qai−1 where q is the integer part of the quotient between ai−2 and ai−1 , we simply
let xi = xi−2 − qxi−1 and yi = yi−2 − qyi−1 . Indeed, we have:
xi a + yi b = (xi−2 − qxi−1 )a + (yi−2 − qyi−1 )b
= ai−2 − qai−1
= ai .
Algebra-2 For example, we have
gcd
x
y
gcd(365, 211)
1
0
gcd(211, 154)
0
1
gcd(154, 57)
1
−1
gcd(57, 40)
−1
2
gcd(40, 17)
3
−5
gcd(17, 6)
−4
7
gcd(6, 5)
11 −19
gcd(5, 1)
−26 45
gcd(1, 0)
37 −64
Here we found that 6 = 11 ∗ 365 − 19 ∗ 211 by subtracting twice the equation 17 = −4 · 365 + 7 ∗ 211
from 40 = 3 ∗ 365 − 5 ∗ 211.
2
Modular Arithmetic
We will now consider algebraic structures. Before going into the general deﬁnitions, we introduce
a very important example called modular arithmetic, which is one of the most intuitive examples
of algebraic structures. In fact, this is the example we shall need for the RSA cryptosystem.
First an easy deﬁnition: for integers a, b, m we say that
a≡b
if a − b is a multiple of m. That is, if
(mod m)
a − b = im
for some integer i ∈ Z. Here, ”mod” is short for ”modulo”.
For any integer n ∈ Z there is a unique integer r in {0, 1, . . . , m − 1} such that n ≡ r mod m.
Then r is called the residue of n modulo m, and by slight abuse of notation we will refer to it as
n mod m. One can ﬁnd the residue of a number n by taking the remainder when dividing by m.
Although we will often us them interchangeably, there is a slight diﬀerence between a = n mod m
and a ≡ n (mod m); in the former case, a is the residue and thus between 0 and m − 1. In later
notes, however, we typically simply write a = n (mod m) and the interpretation is usually clear.
We can then deﬁne binary operations ⊕, ⊗ on the set Zm := {0, 1, . . . , m − 1} as follows. For
a, b ∈ Zm we deﬁne a ⊕ b to be the residue of (a + b) modulo m. Similarly, we deﬁne a ⊗ b to be
the residue of (a × b) modulo m.
Example. In Z5 , one has 3 ⊕ 4 = 2 and 3 ⊗ 4 = 2.
For a ∈ Zm , we denote a the residue of −a modulo m. Here are a few very easy facts that
the reader is invited to check. If a ∈ Zm and d = a then
a ⊕ 0 = 0 ⊕ a = a,
and
a ⊕ d = d ⊕ a = 0.
Algebra-3 Moreover, for all a, b, c ∈ Zm ,
(a ⊕ b) ⊕ c = a ⊕ (b ⊕ c).
In the next section we will deﬁne groups and see that the above relation are precisely the
conditions showing that (Zm , ⊕) is a group. If we consider the operation ⊗, the role of 0 for ⊕ is
now played by 1 since a ⊗ 1 = 1 ⊗ a = a. 1 is the multiplicative identity, in the same way as 0
was the additive identity. However 0 never has a multiplicative inverse (in the same way as a is
playing the role of the additive inverse); the multiplicative inverse of an element a is deﬁned as an
element b such that b ⊗ a = 1. Even if we exclude 0 and consider Zm − {0}, we will see that some
nonzero elements may not have a multiplicate inverse. However, when m is a prime number, we
will see that (Zm − {0}, ⊗) is a group
Considering the general notion of group allows one to prove theorems that are valid for all
groups, instead of doing a proof for each individual example. For instance we will prove Fermat’s
Little Theorem using general results about groups.
3
Groups
We now discuss algebraic structures and their properties. This is presented in more depth than
what we really need at this point.
Given a set G and a binary operation ∗, if each element in the set obeys the following 4
properties, then the set and its operation (G, ∗) is called a group.
(i) Closure. If a, b ∈ G, them a ∗ b ∈ G.
(ii) Associativity. (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ G.
(iii) Existence of an identity element. Suppose e ∈ G is the identity element, then a ∗ e = e ∗ a = a
for all a ∈ G.
(iv) Inverse. For every a ∈ G, there exists an a−1 ∈ G such that a ∗ a−1 = a−1 ∗ a = e.
If, in addition, each pair of elements a, b ∈ G satisﬁes the commutative property, a ∗ b = b ∗ a,
then the group (G, ∗) is called an Abelian group.
Examples:
• The set integers form an (Abelian) group under addition as the rule of composition; and so do
the rational, real, or complex numbers. The identity element e in these cases are the number 0,
and the inverse of a is −a.
• The integers under multiplication, (Z, ∗), is not a group since the integers a = 1, −1 don’t have
an inverse.
• If one leaves out zero, the additive identity element, the rational, real, and complex numbers each
form an (Abelian) group under the operation of multiplication. We need to leave out zero since
this element does not have a multiplicative inverse.
• Let Gln be the set of invertible n × n matrices with the usual matrix product as operation. Then
(Gln , ∗) is a group (with identity the identity matrix) which is not Abelian (as matrix multiplication
is not commutative in general).
Algebra-4 • Remainders formed by dividing by a polynomial do likewise. For example, if we take the remainder after dividing by say x3 + 2x2 + 1, we can get all polynomials of degree 2 as remainders, and
the identity is the 0 polynomial (p(x) = 0 everywhere) and the inverse of p(x) is −p(x).
In the previous section we have checked that for any integer m the set (Zm , ⊕) is a group, with
identity element 0. The additive inverse of an element a = 0 is a = m − a. Now we consider the
more interesting question: for which integer m does (Zm − {0}, ⊗) is a group? First of all closure
and associativity are clear, and 1 is the identity. So the question is to know whether every element
has a (multiplicative) inverse. First observe that 0 cannot have a multiplicative inverse (indeed
b ⊗ 0 = 0 = 1 for all b ∈ Zm ) and that’s why we excluded it from the start. Now suppose a ∈ Zm is
relatively prime with m. In this case, we know by Euclid’s algorithm that there exist integers s, t
such that
as + mt = gcd(a, m) = 1.
Thus 1 = as mod m. Hence taking r ∈ Zm the residue of s modulo m one gets a ⊗ r = 1. We have
just proved that the elements a ∈ Zm − {0} which are relatively prime with m have a multiplicative
inverse. Thus if m is a prime number, every element in Zm − {0} has a multiplicative inverse so
that (Zm − {0}, ⊗) is a group.
Now if m is not a prime. Consider an element a ∈ Zm − {0} which is not relatively prime with
m. Let d = gcd(a, m) = 1. This means that, for any b, ab is an integer multiple of d and thus
cannot give a residue of 1 modulo m. Thus a cannot have a multiplicative inverse. So the elements
not prime with m have no inverse in (Zm , ⊗). But we can still salvage a multiplicative group as we
show now.
For any integer m we denote by Z∗m the remainders that are relatively prime to m. For instance
if m is prime then Z∗m = Zm − {0} while for m = 15 one gets Z∗15 = {1, 2, 4, 7, 8, 11, 13, 14}. The
product of any two element relatively prime to m is still relatively prime to m, and every element
has an inverse so it is easy to see that (Z∗m , ⊗) is a group.
4
Subgroups and cosets
The number of elements in a group G is called the order of G, written |G|.
Example. The order of (ZN , ⊕) is N . The order of (Z∗N , ⊗) is the number of remainders which
are relatively prime to N . If N is a prime then Z∗N = ZN − {0} and the order in N − 1. Now
consider the situation where N = pq, where p and q are primes. In this case the order of (Z∗N , ⊗) is
(p − 1)(q − 1) = N − p − q + 1. To see this, observe that the remainders in {1, . . . , N − 1} that have
a factor in common with N are multiples of either p or q but not of both. There are q − 1 of the
former type, p − 1 of the latter. So the order of Z∗N is (N − 1) − ((p − 1) + (q − 1)) = N − p − q + 1.
For example, for N = 15, we have that the order of Z∗15 is 2 · 4 = 8, and indeed, this is the number
of elements we found.
A group G is said to have a subgroup H if H is a subset of G, and H is also a group (under the
same operation ∗ as G). Check for yourself, that if we know G is a group, and we want to know
if some subset H of G is a subgroup, the only group properties we really have to check are closure
Algebra-5 and inverses. We now are ready to state (and prove) one of the simplest and most fundamental
facts about groups.
Theorem 1 (Lagrange’s Theorem). Suppose |G| is ﬁnite, and H is a subgroup of G. Then |H|
divides |G|.
For example, take for G the multiplicative group Z∗7 with 6 elements, and consider H to be
subgroup consisting of all the distinct powers of 2, that is 2, 21 = 2, 22 = 4, 23 = 1 mod 7. H is
a subgroup (since it is closed and every element has an inverse), and the order of H is 3 which
divides the order of G (equal to 6).
Proof. Suppose H has h elements. We will partition the elements of G into disjoint ’copies’ of H.
Each such copy will be of the form xH, where x is an element of G and xH = {xy : y ∈ H} denotes
the h elements obtained by multiplying (i.e. performing the group operation) x by each element of
H. (In the example above, 3H corresponds to {3 ⊗ 1, 3 ⊗ 2, 3 ⊗ 4} = {3, 6, 5} where ⊗ corresponds
to multiplication in Z∗7 .)
Suppose we have already identiﬁed x1 , · · · , xj such that all xk H are disjoint for k = 1, · · · , j.
So far, these sets cover precisely jh distinct elements of G. To initialize this process, we set x1 = e
(where e is the identity element) and j = 1. Now, either we have all elements of G or we don’t. In
the ﬁrst case, we know that |G| = j|H| and we are done. In the second case, let xj+1 be an element
of G which is not of the form xk q for k ≤ j and some q in H.
We now add to our list the h additional elements xj+1 H of G. We prove that these elements
are all new and distinct: if xj+1 h1 = xj+1 h2 holds then by multiplying on the left of both sides of
1
this equation by x−
j +1 we ﬁnd that h1 and h2 are equal; if xk g = xj+1 h holds for some g and h in
H and k ≤ j, then upon postmultiplying both sides of this equation by h−1 we get
xk gh−1 = xj+1
and therefore xj+1 ∈ xk H (since gh−1 ∈ H by closure and the existence of inverse), a contradiction.
We then increase j by 1 and repeat. If G is ﬁnite, this argument must come to an end which can
only happen when the order of G is jh for some j.
Let x be an element of a ﬁnite group G. The powers of x form a subgroup of G called the group
generated by x, and we deﬁne the order of an element x to be the order of that subgroup. One can
see that the order of x is the smallest positive power k such that xk = 1 (indeed if there were two
indices j, l with j < l ≤ k and xj = xl , then xl−j = 1 contradicting the deﬁnition of k). Hence
for all x ∈ G, we must have xo(x) = 1, where o(x) denote the order of x. If we apply Lagrange’s
theorem to G and x ∈ G, then we see that o(x) divides the order |G| of G, and therefore xo(x) = 1
implies that x|G| = 1 for all x ∈ G. In particular, if we take G = Z∗p with p prime, we get Fermat’s
little theorem (since the order of Z∗p is p − 1):
Theorem 2 (Fermat’s little theorem). If p is prime and a is not divisible by p then ap−1 = 1 mod p.
Also, if we take G = Z∗N with N = pq, p and q being prime, we get that |G| = (p − 1)(q − 1)
and thus x(p−1)(q−1) = 1 for all x relatively prime with pq.
Cosets of normal subgroups.
Let G be a group and H be a subgroup with h elements. For any group element x ∈ G, the h
Algebra-6 elements of G of the form xH form what is called a left coset of the subgroup H; a right coset is
similarly deﬁned as Hx. If the right and left cosets for each element are the same so that for all a
in G we have aH = Ha, then H is said to be normal. In Abelian groups, all subgroups are normal.
If H is a normal subgroup of G, then its cosets form a group under the rule of composition
aHbH = abH; this subgroup is called the factor group G/H of G with respect to H.
For example, if G is the group Z of integers under addition, and H is the subgroup consisting
of those integers divisible by n (which we denote by nZ), then the factor group has elements which
correspond to the remainders upon dividing integers by n. This is called, as we remarked earlier,
Zn , and is often referred to as the integers mod n. Thus Zn can be seen as the factor group Z/nZ
of the group (Z, +).
5
The Chinese Remainder Theorem
This theorem was discovered by the Chinese mathematician Sun Tzu in the 4-th century AD and
written in his book the Sun Tzu Suan Ching. It says the following. If a and b are relatively
prime then there is a bijection between the possible remainders modab and the pairs of possible
remainders moda and modb. In other words, the two numbers (the remainder of x upon dividing
by a and the remainder of x upon dividing by b) uniquely determines the number x upon dividing
by ab, and vice versa. Let’s look at an example. Let a = 7 and b = 13, then ab = 91. Any
arbitrary remainder, say 73 mod 91, is equivalent to the pair (3, 8) = (73 mod 7, 73 mod 13). No
other remainder mod 91 leads to the pair (3, 8).
Theorem 3 (Chinese Remainder Theorem). Let a and b be integers that are relatively prime. Each
pair of remainders (r, s) mod a and b separately corresponds to exactly one remainder t mod ab
such that r = t mod a and s = t mod b.
Moreover, if one adds or multiplies remainders with respect to ab, the corresponding remainders
with respect to a and b separately add or multiply correspondingly.
Proof. In order to show this, ﬁrst note that the number of possible remainders modab is ab, while
the number of pairs of possible remainders mod a and mod b is also ab. To any remainder t mod ab,
there corresponds a pair (t mod a, t mod b) of remainders moda and modb. So we only need to
show that there cannot exist two distinct remainders x and y upon dividing by ab, and that x and
y have the same remainders upon dividing by a and by b. Suppose the contrary. In this case, both
a and b divide the diﬀerence x − y. Since we assumed that a and b are relatively prime, it implies
that ab divides x − y. This implies that x and y have the same remainder upon dividing by ab and
are therefore equal. This is a contradiction. Thus, each remainder of ab corresponds to a unique
pair of remainders for a and b separately. This proves the ﬁrst statement of the theorem. The
second statement is easy to check and is left to the reader.
By the above theorem, we can now describe remainders with respect to ab by the corresponding
pairs of remainders with respect to a and b separately: we let (s, t) represent the remainder that is
s with respect to a and t with respect to b.
Now suppose that we want to ﬁnd the remainder modulo ab that correspond to the pair (s, t).
Let x ∈ {0, 1, . . . , ab − 1} be this remainder. We know that s = x mod a and t = x mod b. In
particular, x = kb + t where 0 ≤ k < a. Moreover s ≡ kb + t (mod a), hence kb ≡ s − t (mod a).
Now recall that b has a multiplicative inverse in the group (Z∗a , ⊗), and that this inverse can be
Algebra-7 found using Euclid’s algorithm. Using this inverse we can compute k = b−1 (s − t) mod a, and then
compute x = kb + t.
Now let us play a little bit with the remainder pairs representations. Let us try for instance
to ﬁnd the solutions of the equation x2 = 1 mod ab where a and b are relatively prime. Since
the remainder 1 mod ab is represented by the remainder pair (1, 1) (where the pair represents the
values modulo a and b), it is easy to see that this equation has four solutions: the remainder pairs
(1, 1), (−1, 1), (1, −1), (−1, −1). (Here, (−1, 1) is a convenient notation for (a − 1, 1).) Why are
these all solutions to x2 = 1 mod ab? For any x ∈ {(1, 1), (−1, 1), (1, −1), (1, 1)}, we have that
x2 gets represented by ((±1)2 , (±1)2 ) = (1, 1) and, by the Chinese remainder theorem, the only
remainder mod(ab) that corresponds to this is 1.
For example, consider a = 3 and b = 5. The remainders r = 1, 4, 11, and 14 are relatively
prime to ab = 15, and correspond to the remainder pairs (1, 1), (1, −1) = (1, 4), (−1, 1) = (2, 1)
and (−1, −1) = (2, 4) respectively. In each case, r ≡ ±1 (mod 3) and r ≡ ±1 (mod 5) so that
r 2 ≡ (±1)2 ≡ 1 (mod 3) and r2 ≡ (±1)2 ≡ 1 (mod 5), so that r2 ≡ 1 (mod 15). And indeed
12 = 1, 42 = 16, 112 = 121, and 142 = 196 are all ≡ 1 mod 15.
6
Fields and algebraic equations
(F, +, ∗) is a ﬁeld if
1. (F, +) is an Abelian group with identity element denoted 0.
2. (F − {0}, ∗) is an Abelian group with identity element denoted 1 = 0.
3. Distributive property. For all a, b, c ∈ F , a ∗ (b + c) = (a ∗ b) + (a ∗ c).
From our previous discussion on Zm it is easy to see that if (Zm , ⊕, ⊗) is a ﬁeld if and only if
m is a prime. The rationals, real or complex numbers (with usual addition and multiplication) are
also ﬁelds.
Fields have the important property that the product of any two non-zero elements is not zero.
Lemma 2. If a and b belong to a ﬁeld F and ab = 0. Then either a = 0 or b = 0.
Proof. Suppose that neither a nor b is 0. Then a−1 and b−1 exist and (b−1 )(a−1 )(ab) = 1. This
implies that ab has an inverse, but this cannot be true since ab = 0. A contradiction.
Lemma 3. If a is a solution of the polynomial equation p(x) = 0 with coeﬃcients in a ﬁeld, then
(x − a) divides p(x).
Proof. We can express p(x) = q(x)(x − a) + r for some polynomial q(x) and remainder r. Since
p(a) = 0, this implies that r = 0.
We can now prove the fundamental theorem of algebra.
Theorem 4. A polynomial of degree d ≥ 1 with coeﬃcients in a ﬁeld F can have at most d roots
in F .
Algebra-8 Proof. We will show this by induction on k. If ax + b = 0, then x = −ba−1 is the unique solution,
so the statement is true for k = 1. Let p(x) be a polynomial of degree d > 1, and let x = a be one
of its roots. By lemma 3, we have p(x) = (x − a)q(x), where q(x) is a polynomial of degree d − 1,
and by the induction hypothesis, q(x) has at most d − 1 roots. Lemma 2 says that every root of
q(x)(x − a) is either a root of q(x) or a root of (x − a). Thus, this implies that p(x) = q(x)(x − a)
has at most (d − 1) + 1 = d roots.
Algebra-9
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