Algebra for Digital Communication Test 1

EPFL - Section de Mathématiques
Algebra for Digital Communication
Prof. E. Bayer Fluckiger
Sections de Systèmes de Communications et Physique
Fall semester 2007
Test 1
Thursday, 25 October 2007
13h15 - 15h15
Nom : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Prénom : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Section : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
No other documents are to be used during the test. The use of calculators is not allowed. Do not
remove the staples from this document. All the calculations and arguments have to be written on
the white sheets of paper. The colored sheets are supposed to be used as scratch paper. Solutions
and arguments on those sheets will not be considered.
Aucun document n’est autorisé. Les calculatrices sont interdites. Ne pas dégrafer le document.
Utiliser les feuilles de couleurs comme brouillons. Tous les calculs et raisonnements doivent figurer
dans le dossier rendu.
Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 5
/22 points
/19 points
/14 points
/22 points
/23 points
Total / 100
Exercise 1 (english, 22 points)
(1) Check whether [587]726 is a unit in Z/726Z. If this is the case, find the
multiplicative inverse of [587]726 in Z/726Z.
(2) Which of the following congruences
(a) 14 · x ≡ 35 (mod 66),
(b) 1437 · x ≡ 12 (mod 2875),
(c) 15 · x ≡ 0 (mod 42),
(d) 49 · x ≡ 30 (mod 151),
has a solution x ∈ Z. Explain for each case why such a solution does or does
not exist. Find all the solutions whenever there exists at least one.
Exercice 1 (français, 22 points)
(1) Est-ce que [587]726 est une unité dans Z/726Z ? Si c’est le cas, trouver l’inverse multiplicatif de [587]726 dans Z/726Z.
(2) Parmi les congruences suivantes,
(a) 14 · x ≡ 35 (mod 66),
(b) 1437 · x ≡ 12 (mod 2875),
(c) 15 · x ≡ 0 (mod 42),
(d) 49 · x ≡ 30 (mod 151),
lesquelles admettent une solition x ∈ Z. Justifier dans chaque cas l’existence
ou l’absence de solution. Donner toutes les solutions lorsqu’il en existe au
moins une.
Solution 1
(1) The element [587]726 is a unit if and only if gcd(587, 726) = 1. So let’s use
the Euclidean algorithm to check that this is true.
726
587
139
31
=
=
=
=
1 · 587 + 139,
4 · 139 + 31,
4 · 31 + 15,
2 · 15 + 1.
Consequently 587 and 726 are coprime and [587]726 is a unit in Z/726Z. To
find its inverse we use the successive Euclidean divisions above :
1 =
=
=
=
31 − 2 · 15
31 − 2 · (139 − 4 · 31) = 9 · 31 − 2 · 139
9 · (587 − 4 · 139) − 2 · 139 = 9 · 587 − 38 · 139
9 · 587 − 38 · (726 − 587) = 47 · 587 − 38 · 726
Therefore [587]−1
726 = [47]726 .
(2) (a) There does not exist any x ∈ Z such that 14 · x ≡ 35 (mod 66). In fact,
since [14]66 is a zero divisor in Z/66Z (gcd(14, 66) = 2), we have that
[14]66 [x]66 is a zero divisor too, for every [x]66 ∈ Z/66Z. But [35]66 is a
unit in Z/66Z.
(b) Notice that 1437·2 ≡ −1 (mod 2875). It follows that [1437]−1
2875 = [−2]2875
and therefore x ∈ Z is a solution of 1437 · x ≡ 12 (mod 2875) if and only
if x ≡ −2 · 12 ≡ −24 (mod 2875), that is
x ∈ {2851 + 2875 · k | k ∈ Z}.
(c) The congruence 15 · x ≡ 0 (mod 42) is satisfied if and only if there exists
y ∈ Z such that 15x = 42y, that is 5x = 14y. Since gcd(5, 14) = 1, it
follows that x ∈ 14Z.
(d) Since 151 is prime, every element of Z/151Z is invertible, so that every
congruence has a solution. Using the Euclidean algorithm we get [49]−1
151 =
[37]151 , so that 49 · x ≡ 30 (mod 151) if and only if x ≡ 37 · 30 ≡ 53
(mod 151).
Exercise 2 (english, 19 points)
Consider the set of matrices
L :=

 a

 −b



b  a,
b
∈
Z/7Z
.

a (1) Show that L is a ring with respect to the usual addition and multiplication
of matrices.
(2) Prove that L is a commutative ring, i.e. for any two matrices A, B ∈ L the
equality AB = BA holds.
(3) Show that a2 + b2 6= 0 for all a, b ∈ Z/7Z such that not both a and b are 0.
(4) Deduce that L is a field.
(5) Show that there exists an injective ring homomorphism Z/7Z → L. Is it
surjective ? Justify your answer.
Exercice 2 (français, 19 points)
On considère l’ensemble de matrices
L :=

 a

 −b



b  a,
b
∈
Z/7Z
.

a (1) Montrer que L est un anneau pour l’addition et la multiplication de matrice
usuelles.
(2) Démontrer que L est un anneau commutatif, i.e. pour toutes matrices A et
B dans L, on a AB = BA.
(3) Montrer que a2 + b2 6= 0 pour tous a, b ∈ Z/7Z tels que a et b ne sont pas
nuls simultanément.
(4) En déduire que L est un corps.
(5) Montrer qu’il existe un homomorphisme d’anneaux injectif Z/7Z → L. Est-il
surjectif ? Justifier la réponse.
Solution 2
(1) proof. First we show that sum and product of two matrices in L both lie
again in L. Let a, b, c, d ∈ Z/7Z. Then



a b   c d

=
−b a −d c


a + c b + d

=
−b − d a + c


a + c b + d

∈ L
−(b + d) a + c
and





ac − bd ad + bc 

−bc − ad −bd + ac
a b   c d

=
−b a −d c


ac − bd ad + bc
= 
∈ L.
−(ad + bc) ac − bd
The addition in L is associative and commutative because the addition of
matrices is defined entry-wise and the addition in Z/7Z is associative and
commutative. By definition the matrix 0 = ( 00 00 ) is contained in L and




a b  0 0

+
=
−b a
0 0



0 0

=
0 0



0 0  a b 

+
0 0
−b a
−b
for arbitrary a, b ∈ Z/7Z. Furthermore, for a, b ∈ Z/7Z, the matrix ( −a
b −a )
is contained in L, and it can easily be checked that it is the additive inverse
a b ). Hence (L, +) is a commutative group.
of ( −b
a
The multiplication in L inherits its associativity from the commutativity of
the multiplication in Z/7Z and the associativity of the usual matrix multiplication. By definition L contains the unit matrix ( 10 01 ) and



a b  1 0 

=
−b a 0 1


a b

=
−b a



1 0  a b 

0 1 −b a
for arbitrary a, b ∈ Z/7Z.
Finally the distributivity of addition and multiplication in L follow from the
commutativity of the multiplication in Z/7Z and the distributivity of the
usual matrix addition and multiplication.
Thus (L, +, ·) satisfies all the axioms of a ring.
(2) proof. Let a, b, c, d ∈ Z/7Z. Then



a b   c d

=
−b a −d c


ac − bd ad + bc 

−bc − ad −bd + ac


ca − db bc + da 
=
= 
−da − cb −db + ca



c d  a b 

.
−d c −b a
(3) proof. In Z/7Z we have
[0]2 = [0], [1]2 = [1], [2]2 = [4], [3]2 = [2], [4]2 = [2], [5]2 = [4] and [6]2 = [1].
So the only squares in Z/7Z are [0], [1], [2] and [4]. It is now clear that adding
two squares that are not both [0] cannot result in [0].
(4) proof. We have to show that every element 6= 0 in L has a multiplicative
a b ) ∈ L. Then not both a and b can be zero.
inverse in L. Let 0 6= A := ( −b
a
Hence det(A) = a2 + b2 6= [0] by part (3). This shows that A is invertible in
M2 (Z/7Z).
It remains to show that the inverse of A also lies in L. But in M2 (Z/7Z) the
inverse of A is given by

−1
A

1 a −b
= 2
a + b2 b a
and we immediately see that A−1 lies in L.
(5) proof. Define a map

ϕ:
Z/7Z −→ L,

a 0
a 7−→ 
.
0 a
For a, a0 ∈ Z/7Z we have
a + a0
0 

ϕ(a + a ) =
=
0
a + a0

a · a0 0 
=
ϕ(a · a0 ) = 
0 a · a0



0
a 0   a0 0 

+
= ϕ(a) + ϕ(a0 )
0
0 a
0 a



and


a 0 a0 0 

= ϕ(a) · ϕ(a0 ).
0 a0
0 a


0
Also ϕ(0) = ( 00 00 ) and ϕ([1]) = ( [1]
0 [1] ), which shows that ϕ is a ring homomorphism. It is furthermore injective since ( a0 a0 ) = 0 implies a = 0. But ϕ is
0 [1]
not surjective since for example ( −[1]
0 ) does not lie in the image of ϕ.
Exercise 3 (english, 14 points)
Let R be a ring. For two ideals I, J ⊂ R we define
I + J := {a + b | a ∈ I, b ∈ J}.
(1) Show that I + J is an ideal of R.
Now let R = Z, and let I = mZ and J = nZ.
(2) Show that mZ + nZ = gcd(m, n)Z.
Exercice 3 (français, 14 points)
Soit R un anneau. Pour deux idéaux I, J ⊂ R on définit
I + J := {a + b | a ∈ I, b ∈ J}.
(1) Montrer que I + J est un idéal de R.
On considère le cas particulier R = Z, I = mZ et J = nZ.
(2) Montrer que mZ + nZ = pgdc(m, n)Z.
Solution 3
(1) proof. Let x, y ∈ I + J. Then x = a + b and y = a0 + b0 with a, a0 ∈ I and
b, b0 ∈ J. Now
b0 ) ∈ I + J.
x + y = (a + b) + (a0 + b0 ) = (a
+ a0 ) + (b| +
{z }
| {z }
∈I
∈J
Furthermore, if λ ∈ R, then
λx = λ(a + b) = (λa
λb ) ∈ I + J.
|{z}) + (|{z}
∈I
∈J
By the course these two conditions suffice to show that I + J is an ideal. (2) proof. From part (1) it follows that I + J = mZ + nZ is an ideal of Z. Let
d = gcd(m, n), and let x = ma + nb ∈ mZ + nZ with ma ∈ mZ and nb ∈ nZ.
Then by definition m = dm0 and n = dn0 . Hence x = d(m0 a + n0 b) ∈ dZ,
which shows mZ + nZ ⊂ dZ.
Now let x = da ∈ dZ. The Bezout identity provides us with the existence of
b, c ∈ Z such that d = mb + nc. Thus x = da = mba + nca ∈ mZ + nZ and
therefore dZ ⊂ mZ + nZ.
Exercise 4 (english, 22 points)
Let G be a finite cyclic group of order n, and let g ∈ G be a generator of G.
Furthermore let d ∈ N with d|n.
(1) Prove that there exists an element h ∈ G of order d.
(2) Deduce that there exists a subgroup H ⊂ G of order d.
(3) Show that a subgroup H ⊂ G of order d contains all elements of h ∈ G with
ord(h) = d, where ord(h) denotes the order of h.
(4) Conclude that for every d ∈ N with d|n there exists exactly one subgroup
H ⊂ G of order d.
Exercice 4 (français, 22 points)
Soit G un groupe cyclique fini d’ordre n et soit g ∈ G un générateur de G. Soit
également d ∈ N un diviseur de n.
(1) Montrer qu’il existe un élément h ∈ G d’ordre d.
(2) En déduire qu’il existe un sous-groupe H ⊂ G d’ordre d.
(3) Montrer qu’un sous-groupe H ⊂ G d’ordre d contient tous les éléments h ∈ G
vérifiant ord(h) = d, où ord(h) désigne l’ordre de h.
(4) Conclure que, pour tout d ∈ N tel que d|n, il existe un sous-groupe H ⊂ G
d’ordre d.
Solution 4
(1) proof. Since G is cyclic of order n, and since g is a generator of G, we know
that g has order n. By our assumption n = dm for some m ∈ N. Then
e = g n = g md = (g m )d .
This shows that ord(g m ) < d.
Assume that ord(g m ) = d0 < d. Then
0
0
e = (g m )d = g md .
But this contradicts the fact that ord(g) = n since md0 < md = n. Hence we
must have ord(g m ) = d.
(2) proof. By part (1) there exist an element h ∈ G of order d. We know from
the course that H := <h> ⊂ G is a cyclic subgroup of order d.
(3) proof. Let n = dm. We know from the proof of part (1) that g m has order d.
Denote by H ⊂ G the subgroup of order d generated by g m , and let h ∈ G
be any element of order d. There exists an s ∈ N such that h = g s . We
must have s ≥ m since otherwise sd < n and (g s )d = g sd 6= e. By Euclidean
division there exist q ∈ N and r ∈ N ∪ {0} with r < m such that s = qm + r.
Now
(g r )d = (g s−qm )d = g sd g qmd = (g s )d = e.
This shows that the g r has an order that divides d. But it follows from r < m
that rd < n. Hence we must have g r = e and r = 0. Therefore s = qm and
g s = (g m )q ∈ H, which shows that every element of order d lies in H.
(4) proof. By part (2) we already know that there exists a subgroup H ⊂ G of
order d. It remains to show that H is unique. Let H 0 ⊂ G be any subgroup of
order d. From the course we know that a subgroup of a cyclic group is again
cyclic. Hence H 0 is cyclic and therefore generated by an element h ∈ H of
order d. But by (3) h lies in H. Since h generates H 0 we must have H 0 ⊂ H.
Since by (3) also H 0 contains all elements of G of order d, H 0 must also
contain the generator of H, whence H ⊂ H 0 . This proves H = H 0 , i.e. H is
the unique subgroup of G of order d.
Exercise 5 (english, 23 points)
Consider the complex numbers C, and denote by i the imaginary unit of C. For
an element z = a + bi ∈ C with a, b ∈ R we define as usual
√
|z| := a2 + b2 .
(1) Show that
S 1 := {z ∈ C∗ | |z| = 1}
is a subgroup of (C∗ , ·).
Let n ∈ N be arbitrary. Recall that every polynomial f ∈ C[X] of degree n
has exactly n roots in C. Define µn ⊂ C∗ to be the set of all the roots of the
polynomial X n − 1 ∈ C[X].
(2) Show that µn is a subgroup of S 1 .
Let m ∈ N. We define the product
µm · µn := {xy | x ∈ µm , y ∈ µn }.
(3) Show that µm · µn is a group.
(4) Prove that there exists an r ∈ N such that µm · µn = µr and give r explicitly.
(5) Show that µn is cyclic.
Exercice 5 (français, 23 points)
Soit C l’ensemble des nombres complexes. On note i l’unité imaginaire de C, i.e.
i2 = −1. On définit de façon habituelle, pour un élément z = a + bi ∈ C avec
a, b ∈ R,
√
|z| := a2 + b2 .
(1) Montrer que
S 1 := {z ∈ C∗ | |z| = 1}
est un sous-groupe de (C∗ , ·).
Soit n ∈ N un entier arbitraire. On rappelle que tout polynôme f ∈ C[X]
de degré n admet exactement n racines dans C. On définit µn ⊂ C∗ comme
l’ensemble des racines du polynôme X n − 1 ∈ C[X].
(2) Montrer que µn est un sous-groupe de S 1 .
Soit m ∈ N. On définit le produit suivant
µm · µn := {xy | x ∈ µm , y ∈ µn }.
(3) Montrer que µm · µn est un groupe.
(4) Démontrer qu’il existe un entier r ∈ N tel que µm · µn = µr et donner
explicitement r.
(5) Montrer que µn est cyclique.
Solution 5
(1) In order to show that S 1 is a subgroup of C∗ , we need to check that S 1 is
stable under multiplication and that every element of S 1 has an inverse in
S 1.
(i) Let z, w ∈ S 1 , then |z · w| = |z| · |w| = 1 · 1 = 1, so that z · w ∈ S 1
(ii) Let z̄ denote the complex conjugate of z ∈ C. Then, if z ∈ S 1 we have
z · z̄ = |z|2 = 1. Consequently z −1 = z̄ ∈ S 1 for every z ∈ S 1 .
(2) First we check easily that µn ⊂ S 1 since |z n | = |z|n for every complex number
z. Moreover µn is a subgroup :
(i) Let z, w ∈ µn , then (z · w)n = z n · wn = 1 · 1 = 1, so that z · w ∈ µn
(ii) Let z ∈ S 1 and let z −1 be its inverse in C∗ . Then (z −1 )n = z −n =
(z n )−1 = 1, therefore z −1 ∈ S 1 .
(3) Let’s show that µm · µn is a group. First, since µm · µn ⊂ C∗ (even ⊂ S 1 ),
associativity and the properties of the neutral element 1, which clearly belongs to µm · µn , come from the multiplication in C∗ . So it remains to show
the stability under multiplication and the existence of inverses :
(i) Let a = z1 · w1 , b = z2 · w2 ∈ µm µn . Then
a · b = (z1 · w1 )(z2 · w2 ) = (z1 · z2 ) (w1 · w2 ),
|
{z } |
∈µm
{z
∈µn
}
so that a · b ∈ µm · µn .
(ii) Let z · w ∈ µm · µn and let z −1 ∈ µm and w−1 ∈ µn be their inverses.
Then z −1 · w−1 ∈ µm · µn is the inverse of z · w.
(4) Let z · w ∈ µm · µn then (z · w)r = z r · wr = 1 if m|r and n|r. It follows that, if
lcm(m, n)|r, then µm · µn ⊂ µr . Conversely, let’s prove that, if r = lcm(m, n),
then µr ⊂ µm · µn . Let z ∈ C be such that z r = 1 and let d = gcd(m, n)
and a, b ∈ Z be such that am + bn = d (a and b exist by Bezout’s theorem).
Then we have
n
m
bn+am
zb d · za d = z d = z
with, since
mn
gcd(m,n) = lcm(m, n),
n m
bd
b mn
d
z
n
=z
m
= z br = 1 and
m n
za d
= 1.
Therefore z b d ∈ µm , z a d ∈ µn and z ∈ µm · µn . This implies µr ⊂ µm · µn .
2πi
(5) Let’s show that ζ = e n generates the group µn . By the fundamental theorem
of algebra we know that the polynomial X n −1 ∈ C[X] has at most n distinct
roots, so that the order of the group µn is ≤ n. Now it is clear that ζ ∈ µn
2πi
has order n, since ζ k = ek n 6= 1, for all k = 1, . . . , n − 1 and ζ n = 1. This
shows that µn is generated by ζ, and so it is cyclic.