Download Algebra for Digital Communication Test 2

EPFL - Section de Mathématiques
Algebra for Digital Communication
Prof. E. Bayer Fluckiger
Sections de Systèmes de Communications et Physique
Winter semester 2006-2007
Test 2
Thursday, 1st February 2007
13h15 - 15h00
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
/25 points
/25 points
/25 points
/25 points
Total / 100
Exercise 1 (english, 25 points)
(1) (a) Find all solutions x ∈ Z of the system of congruences
x ≡ 0 (mod 3),
x ≡ 3 (mod 4),
x ≡ −1 (mod 5).
(b) Why does there exist an isomorphism
ϕ:
Z/60Z −→ Z/3Z × Z/4Z × Z/5Z,
and how can it be described? Determine ϕ−1 (0, 3, 4).
(2) Consider Z/180Z.
(a) How many different ways are there to write Z/180Z as a product of cyclic
groups
Z/180Z ∼
= Z/m1 Z × · · · × Z/mr Z
with m1 ≤ · · · ≤ mr ?
(b) For (Z/180Z)∗ proceed as in (a), i.e. find all r-tuplets (m1 , . . . , mr ) with
m1 ≤ · · · ≤ mr such that
(Z/180Z)∗ ∼
= Z/m1 Z × · · · × Z/mr Z.
Exercice 1 (français, 25 points)
(1) (a) Trouver toutes les solutions x ∈ Z du système de congruences
x ≡ 0 (mod 3),
x ≡ 3 (mod 4),
x ≡ −1 (mod 5).
(b) Pourquoi existe-t-il un isomorphisme
ϕ:
Z/60Z −→ Z/3Z × Z/4Z × Z/5Z,
et comment le décrire? Déterminer ϕ−1 (0, 3, 4).
(2) Considérons le groupe Z/180Z.
(a) De combien de manières différentes peut-on écrire Z/180Z comme produit de groupes cycliques ? En d’autres termes, trouver (avec justification) le nombre de r-uplets (m1 , ..., mr ), avec m1 ≤ ... ≤ mr , tels qu’on
ait un isomorphisme
Z/180Z ∼
= Z/m1 Z × · · · × Z/mr Z.
(b) Même question pour (Z/180Z)∗ . Autrement dit, trouver tous les r-uplets
(m1 , ..., mr ), avec m1 ≤ ... ≤ mr , tels qu’il existe un isomorphisme
(Z/180Z)∗ ∼
= Z/m1 Z × · · · × Z/mr Z.
Solution 1
(1) (a) We employ the first method given in the script of the lecture. Set a1 := 0,
a2 := 3 and a3 := −1. Furthermore let m1 := 3, m2 := 4 and m3 := 5.
We calculate
k1 := m2 m3 = 20,
k2 := m1 m3 := 15 and k3 := m1 m2 = 12.
Now for i = 1, 2, 3 we need to find ri ∈ Z with ki ri ≡ 1 (mod mi ). Since
k1 ≡ 2 (mod m1 ), k2 ≡ 3 (mod m2 ) and k3 ≡ 2 (mod m3 ), we easily
find r1 := 2, r2 := 3 and r3 := 3. Set
x1 := r1 k1 = 40,
x2 := r2 k2 = 45 and x3 := r3 k3 = 36.
Hence
x := a1 x1 + a2 x2 + a3 x3
= 0 · 40 + 3 · 45 − 1 · 36
= 135 − 36 = 99
is a solution of the given system of congruences. As 3· 4· 5 = 60 it follows
that {99 + 60k | k ∈ Z} is the set of all solutions (see also part (b)).
(b) The isomorphism exists, since
gcd(3, 4) = gcd(3, 5) = gcd(4, 5) = 1.
This means that we can apply the Chinese remainder theorem (theorem
6.2 from the script) to get the isomorphism
Z/60Z −→ Z/3Z × Z/4Z × Z/5Z,
[x]60 7−→ ([x]3 , [x]4 , [x]5 ).
Now by part (a) we have
ϕ−1 ([0]3 , [3]4 , [4]5 ) = ϕ−1 ([0]3 , [3]4 , [−1]5 ) = [99]60 .
(2) (a) The prime decomposition of 180 is 2·2·3·3·5. By the Chinese remainder
theorem we need to find tuples (m1 , . . . , mr ) with mi ≥ 2, gcd(mi , mj ) =
1 for i 6= j, and m1 · · · mr = 180. In this case we have
Z/180Z ∼
= Z/m1 Z × · · · × Z/mr Z.
(i)
Any tuple that does not satisfy the above conditions will not give an
isomorphism. For example if m1 · · · mr 6= 180, then the group on the
right side in (i) does not have the right order. If on the other hand
m1 · · · mr = 180 but gcd(mi , mj ) > 1 for some i 6= j, then every element
in Z/mi Z × Z/mj Z has order < mi · mj . Hence every element in the
group on the right side in (i) has order < m1 · · · mr = 180.
Without loss of generality we can assume that m1 ≤ · · · ≤ mr . But the
only such tuples are
(180),
(9, 20),
(5, 36),
(4, 45) and (4, 5, 9).
(b) From part (a) we see know that
Z/180Z ∼
= Z/4Z × Z/5Z × Z/9Z.
Hence
(Z/180Z)∗ ∼
= (Z/4Z)∗ × (Z/5Z)∗ × (Z/9Z)∗ .
Now we must have (Z/4Z)∗ ∼
= Z/2Z, and since Z/5Z is a field we get
∗ ∼
(Z/5Z) = Z/4Z. Consider the element [2]9 ∈ (Z/9Z)∗ . It has neither
order 2 nor 3 so it must have order 6. Therefore (Z/9Z)∗ is cyclic and
we obtain (Z/9Z)∗ ∼
= Z/6Z. Altogether this results in
(Z/180Z)∗ ∼
= Z/2Z × Z/4Z × Z/6Z.
Since gcd(2, 4) = gcd(2, 6) = gcd(4, 6) = 2 > 1, we see that is is impossible to write (Z/180Z)∗ as a product of less than 3 cyclic factors. We
can only use Z/6Z ∼
= Z/2Z × Z/3Z to obtain
(Z/180Z)∗ ∼
= Z/2Z × Z/2Z × Z/3Z × Z/4Z.
Hence the only possible tuples are
(2, 4, 6),
(2, 2, 12) and (2, 2, 3, 4).
Exercise 2 (english, 25 points)
(1) Consider the polynomials f := X 2 − 3X + 2, g := X 3 − 3X 2 + X − 3 and
q := X 2 − 7X + 12 in Z[X], and let
(a) R1 := Q[X]/(q),
(b) R2 := F2 [X]/(q), where q denotes the image of q in F2 [X].
For i = 1, 2 determine whether the images of f and g in Ri lie in the group
of units (Ri )∗ ?
(2) Let K be a field, and let n ∈ N. Consider the ring R := K[X]/(X n ).
(a) Show that R is a K-vector space.
(b) Give a basis B = {v1 , . . . , vm } of R as a K-vector space and prove that
B really is a basis.
(c) Conclude that the dimension of R as a vector space over K is n.
Exercice 2 (français, 25 points)
(1) Considérons les polynômes suivants: f := X 2 −3X +2, g := X 3 −3X 2 +X −3
et q := X 2 − 7X + 12. Ce sont des éléments de Z[X]. Soient
(a) R1 := Q[X]/(q),
(b) R2 := F2 [X]/(q), où q désigne l’image de q dans F2 [X].
Pour i = 1, 2, déterminer si les images de f et de g dans Ri sont des unités.
(2) Soit K un corps, et n ∈ N. Considérons l’anneau R := K[X]/(X n ).
(a) Montrer que R est un K-espace vectoriel.
(b) Donner (en justifiant) une base B = {v1 , . . . , vm } de R en tant que Kespace vectoriel.
(c) Conclure que la dimension de R comme K-espace vectoriel est n.
Solution 2
(1) It is easy to see that f = (X −1)(X −2) and q = (X −3)(X −4). Furthermore
by a bit of trial-and-error we see that g has the root 3, and simple polynomial
division gives us g = (X − 3)(X 2 + 1).
(a) We consider R1 := Q[X]/(q). Since
g · (X − 4) = (X 2 + 1) · q ≡ 0
(mod q),
we see that the image of g in R1 is a zero-divisor and can therefore not
be a unit.
Now since gcd(f, q) = 1 in Q[X], there exist by Bezout’s identity polynomials r, s ∈ Q[X] with rf + sq = 1. Hence
rf = 1 − sq ≡ 0
(mod q),
which shows that the image of f in R1 is a unit.
(b) Now we consider R2 := F2 [X]/(q). We denote the two elements of F2
by 0 and 1. Denote by f and g are the images of f and g in F2 [X]
respectively. If we take into account that X 2 + 1 = (X − 1)2 in F2 [X] we
obtain
f = X(X − 1),
g = (X − 1)3
and q = X(X − 1).
This shows that
f ≡ 0
(mod q),
and therefore the image of f in R2 is not a unit. Furthermore
g · X = (X − 1)2 · q ≡ 0
(mod q),
which means that the image of g in R2 is a zero-divisor and hence not a
unit.
(2) (a) proof. Let m = X n ∈ K[X]. Since R is a ring, it is by definition an
additive group. So it only remains to show, that
K × R −→ R,
(λ, [f ]m ) 7−→ λ[f ]m := [λf ]m ,
(ii)
satisfies the properties of a scalar multiplication. Let λ, µ ∈ K and
f, g ∈ K[X]. Then
(λµ)[f ]m = [λµf ]m = λ[µf ]m = λ(µ[f ]m ),
1[f ]m = [1f ]m = [f ]m ,
λ([f ]m + [g]m ) = λ[f + g]m = [λf + λg]m
= [λf ]m + [λg]m = λ[f ]m + λ[g]m ,
(λ + µ)[f ]m = [(λ + µ)f ]m = [λf + µf ]m
= [λf ]m + [µf ]m = λ[f ]m + µ[f ]m ,
which shows that (ii) actually defines a scalar multiplication. Hence R
is a K-vector space.
(b) We claim that B := {1, [X]m , [X 2 ]m , . . . , [X n−1 ]m } is a basis of R as a
K-vector space.
proof. Since [X n ]m = 0, and since every polynomial of degree ≥ n is
congruent modulo m to a polynomial of degree < n, it follows that B
generates R as a K-vector space. It remains to show, that the elements
of B are linearly independent. Let λ0 , . . . , λn−1 ∈ K with
λ0 + λ1 [X]m + λ2 [X 2 ]m + · · · + λn−1 [X n−1 ]m = 0.
On other words
P := λ0 + λ1 X + λ2 X 2 + · · · + λn−1 X n−1 ≡ 0
(mod (m)).
This means that X n divides P in K[X]. Since deg(P ) < deg(X n ) we
must therefore have P = 0, i.e. λ0 = · · · = λn−1 = 0. Hence the elements
of B are linearly independent and form a K-basis of R.
(c) proof. Since the cardinality of the basis B = {1, [X]m , . . . , [X n−1 ]m } is
n, it follows that the dimension of R as a K-vector space is n.
Exercise 3 (english, 25 points)
(1) Find a polynomial P ∈ F2 [X] such that K := F2 [X]/(P ) is a field with 8
elements. Explain why you chose such P .
(2) Determine all the roots of P in K.
(3) Denote by α the image of X in K, i.e. K = F2 (α). Let ϕ : K → K be an
isomorphism of fields.
(a) Prove that ϕ(y) = y for all y ∈ F2 ⊂ K.
(b) Show that ϕ(α) is a root of P .
(4) Let β be any root of P in K. Show that there exists a unique isomorphism
of fields ψ : K → K with ψ(α) = β.
(5) Conclude that there are exactly three field isomorphisms K → K.
(6) Does there exist a subfield L ⊂ K such that L ∼
= F4 ? Justify your answer.
Exercice 3 (français, 25 points)
(1) Trouver un polynôme P ∈ F2 [X] tel que K := F2 [X]/(P ) soit un corps à 8
éléments. Expliquer le choix de P .
(2) Déterminer toutes les racines de P dans K.
(3) Notons α l’image de X dans K. En particulier, K = F2 (α). Soit ϕ : K → K
un isomorphisme de corps.
(a) Montrer que ϕ(y) = y pour tout y ∈ F2 ⊂ K.
(b) Montrer que ϕ(α) est une racine de P .
(4) Soit β une racine quelconque de P dans K. Montrer qu’il existe un unique
isomorphisme de corps ψ : K → K, avec ψ(α) = β.
(5) Conclure qu’il y a exactement trois isomorphismes de corps K → K.
(6) Existe-t-il un sous-corps L ⊂ K tel que L ∼
= F4 ? Justifier votre réponse.
Solution 3
(1) We claim that for P = X 3 + X + 1 ∈ F2 [X] the ring K := F2 [X]/(P ) is a
field with 8 elements.
proof. Since the polynomial P does not have any root in F2 , it is irreducible
in F2 [X]. Hence K is a field, and by proposition 8.1 from the script of the
lecture K has 8 elements.
(2) If α is the image of X in K, then all elements of K are given by
{0, 1, α, α + 1, α2 , α2 + 1, α2 + α, α2 + α + 1}.
Now either by testing all the elements of K, or by looking at the multiplication table of K, we find that the set of roots of P in K is given by
{α, α2 , α2 + α}.
(3) (a) proof. Since ϕ is a field isomorphism it is in particular a homomorphism
of rings. Hence ϕ(0) = 0 and ϕ(1) = 1 by definition. The claim follows
from the fact that F2 = {0, 1}.
(b) proof. By the properties of a ring homomorphism we obtain
(ϕ(α))3 +ϕ(α)+1 = ϕ(α3 )+ϕ(α)+ϕ(1) = ϕ(α3 +α+1) = ϕ(0) = 0.
Hence ϕ(α) is also a root of P .
(4) proof. Every element of K can be written as λ2 α2 + λ1 α1 + λ0 with unique
λ0 , λ1 , λ2 ∈ F2 . Hence we can define a map
ψ:
K −→ K,
λ2 α2 + λ1 α1 + λ0 7−→ λ2 β 2 + λ1 β1 + λ0 .
Obviously ψ(0) = 0 and ψ(1) = 1. Let µ0 , µ1 , µ2 ∈ F2 , and set x := λ2 α2 +
λ1 α + λ0 and y := µ2 α2 + µ1 α + µ0 . Then
ψ(x + y) = ψ (λ2 + µ2 )α2 + (λ1 + µ1 )α + (λ0 + µ0 )
= (λ2 + µ2 )β 2 + (λ1 + µ1 )β + (λ0 + µ0 )
= (λ2 β 2 + λ1 β + λ0 ) + (µ2 β 2 + µ1 β + µ0 )
= ψ(x) + ψ(y),
ψ(xy) = ψ (λ2 µ2 + λ2 µ0 + λ1 µ1 + λ0 µ2 )α2
+(λ2 µ2 + λ2 µ1 + λ1 µ2 + λ1 µ+ λ1 µ1 )α
+(λ2 µ1 + λ1 µ2 + λ0 µ0 )
= (λ2 µ2 + λ2 µ0 + λ1 µ1 + λ0 µ2 )β 2
+(λ2 µ2 + λ2 µ1 + λ1 µ2 + λ1 µ+ λ1 µ1 )β
+(λ2 µ1 + λ1 µ2 + λ0 µ0 )
= (λ2 β 2 + λ1 β + λ0 0(µ2 β 2 + µ1 β + µ0 )
= ψ(x)ψ(y),
where the equality ψ(xy) = ψ(x)ψ(y) follows from the fact that β is a root
of P . Hence ψ is a ring homomorphism. Since ψ is a homomorphism of
fields, it must be injective. Furthermore since K is finite, ψ must also be
surjective.
It remains to show that ψ is unique. So let ψ 0 : K → K be another isomorphism of fields with ψ 0 (α) = β = ψ(α). Then also ψ 0 (α2 ) = β 2 = ψ(α2 ).
Naturally ψ 0 (1) = 1 = ψ(1). Since K is a vector space of dimension 3 over
F2 with basis {1, α, α2 }, and since ψ 0 and ψ can also be considered as vector
space homomorphism, it follows that ψ 0 (x) = ψ(x) for all x ∈ K. Hence
ψ 0 = ψ.
(5) proof. Every isomorphism of fields K → K sends α to a root of P , and there
are exactly three roots of P in K. By part (4) there exists for every root β
of P a unique field isomorphism ψ : K → K with ψ(α) = β. Thus there are
exactly three isomorphism of fields K → K.
(6) We claim that there does not exist a subfield L ⊂ K with L ∼
= F4 .
proof. Since the multiplicative group of a subfield L ⊂ K must be a subgroup
of the multiplicative group of K, and since the order of K ∗ is 7, the subfield
L either has the multiplicative group K ∗ or {1}. Hence we must have L = K
or L ∼
= F2 .
Exercise 4 (english, 25 points)
(1) Consider the polynomial ring Z[X].
(a) Let P (X), Q(X) ∈ Z[X] be polynomials. Prove: If P (X) divides Q(X),
Then P (X d ) divides Q(X d ) for all d ∈ N.
Hint: Write Q(X) = P (X)R(X) with some R(X) ∈ Z[X].
(b) Show that X − 1 divides X m − 1 for all m ∈ N.
(c) Let m, n ∈ N with m ≤ n. Deduce from (a) and (b) that X m − 1 divides
X n − 1 if m divides n.
(2) Let p be a prime number, and let m, n ∈ N with m ≤ n.
(a) Use (1).(c) to show that pm − 1 divides pn − 1 if m divides n.
m
n
(b) Prove that X p −1 − 1 divides X p −1 − 1 in Z[X] if m divides n by using
(2).(a) and (1).(c).
(3) For a prime number p and m, n ∈ N, we want to show that Fpn contains a
field with pm elements if and only if m divides n.
(a) Let m divide n. Prove that Fpn contains a subfield K with pm elements.
m
Hint: Define K to be the set of roots in Fpn of the polynomial X p − X.
Why do all the roots of this polynomial lie in Fpn ?
(b) Let Fpn contain a subfield K with pm elements. Show that m divides n.
Hint: Recall that Fpn is a vector space over K.
Exercice 4 (français, 25 points)
(1) Considérons l’anneau Z[X].
(a) Soient P (X), Q(X) ∈ Z[X] des polynômes. Montrer que, si P (X) divise
Q(X), alors P (X d ) divise Q(X d ) pour tout d ∈ N.
Indication: Ecrire Q(X) = P (X)R(X) avec R(X) ∈ Z[X].
(b) Montrer que X − 1 divise X m − 1 pour tout m ∈ N.
(c) Soient m, n ∈ N avec m ≤ n. Déduire de (a) et (b) que X m − 1 divise
X n − 1 si m divise n.
(2) Soit p un nombre premier, et m, n ∈ N deux entiers tels que m ≤ n.
(a) Utiliser (1).(c) pour montrer que pm − 1 divise pn − 1 si m divise n.
m
n
(b) Montrer que X p −1 − 1 divise X p −1 − 1 dans Z[X] si m divise n, en
utilisant (2).(a) et (1).(c).
(3) Soit p premier et m, n ∈ N. On veut montrer que Fpn contient un corps à pm
élements si et seulement si m divise n.
(a) Supposons que m divise n. Montrer que Fpn contient un sous-corps K à
pm élements.
Indication: Définir K comme l’ensemble des racines dans Fpn du polym
nôme X p − X. Pourquoi ce polynôme a-t-il toutes ses racines dans
Fpn ?
(b) Supposons que Fpn contienne un sous-corps K à pm élements. Montrer
que m divise n.
Indication: Fpn est un K-espace vectoriel.
Solution 4
(1) (a) proof. As P (X) divides Q(X) there exists an R(X) ∈ Z[X] with Q(X) =
P (X)R(X). But obviously this equality stays an equality if we replace
X by X d , i.e. we have Q(X d ) = P (X d )R(X d ). Hence P (X d ) divides
Q(X d ).
(b) We show by induction on m that
X m − 1 = (X − 1)(X m−1 + X m−2 + · · · + 1).
proof. If m = 1, the claim is trivial. So let m > 1. By Euclidian division
we obtain
X m − 1 = X m−1 (X − 1) + (X m−1 − 1).
The induction hypothesis claims (X m−1 − 1) = (X − 1)(X m−2 + · · · + 1).
Altogether we obtain
X m − 1 = (X − 1)X m−1 + (X − 1)(X m−2 + · · · + 1)
= (X − 1)(X m−1 + X m−2 + · · · + 1).
(c) proof. If m divides n, then there exists a d ∈ N with n = dm. We know
by (1).(b) that X − 1 divides X d − 1. It follows from (1).(a) that X m − 1
divides X dm − 1 = X n − 1.
(2) (a) proof. By (1).(c) there exists some R(X) ∈ Z[X] such that X n − 1 =
(X m − 1)R(X) in Z[X]. This equality still holds if we replace X by p,
i.e. pn − 1 = (pm − 1)R(p). Hence pm − 1 divides pn − 1.
(b) proof. Since pm − 1 divides pm − 1 by (2).(a), it follows from (1).(c) that
m
n
X p −1 − 1 divides X p −1 − 1.
(3) (a) proof. From the lecture we know that Fpn can be regarded as the set of
n
n
all roots of the polynomial X p − X = X(X p −1 − 1). Since by (2).(b)
m
n
m
n
X p −1 − 1 divides X p −1 − 1, it follows that X p − X divides X p − X.
n
Now since X p − X is a product of linear factors in Fpn [X], the same
m
must be true for X p − X. This means that Fpn contains all the roots
m
of the polynomial X p − X. Let K ⊂ Fpn the subset of all the roots of
m
X p − X. Then the cardinality of K is pm .
Obviously 0, 1 ∈ K. Now let a, b ∈ K. Then (a + b)p = ap + bp , since Fpn
m
has characteristic p. Hence by simple induction we obtain (a + b)p =
m
m
ap + bp = a + b. This shows that also a + b lies in K. Furthermore
m
m
m
m
m
from ap = a and bp = b it follows that (ab)p = ap bp = ab. Hence ab
lies in K as well.
Let a ∈ K. Then −a = [p − 1]p a. As [p − 1]p ∈ Fp ⊂ K, and since we
have already shown that K contains products of two elements in K, it
follows that also −a ∈ K. Consider again an arbitrary p and an a ∈ K
m
m
with a 6= 0. Then (a−1 )p = (ap )−1 = a−1 , which shows that K contains
the multiplicative inverse of all its nonzero elements. We conclude that
K is a field.
(b) proof. If Fpn contains a subfield K with pm elements, then of course
Fpn is a K-vector space. Let d be the dimension of Fpn over K. Then
there exists a basis {v1 , . . . , vd } of Fpn over K. This means that for every
element x ∈ Fpn there exists a unique vector (λ1 , . . . , λd ) ∈ K d such that
x = λ1 v1 + . . . λd vd . Also if (λ1 , . . . , λd ) ∈ K d is arbitrary then naturally
λ1 x1 + · · · + λd vd ∈ Fpn . In other words we have an isomorphism of
K-vector spaces Fpn ∼
= K d . Now K d has exactly (pm )d = pmd elements.
Since Fpn and K d are finite they must both have the same cardinality.
Hence pn = pmd which implies n = md, i.e. m divides n.