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/m 1 Z Z/m r Z with m 1 m r? (b) For (Z/180Z) proceed as in (a), i.e. find all r-tuplets (m 1,..., m r ) with m 1 m r such that (Z/180Z) = Z/m1 Z Z/m r 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 (m 1,..., m r ), avec m 1... m r, tels qu on ait un isomorphisme Z/180Z = Z/m 1 Z Z/m r Z.
(b) Même question pour (Z/180Z). Autrement dit, trouver tous les r-uplets (m 1,..., m r ), avec m 1... m r, tels qu il existe un isomorphisme (Z/180Z) = Z/m1 Z Z/m r Z. Solution 1 (1) (a) We employ the first method given in the script of the lecture. Set a 1 := 0, a 2 := 3 and a 3 := 1. Furthermore let m 1 := 3, m 2 := 4 and m 3 := 5. We calculate k 1 := m 2 m 3 = 20, k 2 := m 1 m 3 := 15 and k 3 := m 1 m 2 = 12. Now for i = 1, 2, 3 we need to find r i Z with k i r i 1 (mod m i ). Since k 1 2 (mod m 1 ), k 2 3 (mod m 2 ) and k 3 2 (mod m 3 ), we easily find r 1 := 2, r 2 := 3 and r 3 := 3. Set Hence x 1 := r 1 k 1 = 40, x 2 := r 2 k 2 = 45 and x 3 := r 3 k 3 = 36. x := a 1 x 1 + a 2 x 2 + a 3 x 3 = 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 ([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 (m 1,..., m r ) with m i 2, gcd(m i, m j ) = 1 for i j, and m 1 m r = 180. In this case we have Z/180Z = Z/m 1 Z Z/m r Z. (i)
Any tuple that does not satisfy the above conditions will not give an isomorphism. For example if m 1 m r 180, then the group on the right side in (i) does not have the right order. If on the other hand m 1 m r = 180 but gcd(m i, m j ) > 1 for some i j, then every element in Z/m i Z Z/m j Z has order < m i m j. Hence every element in the group on the right side in (i) has order < m 1 m r = 180. Without loss of generality we can assume that m 1 m r. 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 Hence Z/180Z = Z/4Z Z/5Z Z/9Z. (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) R 1 := Q[X]/(q), (b) R 2 := F 2 [X]/(q), where q denotes the image of q in F 2 [X]. For i = 1, 2 determine whether the images of f and g in R i lie in the group of units (R i )? (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 = {v 1,..., v m } 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) R 1 := Q[X]/(q), (b) R 2 := F 2 [X]/(q), où q désigne l image de q dans F 2 [X]. Pour i = 1, 2, déterminer si les images de f et de g dans R i 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 = {v 1,..., v m } de R en tant que K- espace 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 R 1 := Q[X]/(q). Since g (X 4) = (X 2 + 1) q 0 (mod q),
we see that the image of g in R 1 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 R 1 is a unit. (b) Now we consider R 2 := F 2 [X]/(q). We denote the two elements of F 2 by 0 and 1. Denote by f and g are the images of f and g in F 2 [X] respectively. If we take into account that X 2 + 1 = (X 1) 2 in F 2 [X] we obtain This shows that f = X(X 1), g = (X 1) 3 and q = X(X 1). f 0 (mod q), and therefore the image of f in R 2 is not a unit. Furthermore g X = (X 1) 2 q 0 (mod q), which means that the image of g in R 2 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 ) λ[f] m := [λf] m, (ii) satisfies the properties of a scalar multiplication. f, g K[X]. Then Let λ, µ K and (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 λ 0,..., λ n 1 K with On other words λ 0 + λ 1 [X] m + λ 2 [X 2 ] m + + λ n 1 [X n 1 ] m = 0. P := λ 0 + λ 1 X + λ 2 X 2 + + λ n 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. λ 0 = = λ n 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 F 2 [X] such that K := F 2 [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 = F 2 (α). Let ϕ : K K be an isomorphism of fields. (a) Prove that ϕ(y) = y for all y F 2 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 = F 4? Justify your answer. Exercice 3 (français, 25 points) (1) Trouver un polynôme P F 2 [X] tel que K := F 2 [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 = F 2 (α). Soit ϕ : K K un isomorphisme de corps. (a) Montrer que ϕ(y) = y pour tout y F 2 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 = F 4? Justifier votre réponse. Solution 3 (1) We claim that for P = X 3 + X + 1 F 2 [X] the ring K := F 2 [X]/(P ) is a field with 8 elements. proof. Since the polynomial P does not have any root in F 2, it is irreducible in F 2 [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, α 2, α 2 + 1, α 2 + α, α 2 + 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, α 2, α 2 + 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 F 2 = {0, 1}. (b) proof. By the properties of a ring homomorphism we obtain (ϕ(α)) 3 +ϕ(α)+1 = ϕ(α 3 )+ϕ(α)+ϕ(1) = ϕ(α 3 +α+1) = ϕ(0) = 0. Hence ϕ(α) is also a root of P. (4) proof. Every element of K can be written as λ 2 α 2 + λ 1 α 1 + λ 0 with unique λ 0, λ 1, λ 2 F 2. Hence we can define a map ψ : K K, λ 2 α 2 + λ 1 α 1 + λ 0 λ 2 β 2 + λ 1 β 1 + λ 0. Obviously ψ(0) = 0 and ψ(1) = 1. Let µ 0, µ 1, µ 2 F 2, and set x := λ 2 α 2 + λ 1 α + λ 0 and y := µ 2 α 2 + µ 1 α + µ 0. Then ψ(x + y) = ψ ( (λ 2 + µ 2 )α 2 + (λ 1 + µ 1 )α + (λ 0 + µ 0 ) ) = (λ 2 + µ 2 )β 2 + (λ 1 + µ 1 )β + (λ 0 + µ 0 ) = (λ 2 β 2 + λ 1 β + λ 0 ) + (µ 2 β 2 + µ 1 β + µ 0 ) = ψ(x) + ψ(y), ψ(xy) = ψ ( (λ 2 µ 2 + λ 2 µ 0 + λ 1 µ 1 + λ 0 µ 2 )α 2 +(λ 2 µ 2 + λ 2 µ 1 + λ 1 µ 2 + λ 1 µ + λ 1 µ 1 )α +(λ 2 µ 1 + λ 1 µ 2 + λ 0 µ 0 ) ) = (λ 2 µ 2 + λ 2 µ 0 + λ 1 µ 1 + λ 0 µ 2 )β 2 +(λ 2 µ 2 + λ 2 µ 1 + λ 1 µ 2 + λ 1 µ + λ 1 µ 1 )β +(λ 2 µ 1 + λ 1 µ 2 + λ 0 µ 0 ) = (λ 2 β 2 + λ 1 β + λ 0 0(µ 2 β 2 + µ 1 β + µ 0 ) = 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 : K K be another isomorphism of fields with ψ (α) = β = ψ(α). Then also ψ (α 2 ) = β 2 = ψ(α 2 ). Naturally ψ (1) = 1 = ψ(1). Since K is a vector space of dimension 3 over F 2 with basis {1, α, α 2 }, and since ψ and ψ can also be considered as vector space homomorphism, it follows that ψ (x) = ψ(x) for all x K. Hence ψ = 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 = F 4. 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 = F 2.
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 p m 1 divides p n 1 if m divides n. (b) Prove that X pm 1 1 divides X pn 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 F p n contains a field with p m elements if and only if m divides n. (a) Let m divide n. Prove that F p n contains a subfield K with p m elements. Hint: Define K to be the set of roots in F p n of the polynomial X pm X. Why do all the roots of this polynomial lie in F p n? (b) Let F p n contain a subfield K with p m elements. Show that m divides n. Hint: Recall that F p n 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 p m 1 divise p n 1 si m divise n. (b) Montrer que X pm 1 1 divise X pn 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 F p n contient un corps à p m élements si et seulement si m divise n.
(a) Supposons que m divise n. Montrer que F p n contient un sous-corps K à p m élements. Indication: Définir K comme l ensemble des racines dans F p n du polynôme X pm X. Pourquoi ce polynôme a-t-il toutes ses racines dans F p n? (b) Supposons que F p n contienne un sous-corps K à p m élements. Montrer que m divise n. Indication: F p n 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. p n 1 = (p m 1)R(p). Hence p m 1 divides p n 1. (b) proof. Since p m 1 divides p m 1 by (2).(a), it follows from (1).(c) that X pm 1 1 divides X pn 1 1. (3) (a) proof. From the lecture we know that F p n can be regarded as the set of all roots of the polynomial X pn X = X(X pn 1 1). Since by (2).(b)
X pm 1 1 divides X pn 1 1, it follows that X pm X divides X pn X. Now since X pn X is a product of linear factors in F p n[x], the same must be true for X pm X. This means that F p n contains all the roots of the polynomial X pm X. Let K F p n the subset of all the roots of X pm X. Then the cardinality of K is p m. Obviously 0, 1 K. Now let a, b K. Then (a + b) p = a p + b p, since F p n has characteristic p. Hence by simple induction we obtain (a + b) pm = a pm + b pm = a + b. This shows that also a + b lies in K. Furthermore from a pm = a and b pm = b it follows that (ab) pm = a pm b pm = ab. Hence ab lies in K as well. Let a K. Then a = [p 1] p a. As [p 1] p F p 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 with a 0. Then (a 1 ) pm = (a pm ) 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 F p n contains a subfield K with p m elements, then of course F p n is a K-vector space. Let d be the dimension of F p n over K. Then there exists a basis {v 1,..., v d } of F p n over K. This means that for every element x F p n there exists a unique vector (λ 1,..., λ d ) K d such that x = λ 1 v 1 +... λ d v d. Also if (λ 1,..., λ d ) K d is arbitrary then naturally λ 1 x 1 + + λ d v d F p n. In other words we have an isomorphism of K-vector spaces F p n = K d. Now K d has exactly (p m ) d = p md elements. Since F p n and K d are finite they must both have the same cardinality. Hence p n = p md which implies n = md, i.e. m divides n.