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EPFL - Section de Mathématiques Algebra for Digital Communication Prof. E. Bayer Fluckiger Fall semester 2007 Exercise sheet and solutions 4 11.10.07 Exercise 1. Let R = {[0]12 , [3]12 , [6]12 , [9]12 } be a subset of Z/12Z. (1) Show that R is a ring. In particular, find 1R . (2) Is R a subring of Z/12Z? Is it an ideal? (3) Construct homomorphisms f : Z/4Z −→ R and g : Z/12Z −→ R. (4) Are f and g injective and/or surjective? Solution 1. (1) To show that R is a ring, there are 7 points to verify. Most of them are consequences of the fact that R is a subset of a ring. (i) (ii) (iii) (iv) (v) (vi) Associativity of addition in Z/12Z yields associativity in R. Commutativity in Z/12Z yields commutativity in R. [0]12 ∈ R =⇒ 0R = [0]12 . [3] + [9] = [0] and [6] + [6] = [0], so every element of R has an additive inverse. Associativity of multiplication in Z/12Z yields associativity in R. To find 1R , we compute the multiplication table of R : [0]12 [3]12 [6]12 [9]12 [0]12 [0]12 [0]12 [0]12 [0]12 [3]12 [0]12 [9]12 [6]12 [3]12 [6]12 [0]12 [6]12 [0]12 [6]12 [0]12 [0]12 [3]12 [6]12 [9]12 This yields 1R = [9]12 . (vii) Distributivity is inherited from Z/12Z. (2) The ring R is not a subring of Z/12Z, since 1Z/12Z = [1]12 is not contained in R. On the other hand it is an ideal. We only need to check that, for all a ∈ Z/12Z and all 1 of 4 Algebra for Digital Communication Exercise sheet and solutions 4 r ∈ R,we have ar ∈ R. We can write R = {3 · b | b ∈ Z/12Z}, so, for all r = [3b]12 ∈ R, and all [a] ∈ Z/12Z, we have, [a]12 · [3b]12 = [3ab]12 ∈ R . (3) Let’s give explicit descriptions of these two homomorphisms, constructed, as usual, by sending [1] (in Z/4Z or Z/12Z) on 1R = [9]12 . Then using additivity, the only possibility is: f ([x]4 ) = f (x · [1]4 ) = x · f ([1]4 ) = x · [9]12 = [9x]12 , and g([x]12 ) = [9x]12 . We can then verify that they are well-defined and are homomorphisms. This is done in the same way for both f and g, so we’ll do it only for f . The application is well-defined since : f ([x + 4y]4 ) = [9(x + 4y)]12 = [9x]12 + [36y]12 = [9x]12 . Moreover, it is clearly additive by construction, and multiplicative since : f ([x]4 )f ([y]4 ) = [9x]12 [9y]12 = [9]12 [9xy]12 = [9xy]12 = f ([xy]4 ) = f ([x]4 [y]4 ). (4) An explicit computation of images of the 4 elements of Z/4Z by f yields four discinct elements of R, so f is injective. Moreover, as both Z/4Z and R contain 4 elements, f is surjective as well. The homomorphism f is not injective, since, for example, both [0]12 and [4]12 are sent to [0]12 . On the other hand it is surjective. Indeed the first 4 elements of Z/12Z are sent on 4 different elements of R. Exercise 2. Consider the ring R = Z × Z with addition and multiplication given by (n1 , n2 ) + (m1 , m2 ) = (n1 + m1 , n2 + m2 ) (n1 , n2 ) · (m1 , m2 ) = (n1 · m1 , n2 · m2 ). (1) Give 0R , 1R , zero divisors and units of R. (2) Which of the following sets A, B and C are subrings of R? Which are ideals? A = {(n, n) | n ∈ Z} B = {(0, n) | n ∈ Z} C = {(3n, 2m) | m, n ∈ Z} Solution 2. (1) - 0R = (0, 0). - 1R = (1, 1). - To get (n1 , n2 ) · (m1 , m2 ) = (n1 · m1 , n2 · m2 ) = (0, 0), we need to have n1 = 0 or m1 = 0 and n2 = 0 or m2 = 0. So the set of zero divisors is : {(n, 0)|n ∈ Z} ∪ {(0, m)|m ∈ Z}. - Since the only units of Z are ±1, the units of R are {(1, 1), (−1, 1), (1, −1), (−1, −1)}. 2 of 4 Algebra for Digital Communication (2) Exercise sheet and solutions 4 - We can check that A is a subring of R, and so it is not an ideal (A R and 1R ∈ A). - Since 1R is not contained in B and C, these sets cannot be subrings of R. - It is easily checked that B and C are ideals. Exercise 3. Let A be a commutative ring of characteristic 2. (1) Show that a = −a for all a ∈ A. (2) Let ϕ : R −→ A be given by ϕ(x) = x2 . Prove that ϕ is an endomorphism of A, i.e. a homomorphism from A to itself. Solution 3. (1) Since A has characteristic 2, 2 · a = a + a = 0 for each a ∈ A. So a = −a. (2) There are three conditions to check : - ϕ(a + b) = (a + b)2 = a2 + 2ab + b2 = a2 + b2 = ϕ(a) + ϕ(b). - ϕ(a · b) = (a · b)2 = a2 · b2 = ϕ(a) · ϕ(b). - ϕ(1) = 12 = 1. Exercise 4. Consider the ring Z/2Z × Z/3Z with operations defined componentwise, as in exercise 2. Let f : Z −→ Z/2Z × Z/3Z be given by f (x) = ([x]2 , [x]3 ). Prove the following : (1) f is a homomorphism and 6Z is contained in ker(f ). (2) The induced homomorphism f¯ : Z/6Z −→ Z/2Z × Z/3Z is an isomorphism. [x]6 7−→ ( [x]2 , [x]3 ) What is the characteristic of the ring Z/2Z × Z/3Z ? Solution 4. (1) The map f is additive and multiplicative, for example for addition (same with multiplication) : f (a + b) = ([a + b]2 , [a + b]3 ) = ([a]2 + [b]2 , [a]3 + [b]3 ) = ([a]2 , [a]3 ) + ([b]2 , [b]3 ) = f (a) + f (b). Moreover f (1) = ([1]2 , [1]3 ) = 1Z/2Z×Z/3Z . So f is a homomorphism. Let a be an element of 6Z. It can be written as a = 6n for an integer n. Then, f (a) = f (6n) = ([6n]2 , [6n]3 ) = ([0]2 , [0]3 ). So the ideal 6Z is included in the kernel of f . (2) The induced homomorphism f¯ is injective since 6Z ⊂ ker(f ) (theorem 3.42). It remains to show that f is onto. This is obvious using injectivity and the fact that both Z/6Z and 3 of 4 Algebra for Digital Communication Exercise sheet and solutions 4 Z/2Z × Z/3Z have 6 elements. An alternative proof is an explicit description of f : f¯ : Z/6Z [0]6 [1]6 [2]6 [3]6 [4]6 [5]6 −→ Z/2Z × Z/3Z 7 → ( [0]2 , [0]3 ) − 7−→ ( [1]2 , [1]3 ) 7−→ ( [0]2 , [2]3 ) 7−→ ( [1]2 , [0]3 ) 7−→ ( [0]2 , [1]3 ) 7−→ ( [1]2 , [2]3 ) Using (2), we have ker(f ) = 6Z, so characteristic of Z/2Z × Z/3Z is 6. 4 of 4