Download Algebra for Digital Communication

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
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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)}.
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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
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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.
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