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【国外通信教程】Solution Manual.Error Control Coding 2nd·Lin Shu

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【国外通信教程】Solution Manual.Error Control Coding 2nd·Lin Shu 著名教授Lin Shu and Costello编写的《Error Control Coding 2nd》的习题解答，详细内容见附件

Chapter 2

2.3 Since m is not a prime, it can be factored as the product of two integers a and b,

m = a · b

with 1 < a, b < m. It is clear that both a and b are in the set {1, 2, · · · , m − 1}. It follows

from the deﬁnition of modulo-m multiplication that

a ¡ b = 0.

Since 0 is not an element in the set {1, 2, · · · , m−1}, the set is not closed under the modulo-m

multiplication and hence can not be a group.

2.5 It follows from Problem 2.3 that, if m is not a prime, the set {1, 2, · · · , m − 1} can not be a

group under the modulo-m multiplication. Consequently, the set {0, 1, 2, · · · , m − 1} can not

be a ﬁeld under the modulo-m addition and multiplication.

2.7 First we note that the set of sums of unit element contains the zero element 0. For any 1 ≤

` < λ,

i=1

1 +

λ−`

i=1

1 =

i=1

1 = 0.

Hence every sum has an inverse with respect to the addition operation of the ﬁeld GF(q). Since

the sums are elements in GF(q), they must satisfy the associative and commutative laws with

respect to the addition operation of GF(q). Therefore, the sums form a commutative group

under the addition of GF(q).

Next we note that the sums contain the unit element 1 of GF(q). For each nonzero sum

i=1

with 1 ≤ ` < λ, we want to show it has a multiplicative inverse with respect to the multipli-

cation operation of GF(q). Since λ is prime, ` and λ are relatively prime and there exist two

integers a and b such that

a · ` + b · λ = 1, (1)

where a and λ are also relatively prime. Dividing a by λ, we obtain

a = kλ + r with 0 ≤ r < λ. (2)

Since a and λ are relatively prime, r 6= 0. Hence

1 ≤ r < λ

Combining (1) and (2), we have

` · r = −(b + k`) · λ + 1

Consider

i=1

1 ·

i=1

1 =

`·r

i=1

1 =

−(b+k`)·λ

i=1

= (

i=1

1)(

−(b+k`)

i=1

1) + 1

= 0 + 1 = 1.

Hence, every nonzero sum has an inverse with respect to the multiplication operation of GF(q).

Since the nonzero sums are elements of GF(q), they obey the associative and commutative

laws with respect to the multiplication of GF(q). Also the sums satisfy the distributive law.

As a result, the sums form a ﬁeld, a subﬁeld of GF(q).

2.8 Consider the ﬁnite ﬁeld GF(q). Let n be the maximum order of the nonzero elements of GF(q)

and let α be an element of order n. It follows from Theorem 2.9 that n divides q − 1, i.e.

q − 1 = k · n.

Thus n ≤ q − 1. Let β be any other nonzero element in GF(q) and let e be the order of β.

Suppose that e does not divide n. Let (n, e) be the greatest common factor of n and e. Then

e/(n, e) and n are relatively prime. Consider the element

(n,e)

This element has order e/(n, e). The element

αβ

(n,e)

has order ne/(n, e) which is greater than n. This contradicts the fact that n is the maximum

order of nonzero elements in GF(q). Hence e must divide n. Therefore, the order of each

nonzero element of GF(q) is a factor of n. This implies that each nonzero element of GF(q)

is a root of the polynomial

− 1.

Consequently, q − 1 ≤ n. Since n ≤ q − 1 (by Theorem 2.9), we must have

n = q − 1.

Thus the maximum order of nonzero elements in GF(q) is q-1. The elements of order q − 1

are then primitive elements.

2.11 (a) Suppose that f(X) is irreducible but its reciprocal f

∗

(X) is not. Then

∗

(X) = a(X) · b(X)

where the degrees of a(X) and b(X) are nonzero. Let k and m be the degrees of a(X) and

b(X) respectivly. Clearly, k + m = n. Since the reciprocal of f

∗

(X) is f(X),

f(X) = X

∗

(

) = X

) · X

This says that f (X) is not irreducible and is a contradiction to the hypothesis. Hence f

∗

(X)

must be irreducible. Similarly, we can prove that if f

∗

(X) is irreducible, f(X) is also

irreducible. Consequently, f

∗

(X) is irreducible if and only if f(X) is irreducible.

(b) Suppose that f(X) is primitive but f

∗

(X) is not. Then there exists a positive integer k less

than 2

− 1 such that f

∗

(X) divides X

+ 1. Let

+ 1 = f

∗

(X)q(X).

Taking the reciprocals of both sides of the above equality, we have

+ 1 = X

∗

(

)q(

)

= X

∗

(

) · X

k−n

)

= f(X) · X

k−n

This implies that f(X) divides X

+ 1 with k < 2

− 1. This is a contradiction to the

hypothesis that f(X) is primitive. Hencef

∗

(X) must be also primitive. Similarly, if f

∗

(X) is

primitive, f(X) must also be primitive. Consequently f

∗

(X) is primitive if and only if f(X)

is primitive.

2.15 We only need to show that β, β

, · · · , β

e−1

are distinct. Suppose that

= β

for 0 ≤ i, j < e and i < j. Then,

(β

j−i

−1

)

= 1.

Since the order β is a factor of 2

− 1, it must be odd. For (β

j−i

−1

)

= 1, we must have

j−i

−1

= 1.

Since both i and j are less than e, j − i < e. This is contradiction to the fact that the e is the

smallest nonnegative integer such that

−1

= 1.

Hence β

6= β

for 0 ≤ i, j < e.

2.16 Let n

be the order of β

. Then

(β

)

= 1

Hence

(β

)

= 1. (1)

Since the order n of β is odd, n and 2

are relatively prime. From(1), we see that n divides n

and

= kn. (2)

Now consider

(β

)

= (β

)

= 1

This implies that n

(the order of β

) divides n. Hence

n = `n

(3)

From (2) and (3), we conclude that

= n.

2.20 Note that c · v = c · (0 + v) = c · 0 + c · v. Adding −(c · v) to both sides of the above equality,

we have

c · v + [−(c · v)] = c · 0 + c · v + [−(c · v)]

0 = c · 0 + 0.

Since 0 is the additive identity of the vector space, we then have

c · 0 = 0.

2.21 Note that 0 · v = 0. Then for any c in F ,

(−c + c) · v = 0

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