【国外通信教程】SolutionManual.ErrorControlCoding2nd·LinShu - CSDN文库

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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,

`

X

i=1

1 +

λ−`

X

i=1

1 =

λ

X

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

`

X

i=1

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

1

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