Solutions to Homework 9

MATH 291T Homework 9 Solutions
1. Find the idempotent generator of each of the minimal binary cyclic codes of length 9.
Solution : Recall that in HW 8 #4, we listed the generator polynomial and idempotent generator for each binary
cyclic code of length 9:
idempotent generator e(x)
generator polynomial g(x)
0 0
1 1
x + x 2 + x 4 + x 5 + x 7 + x 8 (x + 1)(x 6 + x 3 + 1)
x 3 + x 6 (x + 1)(x 2 + x + 1)
1 + x + x 2 + x 4 + x 5 + x 7 + x 8 x 2 + x + 1
1 + x 3 + x 6 x 6 + x 3 + 1
x + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 + x 8 x + 1
1 + x + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 + x 8 (x 6 + x 3 + 1)(x 2 + x + 1)
We know the generator polynomials of all minimal codes: they are of the form x9 − 1
where m(x) is an irreducible
m(x)
factor of x 9 − 1. Since
x 9 − 1 = (x + 1)(x 2 + x + 1)(x 6 + x 3 + 1)
is the factorization of x 9 − 1 into irreducible factors over GF (2) is, we have the following minimal generator
polynomials:
x 9 − 1
x + 1 = (x6 + x 3 + 1)(x 2 + x + 1) ,
x 9 − 1
x 6 + x 3 + 1 = (x + 1)(x2 + x + 1) and
So the idempotent generators of the minimal binary cyclic codes of length 9 are
x 9 − 1
x 2 + x + 1 = (x + 1)(x6 + x 3 + 1)
1 + x + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 + x 8 , x 3 + x 6 and x + x 2 + x 4 + x 5 + x 7 + x 8
2. Let r and n be natural numbers such that n is odd and x n − 1 has exactly two irreducible factors over GF (2 r ).
(a) Write down the idempotent generator and the generator polynomial for each cyclic code of length n over
GF (2 r ). Note that we are not working binary!
(b) Identify the idempotent generators of the minimal cyclic codes of length n over GF (2 r ).
Hint: Is an idempotent over GF (2) an idempotent over GF (2 r )?
Solution : (a) We get that there are two cyclotomic cosets for n and q = 2 r :
{0} and {1, 2, . . . , n − 1}
But a coset for q = 2 r is part of a coset for q = 2. Hence there are two cyclotomic cosets for q = 2.
Another way of seeing this: we know that
x n − 1 = (x − 1)(x n−1 + x n−2 + · · · + x + 1)
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Since there are two cyclotomic cosets for q = 2 r , we have that x n−1 + x n−2 + · · · + x + 1 is irreducible over
GF (2 r ). Hence it is also irreducible over GF (2).
But we can use cyclotomic cosets to write down the binary idempotents. There are four binary idempotents:
0 , 1 , x + x 2 + · · · + x n−2 + x n−1 and 1 + x + · · · + x n−2 + x n−1
A binary idempotent is of course also an idempotent for q = 2 r . Since there are four idempotents for q = 2 r , we
have that the idempotents for q = 2 r are the binary idempotents.
Since it is given that x n−1 + x n−2 + · · · + x + 1 is irreducible over GF (2 r ), we can write down the generator
polynomial associated with each idempotent. We get the following table:
idempotent generator generator polynomial
0 0
1 1
1 + x + · · · + x n−2 + x n−1 1 + x + · · · + x n−2 + x n−1
x + x 2 + · · · + x n−2 + x n−1 x + 1
(b) The generator polynomials for the minimal cyclic codes over GF (2 r ) are
x n − 1
x + 1 = 1 + x + · · · + xn−2 + x n−1
and
x n − 1
1 + x + · · · + x n−2 + x n−1 = x + 1
Hence the idempotent generators of the minimal cyclic codes of length n over GF (2 r ) are
1 + x + · · · + x n−2 + x n−1 and x + x 2 + · · · + x n−2 + x n−1
3. Let C be a cyclic code of length n over GF (q) with generator polynomial g(x) and idempotent generator e(x).
Let β be a primitive n-th root of unity. Prove the following:
(a) e(β j ) ∈ {0, 1} for 0 ≤ j ≤ n − 1.
(b) For 0 ≤ j ≤ n − 1, we have that e(β j ) = 0 ⇐⇒ g(β j ) = 0.
(c) dim C = |{0 ≤ j ≤ n − 1 : g(β j ) ≠ 0}|
(d) dim C = |{0 ≤ j ≤ n − 1 : e(β j ) = 1}|
Proof :
Note that
So the roots of x n − 1 are {β j : 0 ≤ j ≤ n − 1}.
x n − 1 = (x − β 0 )(x − β 1 )(x − β 2 ) · · · (x − β n−2 )(x − β n−1 )
Let h(x) be the check polynomial. So x n − 1 = g(x)h(x). Note that gcd(g(x), h(x)) = 1 as g(x) and h(x) have
no roots in common. So
1 = a(x)g(x) + b(x)h(x) (∗)
for some polynomials a(x), b(x) ∈ GF (q)[x]. We proved in class that
e(x) ≡ a(x)g(x) mod (x n − 1)
Hence
e(x) = a(x)g(x) + f(x)(x n − 1) for some polynomial f(x) ∈ GF (q)[x] (∗∗)
(a)(b) Pick 0 ≤ j ≤ n − 1. Then β j is a root of x n − 1 = g(x)h(x).
(a) Suppose first that g(β j ) = 0. Then it follows from (**) that e(β j ) = 0.
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Suppose next that g(β j ) ≠ 0. Then h(β j ) = 0 since x n − 1 = g(x)h(x). It follows from (*) that a(β j )g(β j ) = 1.
Hence it follows from (**) that e(β j ) = 1. This proves (a).
Suppose that e(β j ) = 0. Then it follows from (**) that a(β j )g(β j ) = 0. Hence it follows from (*) that
b(β j )h(β j ) = 1. In particular, h(β j ) ≠ 0. Since β j is a root of x n − 1 = g(x)h(x), we have that g(β j ) = 0. This
proves (b).
(c) For 0 ≤ j ≤ n − 1, we have that β j is either a root of g(x) or a root of h(x), but not a root of both. So
|{0 ≤ j ≤ n − 1 : g(β j ) ≠ 0}| = |{0 ≤ j ≤ n − 1 : h(β j ) = 0}|
= deg(h(x))
= n − deg(g(x))
= dim C
(d) Let 0 ≤ j ≤ n − 1. Using (b) and (a), we get
Hence it follows from (c) that
g(β j ) ≠ 0 ⇐⇒ e(β j ) ≠ 0 ⇐⇒ e(β j ) = 1
dim C = |{0 ≤ j ≤ n − 1 : g(β j ) ≠ 0}| = |{0 ≤ j ≤ n − 1 : e(β j ) = 1}|
✷
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