MATH32031: Coding Theory
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Part 13: Cyclic Codes
Rings of polynomials and quotient rings
We begin by recalling some standard facts about rings of polynomials.
As usual p is a prime number and Fp [x] is the ring of polynomials in variable x with
coefficients from Fp ,
Fp [x] = { a0 + a1 x + · · · + an−1 xn−1 + an xn }
where ai ∈ Fp and n = 0, 1, 2, . . ..
If
f (x) = a0 + a1 x + · · · + an−1 xn−1 + an xn
is a polynomial and an 6= 0 then n is called the degree of f and is denoted by deg(f ).
Definition. An ideal in a commutative ring R is a subset I ⊆ R which is closed under
addition and is closed under multiplication by elements from R; that is,
x + y ∈ I,
rx ∈ I
for all x, y ∈ I, r ∈ R.
The quotient ring R/I has as elements the cosets as an abelian group r + I. Addition
and multiplication are given by
(r + I) + (s + I) = (r + s) + I,
(r + I)(s + I) = rs + I
for all r, s ∈ R.
Recall that the ideal generated by a polynomial f ∈ Fp [x], which we denote by (f ), is
the set of all multiples of f :
(f ) = { f (x)g(x) | g(x) ∈ Fp [x] } = f Fp [x]).
Let f ∈ Fp [x], f 6∈ Fp (that is, f does not have degree 0). We define Rf to be the
quotient ring
Rf = Fp [x]/(f );
Proposition 28 Let f ∈ Fp [x] be of degree deg f = m. Then Rf is an Fp -vector space
with basis
1, x, . . . , xm−1 .
In particular Rf has pm elements.
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Proof. A basis is a generating set which is linearly independent.
• Generation: Given any g ∈ Fp [x], write g = hf + r with either deg r < m or r = 0.
Then g ≡ r (mod f ) and obviously
r∈
m−1
X
F p xi .
i=0
• Linear independence: Suppose for contradiction that
m−1
X
λ i xi ≡ 0
(mod f )
i=0
with not all λi = 0. Notice that the sum cannot actually be equal to zero in Fp [x].
Thus in Fp [x] we have
m−1
X
λi xi = hf
i=0
for some non-zero h. Now compare the degrees:
deg(LHS) < m;
deg(RHS) = deg h + deg f
≥ deg f = m.
It follows that each element of Rf can be represented uniquely by a polynomial of
degree at most m.
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Cyclic codes
With the same notation as before we set fn = xn − 1, and we write Rn for
Rfn = Fp [x]/(fn ).
Note that for n > 1, Rn is never a field, since fn always has the linear factor x − 1 and
thus Rn has zero divisors.
For a(x) ∈ Fp [x] we write a for its image in Rn (with n always clear from the context).
(n)
We will identify Rn with Fp by means of the distinguished basis
{1, x, . . . , xn−1 }
(as we can, from Proposition 28).
In other words, we identify
(a0 , a1 , . . . , an−1 ) ∈ Fp(n)
with
a(x) = a0 + a1 x + . . . + an−1 xn−1 in Rn .
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Throughout, we speak of elements of Rn as polynomials of degree less then n; technically speaking, of course, we mean classes represented by such polynomials. If
a(x) = a0 + a1 x + . . . + an−1 xn−1 in Rn
= (a0 , a1 , . . . , an−1 )
under the above identification, then
x · a(x) = a0 x + a1 x2 + . . . + an−1 xn .
But xn = 1 in Rn , so
x · a(x) = an−1 + a0 x + . . . + an−2 xn−1
= (an−1 , a0 , a1 , . . . , an−2 ).
Thus multiplication by x in Rn corresponds to a cyclic rotation of the corresponding
(n)
vector in Fp . This motivates the next definition.
(n)
Definition of cyclic codes. A cyclic code in Fp is an Fp -linear code such that whenever
(a0 , a1 , . . . , an−1 ) ∈ C
then
(an−1 , a0 , a1 , . . . , an−2 ) ∈ C.
Example The binary code
{ (0, 0, 0), (1, 0, 1), (0, 1, 1), (1, 1, 0) }
is a cyclic code.
The following result reduces cyclic codes down to a very standard and familiar algebraic object.
Theorem 29 A code C ⊂ Rn is a cyclic code if and only if C is an ideal in Rn .
Remark. In concrete terms this means that length n cyclic codes correspond to ideals
in Rn .
Proof. If C is an ideal in Rn , then C closed under addition and it is certainly closed
under multiplication by elements of Fp . Thus C forms a vector subspace and so is linear.
Moreover, xC ⊆ C, and so, by the discussion above, C is a cyclic code.
Conversely, let C be a cyclic code. We wish to prove:
(a) C is an additive subgroup of Rn .
(b) a(x) · C ⊂ C for any a(x) ∈ Rn .
But (a) follows at once from the fact that C is a cyclic code and hence linear. The
fact that
is cyclic implies that x · C ⊂ P
C and so xi · C ⊂ C for every i ≥ 0. If
PC
a(x) = ni=0 ai xi and c ∈ C then a(x)c = ni=0 ai xi c. Since each xi c ∈ C we see that
a(x)c ∈ C, so (b) follows.
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