convert a cyclic code $C$ of length $n$ over a field $\mathbb{F}_{q^m}$ to a code $D$ of length $mn$ over $\mathbb{F_q}$

savecancel

If i correctly understood the task, than the following is a possibility.

First of all, let us generate the given code in sage, and print some information on its basis. Adapted sage code:

S.<A> = PolynomialRing( GF(2) )
K.<a> = GF(2^2, modulus=A^2+A+1)
R.<x> = PolynomialRing( K )

V = x^3 + x^2 + a*x + a + 1;
L.<b> = K.extension( V )

delta = 28
# the bigger this delta (=28 here), the bigger g, so the smaller our code
g = lcm( [ (b^k).minpoly() for k in [1..28] ] )
# note that g divides x^63-1 since each minimal poly in the list above does it
C = codes.CyclicCode( generator_pol = g, length = 2^6-1 )

print "The code C is:\n%s\n" % C
print "The basis of C is...\n"
for v in C.basis():
    print v

Results:

The code C is:
[63, 14] Cyclic Code over GF(4)

The basis of C is...

(1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, 1, a + 1, 0, 1, 1, a + 1, a,  ...
(0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, 1, a + 1, 0, 1, 1, a + 1,  ...
(0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, 1, a + 1, 0, 1, 1,      ...
(0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, 1, a + 1, 0, 1, 1,   ...
(0, 0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, 1, a + 1, 0, 1,   ...
(0, 0, 0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, 1, a + 1, 0,   ...
(0, 0, 0, 0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, 1, a + 1,   ...
(0, 0, 0, 0, 0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, 1,       ...
(0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, 1, a  ...
(0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, 1, ...
(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, 1, ...
(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, 0, ...
(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, a, ...
(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, a + 1, a + 1, a, a, a, 0, 1, ...

The results for the basis vectors were manually truncated.

Now, if i understand the construction of the new code from this one, we build a new basis, replacing the above entries as follows:

• $0$ by $(0,0)$
• $1$ by $(1,0)$
• $a$ by $(0,1)$
• $a+1$ by $(1,1)$.

For this, we construct the matrix $E$ from the above list of vectors, as rows, using the natural lift. Code:

def get_lift(c):
    """
    for an element c in K, lift it to GF(2)^2
    """
    lift = S(c).coefficients(sparse=False)
    return lift + (2-len(lift))*[ GF(2)(0) ]

MS = MatrixSpace( GF(2), 14, 2*63 )
E  = []    # and we append
for v in C.basis():
    w = []    # and we append
    for entry in v:
        w += get_lift( entry )
    E.append( w )

D = LinearCode( MS( E ) )
print D

I think, D is the needed code.