Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-3/3g/solution

Identify a binary word with its coefficient polynomial in . A cyclic code is exactly an ideal of : closure under multiplication by is closure under cyclic shift. Its inverse image in the principal ideal domain has a unique monic generator polynomial of a cyclic code dividing . Put
The cubics have no root in , hence are irreducible polynomials. The eight monic divisors yield all binary cyclic codes of length seven. A code generated by has basis and dimension , with the zero code treated separately.
For the dual of a cyclic code, let . Orthogonality of coefficient vectors to all cyclic shifts gives the generator . Here . Thus the complete pairing, with the respective dimensions, is
These are respectively the whole-space/zero, even-weight/repetition, and two Hamming code/binary simplex code pairs. The reciprocal formula follows because the product of a word polynomial and the reversed polynomial of an orthogonal word has zero coefficients in all cyclic positions; the resulting inclusion has the correct complementary dimension, hence is equality.

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