Cyclic Codes over $GF(q)$

Question

Does the set of cyclic codewords / codeword polynomials themselves form a field ? I think they donot because the modulo operation is with respect to $x^n-1$ which is not a prime polynomial. Also the generator need not be prime polynomial

Answer 1

Your impression is correct (in general). A cyclic (linear) code is an ideal $I$ of the quotient ring $$ R=GF(q)[x]/\langle x^n-1\rangle, $$ and in general an ideal is not a field.

Fields do emerge here in a special case. Assume that $\gcd(n,q)=1$. Then it follows that in the prime factorization in the polynomial ring $GF(q)[x]$ $$ x^n-1=p_1(x)p_2(x)\cdots p_k(x) $$ the irreducible factors $p_i(x), i=1,2,\ldots,k,$ are all distinct. By Chinese Remainder Theorem we have $$ R\simeq \bigoplus_{i=1}^kGF(q)[x]/\langle p_i(x)\rangle.\qquad(*) $$ By irreducibility of the factors the summands on the right hand side are fields. We can view a single summand, $C_i=GF(q)[x]/\langle p_i(x)\rangle$ as a minimal ideal of $R$ though. On the r.h.s. of $(*)$ the elements of $C_i$ are those that have an arbitrary $i$th component and the rest are zeros. As an ideal $C_i$ is not a subring of $R$, because the multiplicative identity of $R$ is not an element of $C_i$. But the idempotent $e_i$ of $C_i$, that is the component $e_i$ in the direct sum presentation $1_R=(e_1,e_2,\ldots,e_k)$, can serve in the role of a multiplicative identity in the ideal $C_i$. It is then easy to see that w.r.t. to the operations of $R$, but using $e_i$ as the multiplicative neutral element, the ideal $C_i$ is isomorphic to the field $GF(q)[x]/\langle p_i(x)\rangle$.

What this translates to in the language of cyclic codes is as follows: The minimal non-zero cyclic codes are fields, when we use their idempotents as multiplicative neutral elements.