The direct sum of irreducible cyclic codes is still a cyclic code

Question

Let $q$ be a prime power and $\gcd(n,q) = 1$. Let $h_1(x), h_2(x) \in \mathbb F_q [x]$, $\gcd(h_1(x), h_2(x)) = 1$, and $h_1(x) \cdot h_2(x) | x^n - 1$. Denote by $C, C_1, C_2$ the $q$-ary cyclic codes with length $n$ generated by $\frac{x^n-1}{h_1(x) \cdot h_2(x)}$, $\frac{x^n-1}{h_1(x)}$, $\frac{x^n-1}{h_2(x)}$ respectively. Prove every codeword $c(x) \in C$ can be uniquely represented by $$ c(x) = c_1(x) + c_2(x), c_1(x) \in C_1, c_2(x) \in C_2. $$

Answer 1

Let $g(x)=(x^n-1)/(h_1(x)h_2(x))$. Then $g(x), g(x)h_2(x), g(x)h_1(x)$ are the generator polynomials of $C, C_1, C_2$, respectively. Since $h_1(x), h_2(x)$ are coprime, there exist polynomials $p(x), q(x)$ such that $$ 1=h_2(x)p(x)+h_1(x)q(x). $$ For every $c(x)\in C$, $c(x)\equiv b(x)g(x)\pmod{(x^n-1)}$. Hence $$ c(x)\equiv b(x)p(x)g(x)h_2(x)+b(x)q(x)g(x)h_1(x)\pmod{(x^n-1)}. $$ Let $c_1(x)\equiv b(x)p(x)g(x)h_2(x)\pmod{(x^n-1)}, c_2(x)\equiv b(x)q(x)g(x)h_1(x)\pmod{(x^n-1)}$, then $c_1(x)\in C_1, c_2(x)\in C_2$. Hence the decomposition is always possible.

For the uniqueness, suppose that $$ 0\equiv c_1(x)+c_2(x)\pmod{(x^n-1)}, $$ where $c_1(x)\equiv s(x)g(x)h_1(x)\pmod{(x^n-1)}\in C_1, c_2(x)\equiv t(x)g(x)h_2(x)\pmod{(x^n-1)}\in C_2$. Then we have $$ g(x)(h_1(x)s(x)+h_2(x)t(x))=(x^n-1)r(x)=g(x)h_1(x)h_2(x) $$ for some $r(x)\in\mathbb{F}_q[x]$. Since $h_1(x), h_2(x)$ are coprime, $h_1(x)\mid t(x), h_2(x)\mid s(x)$. Hence $c_1(x)\equiv c_2(x)\equiv 0\pmod {(x^n-1)}$.