BCH code or not

abstract-algebracoding-theoryfinite-fields

Problem

Consider the field $\,\mathbb{F}_2\,$ and let $f(x)=x^2+x+1,\,g(x)=x^3+x+1\in\mathbb{F}_2[x]$.
If $\,\mathcal{C}=\langle fg\,\rangle$ is a cyclic code of length $n$, is it a BCH code?

Attempt

First I try to determine $n$. Since $fg\,$ is the generator polynomial, the length is the minimum $n\in\mathbb{N}$ such that $fg\,|\,x^n-1$. We deduce both $f\,$ and $\,g\,$ must divide $\,x^n-1$.

I create the finite fields $\,\mathbb{F}_{2^2}\,$ and $\,\mathbb{F}_{2^3}$ by the Kronecker method
$$\mathbb{F}_{2^2}=\frac{\mathbb{F}_2[x]}{\langle f\,\rangle}\qquad\qquad\mathbb{F}_{2^3}=\frac{\mathbb{F}_2[x]}{\langle\,g\,\rangle}$$
to calculate the order of $\,[x]_f\,$ and $\,[x]_g\,$ in the corresponding cyclic groups $\,\mathbb{F}_{2^2}^*\,$ and $\,\mathbb{F}_{2^3}^*$.
I obtain
$$x^3\equiv1\;(\textrm{mod}\; f)\qquad\qquad x^7\equiv1\;(\textrm{mod}\; g)$$
so, $\,n=\textrm{lcm}(7,3)=21$.

Now, from the BCH codes theory:

  • Let $\,\alpha\,$ be a $n$-th primitive root of unity in the field $\,\mathbb{F}_{p^m}$, where $\,m$ is the multiplicative order of $\,p\;(\textrm{mod}\; n)$, i.e.
    $$p^m\equiv1\;(\textrm{mod}\; n)$$

  • The generator polynomial is
    $$\textrm{lcm}\left\{M_{\alpha^b}(x),\,M_{\alpha^{b+1}}(x),\dots,\,M_{\alpha^{b+d-2}}(x)\right\}\qquad b\ge0,\; d=\textrm{dist}(\mathcal{C})$$

It's easy to see that if the choose $\,p=2$, then
$$2^6\equiv1\;(\textrm{mod}\; 21)$$

and if we find an irreducible polynomial $\,h\in\mathbb{F}_2[x]$ with deg$(h)=6$, we know it exists $\alpha\in\mathbb{F}_{2^6}\,$ which is a $\,21^{\textrm{th}}\,$ root of unity.

I'm stuck right here, cause I have no idea how to proceed.
I've only thought about this: since we have

  • $f\in\mathbb{F}_2[x]$ is irreducible;
  • $\mathbb{F}_2\subset\mathbb{F}_{2^6}\,$ and deg$(f)\,|\dim_{\mathbb{F}_2}\mathbb{F}_{2^6}$;
  • $[x]_f\,$ is a root of $f$ and it's contained in $\mathbb{F}_{2^6}$

then $f\,$ is a product of linear factors in $\,\mathbb{F}_{2^6}$. The same could be applied to $g$.
With these hypothesis, maybe I could determine $\alpha$ but I'm not sure.
Any kind of help would be appreciated.

Best Answer

It won't be a narrow sense BCH-code because none of the zeros of $fg$ have order $21$. However, it does fit into the more general definition of a BCH-code that you gave.

You correctly deduced that the zeros of $f$ are roots of unity of order three. If one of them is $\omega$ the other is $\omega^2$. Similarly, you correctly figured out that the zeros of $g$ are roots of unity of order seven. Again, by applying the Frobenius automorphism, if $\beta$ is one of those zeros, the other two are $\beta^2$ and $\beta^4$.

To decide whether the resulting code is a BCH-code or not we need to write those zeros in terms of a root of unity of order $21$. It is straightforward to show that $\alpha=\omega\beta$ will have order $21$. You may recall from a course in group theory that in an abelian group the product of elements of orders $m,n$, $\gcd(m,n)=1$, will have order $mn$.

The rest is easy. We only need to write the zeros of $fg$ as powers of $\alpha$:

  • $\alpha^7=\omega^7\beta^7=\omega$, consequently $\alpha^{14}=\omega^2$,
  • $\alpha^9=\omega^9\beta^9=\beta^2$, consequently $\alpha^{18}=\beta^4$, and $\beta=\beta^8=(\alpha^{18})^2=\alpha^{36}=\alpha^{15}$.

We spot right away that $f(x)$ is the minimal polynomial of $\alpha^{14}$ and $g(x)$ is the minimal polynomial of $\alpha^{15}$. Therefore $$ f(x)g(x)=\operatorname{lcm}\{M_{\alpha^{14}}(x),M_{\alpha^{15}}(x)\}, $$ and you can conclude that this is a BCH-code, $n=21, b=14, d=3$.

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