I am looking over an example from some notes I found online:
Given a binary BCH(15,7,5) codeword r = 100000001000000, we have r(x) = 1 + x8
When calculating the syndromes for this codeword polynomial, you get the following syndromes:
S1 = r($\alpha$) = 1 + $\alpha$8 = $\alpha$2
and
S3 = r($\alpha$3) = 1 + $\alpha$24 = $\alpha$7
I am confused as to how the syndromes were "reduced" in the final step.
The conjugacy classes are setup as follows:
let $\alpha$ be a root of $x^4 + x + 1$
{0} => $x$
{1} => $x + 1$
{$\alpha, \alpha^2, \alpha^4, \alpha^8$} => $x^4 + x + 1$
{$\alpha^3, \alpha^6, \alpha^9, \alpha^{12}$} => $x^4 + x^3 + x^2 + x + 1$
{$\alpha^5, \alpha^{10}$} => $x^2 + x + 1$
{$\alpha^7, \alpha^{11}, \alpha^{13}, \alpha^{14}$} => $x^4 + x^3 + 1$
Where I am confused is:
- How $1 +\alpha^{8}$ is reduced to $\alpha^2$. Is it because $\alpha^2$ and $\alpha^8$ are in the same conjugacy class? And if so, then:
- How is $1 +\alpha^{24}$ is reduced to $\alpha^7$. $\alpha^{24}$ has a degree higher than any polynomial GF(16). How do you deal with that?
I figured it out.
Apparently, I need to go back to algebra.
$\alpha^4 + \alpha + 1 = 0$
Therefore $\alpha^4 = \alpha + 1$
And then it's easy $1 + \alpha^8 = 1 + (\alpha^4)^2 = 1 + (\alpha + 1)^2 = \alpha^2$
Similar logic follows for my second question