Let $\alpha $ be a root of $X^4 + X + 1$, and let C be the BCH code of length 15 with defining set the first four powers of $\alpha $. Determine the error position of the following received word: $1+X^6 + X^7 + X^8$.
I have calculated that the generator polynomial for C is $f(x) = X^8 + X^7 + X^6 + X^4 + 1$. So it seems obvious that there is one error, which occurs at $X^4$. However when I try to get this result algorithmically, it doesn't seem to work.
I am using the following algorithm: Let $\sigma(X)$ be the error locator polynomial = $\prod_{i\in \xi}(1-\alpha^iX)$, where $\xi $ are the indices of the errors. Let t be the number of errors we can correct. Then $\sigma(X)\sum_{j=1}^{2t} r(\alpha^i)X^j \equiv w(X)$, where r(X) is the polynomial we received, and w(X) has degree less than or equal to t.
So we compute the product, and by noting that powers of X greater than t must be zero, we can calculate $\sigma(X)$.
However when I actually do the calculation, I get
$\sigma(X)\sum_{j=1}^{2t} r(\alpha^i)X^j = \sigma(X)[(\alpha+1)X + (\alpha^2+1)X^2 + (1+\alpha + \alpha^2 + \alpha^3)X^3 + \alpha X^4]$.
Plugging in the expected form of the answer $\sigma(X) = (1+(\alpha+1)X)$ doesn't seem to work, and I can't see why.
EDIT NOTE: This comes from my lecture notes not a book, so the formula itself may be wrong. If anyone knows where this occurs in the literature, that would also be appreciated.
OK, I think I've figured out what the answer is by trying a load of examples. I think that the equation should read:
$\sigma(X)\sum_{j=1}^{2t} r(\alpha^i)X^j \equiv w(X) \mod X^{2t+1}$.
This then gives the right answer for all of the examples I've tried. Adding in this mod also seems to be the only way to remove the highest order term.