If I have a generator polynomial, g(x) which divides x9 + 1 but does not have x+1 as a factor and I have to prove that 111111111 is a valid codeword after detecting by the code.
Here, Since x9 + 1 = (x3 + 1)(x6 + x3 + 1 ) = (x+1)(x2 + x + 1)(x6 + x3 + 1 ) . Here, (x2 + x + 1) can't be broken into factors of (x+1). If it were (x2+1), then I can write x2+1 = (x+1)(x+1) but it is not the case here. Now, Since, g(x) divides it, it means, x9 + 1 = g(x)z(x), where z(x) is some polynomial in 'x'.
So, (x9 + 1) = (x+1)(x2 + x + 1)(x6 + x3 + 1 ) = g(x)z(x).
As g(x) does not have (x+1) as a factor, the possible choices for g(x) are :
- (x2 + x + 1)
- (x6 + x3 + 1 )
- (x2 + x + 1)*(x6 + x3 + 1 )
In all these 3 cases, g(x) divides 111111111 at receiver side, So, 111111111 is a valid codeword.
Can anyone please check it and please correct me if I am wrong. Any help would be appreciated.