Does the Factor (x+1) Affect Polynomial Coset Uniqueness in F_2[x]?

[gonzo]

Okay, I have this problem in my book, an I'm pretty sure I solved it, but I there is something that is confuses me in the way the problem was asked.

Assume all polynomials are in the ring
$${F_2[x] \over x^n-1}$$
where $n=2^m-1$ and m>2

Let g(x) be the minimal polynomial of a primitive element of $F_{2^m}$

We want to look at the ideal generated by (x+1)g(x). The problem is to prove that for $i \ne j$ the two polynomials $x^i+x^{i+1}$ and $x^j+x^{j+1}$ can't be in the same coset of this ideal.

My problem is that my proof seems to make the factor of (x+1) in the ideal unnecessary and that the way the problem is worded seems to imply that this factor has some importance to this property.

My proof is relatively simple. Assume j>i and j<n, and assume they are in the same coset, then we can find a pair of q's such that:

$$(x+1)g(x)q_1+x^i+x^{i+1}=(x+1)g(x)q_2+x^j+x^{+1} (mod(x^n-1))$$

Rearranging we get:

$$(x+1)g(x)(q_1+q_2)+(x+1)(x^i)(1+x^{j-i})=0 (mod(x^n-1))$$

Since g divides the sum of the left side and it divides the left hand term, it has to divide the right hand term on the left side as well (g is irreducible). It can't divide (x+1) and it can't divide $x^i$ so it has to divide the remaining factor $(1+x^{j-i})$ but it can't divide this factor because it is the minimal polynomial of a primitive element of $F_{2^m}$. Thus they can't be in the same coset since we have hit a contradiction.

However, none of this depends on the (x+1) factor in the ideal as far as I can see.

Am I missing something, or just totally off on the wrong track?=

 

[fresh_42]

I can't follow the "rearranging". I get modulo $x^n-1$
\begin{align*}
x^i +x^{i+1}&=x^j+x^{j+1}+(x+1)g(x)q(x)\\
(x+1)x^i -(x+1)x^j &= (x+1)g(x)q(x)\\
x^j(x^{i-j}-1) &= g(x)q(x)\\
x^j \,&|\,g(x)q(x)\\
x^j\,&|\,q(x)\\
(x^{i-j}-1) \,&|\,g(x)q'(x)
\end{align*}
where I used the factor $x+1$.