How to show that the extended code of cyclic binary hamming code can fix single and two adjacent positions errors? I need to prove the following:
Let C be a binary cyclic Hamming code; we add a parity check at the end of every code word and get an extended (not necessarily cyclic) code.
We need to prove that this code (the extended one) can fix any single error, and any double error occurring in two adjacent position, excluding the final (parity) digit.
I've been trying to prove it for a long time but i couldn't find the solution, and my problem here is that the extended code is not cyclic at all, so I can't talk about its generator polynomial. I found a similar proof, which can show that if hamming code is generated by polynomial $g(x)$, than the code $<(x+1)g(x)>$ is double adjacent ( fix any single error, and any double error occurring in two adjacent position, excluding the final (parity) digit), but I can't use this proof here because the proof rely on the fact that there is no vector in the code in the form of:
$$
(x^i + x^{i+1})+(x^j+x^{j+1})=(x+1)(x^i+x^j)
$$
And if there is, so this polynomial is multiple of $(x+1)g(x)$, meaning that $(x^i+x^j)$ is  multiple of $g(x)$, meaning $(x^i+x^j)$ is in the code $<g(x)>$, contradicting the hamming code minimal distance 3.
Obviously this kind of proof won't work here. What other way should I use to prove it? Thanks.
 A: Let $c(x) = \sum_{i=0}^{N-1} c_i x^i$ denote the polynomial corresponding
to the codeword $\mathbf c = [c_0, c_1, \ldots, c_{N-1}]$ of a cyclic binary Hamming
code with generator polynomial $g(x)$. Note that $N = 2^m-1$ for some
positive integer $m$ and that $g(x)$ is a primitive
polynomial of degree $m$. Let $\alpha \in \mathbb F_{2^m}$ be a root of
$g(x)$ and note that $\alpha$ has order $2^m-1$.
The parity check matrix $H$ of this cyclic binary Hamming code can be viewed as
$[1, \alpha, \cdots, \alpha^{N-1}]$.  If you are unfamiliar with this notion, then
express each $\alpha^i$ as a column vector to get the
usual binary representation. But, with this formulation,
$H\mathbf c^T = \mathbf0$ is equivalent to $c(\alpha) = 0$ which is
exactly what the polynomial formulation says in other terms: each
codeword polynomial is a multiple of $g(x)$ which happens to have
$\alpha$ as a root, and so $c(\alpha) = 0$.
Let $\mathbf c$ be transmitted and $\mathbf r = \mathbf c + \mathbf e$ denote
the received word. In terms of polynomials, $r(x) = \sum_i r_ix^i$
is the received word polynomial where $r_i = c_i + e_i$ with
$e(x)$ denoting the error word polynomial corresponding to the error
word $\mathbf e$. Note that $r(x) = c(x) + e(x)$.  The receiver computes 
$$H\mathbf r^T = H(\mathbf c + \mathbf e)^T =
H\mathbf c^T + H\mathbf e^T = \mathbf 0 + H\mathbf e^T = H\mathbf e^T.$$
In terms of polynomials, computing $H\mathbf r^T$ is the same as
evaluating $r(x)$ at $\alpha$, andso the above equation can be
expressed as 
$$r(\alpha) = c(\alpha) + e(\alpha) = 0 + e(\alpha) = e(\alpha).$$
If a single error has occurred, then $\mathbf e$ has Hamming weight $1$,
and $e(x)$ is a monomial $x^i$ for some $i$, $0 \leq i \leq N-1$.
Thus, the bit in the $i$-th position has been received incorrectly.
In terms of polynomials, the syndrome is 
$e(\alpha) = x^i\bigr |_{x=\alpha^i} = \alpha^i$. If two errors
have occurred, the syndrome is $\alpha^i + \alpha^{i^\prime} = \alpha^j$.
The decoder for the cyclic Hamming code cannot tell whether one
or two errors have occurred: if the symdrome equals $\alpha^\ell$,
it simply inverts $r_\ell$ to correct the error. 
The extended Hamming code has an overall parity bit $c_N$
whose value is chosen to ensure that the extended codeword 
$\hat{\mathbf c} = (\mathbf c, c_N)$ has even Hamming weight.
Suppose that an extended Hamming code is being used and
it is known that if there are two channel errors,
then they are adjacent errors not involving the overall
parity bit.
Let $\hat{\mathbf r} = (\mathbf r, r_N)$ be the received word for
the extended Hamming code. If $\hat{\mathbf r}$ has odd Hamming
weight, then the decoder assumes that a single error has
occurred, and if $r(\alpha) = \alpha^\ell$, the decoder inverts
$r_\ell$.
But if $\hat{\mathbf r}$ has even Hamming weight and 
$r(\alpha) = \alpha^\ell \neq 0$, then the decoder assumes that
$e(x) = x^i + x^{i+1}$ (double adjacent error) and so
$r(\alpha) = e(\alpha) = \alpha^i + \alpha^{i+1} = \alpha^\ell.$
Since $\alpha^i = \alpha^\ell(1+\alpha)^{-1}$, the
decoder can find the value of $i$ and then invert $r_i$ and $r_{i+1}$.
In this way, single errors and double adjacent errors not involving
the overall parity bit can be corrected with an extended Hamming code.
