Prove that $g(\alpha)$=0 if and only if $g'(\alpha)$=0
$g(t)=t^{11}+t^{10}+t^6+t^5+t^4+t^2+1$
$g'(t)=t^{11}+t^9+t^7+t^6+t^5+t+1$
where $\alpha \in F[t]$. We are working in standard binary space.
This is the answer:
$\alpha^{11}+\alpha^{10}+\alpha^6+\alpha^5+\alpha^4+\alpha^2+1=0$
if and only if
$1+\alpha^{-1}+\alpha^{-5}+\alpha^{-6}+\alpha^{11}+\alpha^{-7}+\alpha^{-9}+\alpha^{-11}=0$
I can't seem to get to it??
It seems to me that $g(t)$ and $g'(t)$ are reciprocals of each other: $$ t^{11} g(1/t)=t^{11}(t^{-11}+t^{-10}+t^{-6}+t^{-5}+t^{-4}+t^{-2}+1)=1+t+t^5+t^6+t^7+t^9+t^{11}=g'(t). $$ This does mean that if one of them has a zero in some field, then so does the other. However, you seem to claim that the zeros coincide (you denote the common zero by $\alpha$). This is false in general. It could be true in some particular case, if the minimal polynomial of $\alpha$ should be a factor of both $g$ and $g'$. That is not the case here, as a quick run of Euclid's algorithm shows that the two polynomials have no common factors.
[Edit] A correct claim is that if $g(\alpha)=0$, then $g'(1/\alpha)=0$. As $g(t)$ is irreducible in $F_2[t]$ so is $g'(t)$. They both have eleven zeros in the field $GF(2048)$, but none of them are common because $\gcd(g(t),g'(t))=1$. [/Edit]