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Prove that $g(\alpha)$=0 if and only if $g'(\alpha)$=0


where $\alpha \in F[t]$. We are working in standard binary space.

This is the answer:

if and only if

I can't seem to get to it??

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$\alpha^{11}+\alpha^{10}+\alpha^6+\alpha^5+\alpha^4+\alpha^2+1$ and $1+\alpha^{-1}+\alpha^{-5}+\alpha^{-6}+\alpha^{11}+\alpha^{-7}+\alpha^{-9}+\alph‌​a^{11}$ aren't statements, so it makes no sense to put if and only if between them. Something must be missing here. – Brian M. Scott May 2 '12 at 4:52
Sure. But is there any connection between these two 'equations'(non statements) ? – Kiv Efehe May 2 '12 at 4:54
The latter is the former divided by $\alpha^{11}$. Not sure how that's an answer though. – anon May 2 '12 at 4:55
@Kiv, they are not equations. An equation has two sides separated by an equality sign. – Jyrki Lahtonen May 2 '12 at 4:55
Ok, feel free to delete this post.... as it's probably no help to anyone but me and this practice paper for algebraic coding theory... i fail to see where the $\alpha^{11}+\alpha^{-7}+\alpha^{-9}+\alpha^{-11}$ bit comes from ... – Kiv Efehe May 2 '12 at 4:58
up vote 1 down vote accepted

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]

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Thank you @JyrkiLahtonen for the detailed answer ! i can see that they are recipricols. If the zeroes coincide because the minimal polynomial is a factor of both - I still can't see the connection between my two equations in the "solution" – Kiv Efehe May 2 '12 at 5:21
@Kiv, the concensus seems to be that your latter equation has an extraneous $\alpha^{11}$. Please check your book! – Jyrki Lahtonen May 2 '12 at 5:23
argh ! I get completely ruined by typos. Sorry everyone ! But thank you in any case – Kiv Efehe May 2 '12 at 5:31

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