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If $$A = \bigg\{\sum_{i=0}^4a_i x^{i}+b_i x^{i}y \ \big| \bigg(\sum_{i=0}^4a_iw^i\bigg)\bigg(\sum_{i=0}^4a_iw^{4i}\bigg) = 0: a_i, b_i \in \mathbb{F}_{2^k}\bigg\}$$ where $\mathbb{F}_{2^k}$ is a field with $2^k$ elements for some positive integer $k$ and $w$ is primitive $5^{th}$ root of unity, $x, y$ are generators of dihederal group of order $10$.

I want to prove that cardinality of $A$ is $2^{4k+1}-2^{3k}$. How to think?

PAMG
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  • Does your condition involve $b_i$ as well? So maybe $\sum_i b_i w^{4i}$ is the second factor? – kodlu Mar 05 '18 at 06:44
  • No this is not involved in the product – PAMG Mar 05 '18 at 06:45
  • So i think $2^{5k}$ are the possibilities for $b_i$'s as these are arbitrary. – PAMG Mar 05 '18 at 06:46
  • OK, then expand that product in the condition, using $w^4=w^{-1}$ and that should help. – kodlu Mar 05 '18 at 06:54
  • I am getting $5$ equations. Don't know how to proceed further, – PAMG Mar 05 '18 at 08:37
  • @ see the edits. – PAMG Mar 05 '18 at 12:36
  • So $A$ is a subset of the group algebra $\Bbb{F}{2^k}[D_5]$? If so, why keep that a secret? Anyway, we are still free to choose the $b_i$s any which way we want, and those elements of the group algebra are linearly independent over $\Bbb{F}{2^k}$. So the answer is $2^{5k}$ times the number of choices for the vector $(a_0,a_1,a_2,a_3,a_4)$. The constraint equation is satisfied if and only if the polynomial $a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ has either $w$ or $w^{-1}$ as its zero. The formula depends on the residue class of $k$ modulo $4$. This is because $w\in\Bbb{F}_{2^4}$. – Jyrki Lahtonen Mar 05 '18 at 21:21
  • (cont'd) So if $K=\Bbb{F}_{2^k}$ then $w$ is an element of $K$, iff $4\mid k$. $[K(w):K]=2$, if $k\equiv2\pmod 4$, and $[K(w):K]=4$, if $k$ is odd. The minimal polynomial $m(x)$ of $w$ is in the respective cases thus $x+w$, $x^2+\omega x+1$ with $\omega$ a third root of unity, and $x^4+x^3+x^2+x+1$. In the latter two cases $w$ and $w^{-1}$ share a minimial polynomial, so in those cases the second factor of the constraint is superfluous. – Jyrki Lahtonen Mar 05 '18 at 21:30
  • Looks like this is a variant of either this, this, or this. I'm not sure what's going on, but this is not good for site hygiene. Mittal, care to explain what are you trying to achieve here? – Jyrki Lahtonen Mar 07 '18 at 04:57

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