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Let $\mathbb{F}$ be a finite field with $q$ elements and $\mathbb{E}$ an extension field of degree $n$ of $\mathbb{F}$.

How I will be able to demonstrate that $$\sum_{k=0}^{n-1}t_k\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}(g_{ij})^{q^k}x^{q^{i+k}+q^{j+k}}= \sum_{k=0}^{n-1}t_k\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}(g_{i-k,j-k})^{q^k}x^{q^{i}+q^{j}}$$

applying Cyclic Rotation?

$i-k$ and $j-k$ are calculated $\mod n$. I'm trying to use variable change, for example for $i$, $m = i+k \mod n$, but I can't get change the índice sumation $i$

juaninf
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  • There's a little confusion in the statement of this question. First of all, note that both of these expressions are automatically in $\mathbb{E}[x]$ (assuming that $t_k$ and $g_{ij}$ are in $\mathbb{E}$ or $\mathbb{F}$) by dint of being polynomials with coefficients in $\mathbb{E}$. Secondly, your variable change is correct - note that when choosing $m=i+k$, then the sum from $i=0$ to $n-1$ is the sum from $m=k$ to $n+k-1$, and this splits naturally into a sum from $m=k$ to $n-1$ and a sum from $m=n$ to $n+k-1$, with the latter being from $0$ to $k-1$ after modding out by $n$. – Steven Stadnicki Feb 13 '14 at 21:11
  • A bit more explanation of what you're trying to do in the abstract (is this a lemma as part of some larger exercise or proof?) might go a long way here. – Steven Stadnicki Feb 13 '14 at 21:12
  • yes, the second equation in the pag 8 of this paper http://www.minrank.org/hfesubreg.pdf –  Feb 13 '14 at 21:15
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    One (not too serious) protest. The stated identity is false in the polynomial ring $\Bbb{E}[x]$. This is because the polynomial on the left is of degree $2q^{2(n-1)}$ (the term with $i=j=k=n-1$). OTOH the polynomial on the right is of degree $\le 2q^{n-1}$. The identity does hold for all $x\in\Bbb{E}$, because for any $x\in\Bbb{E}$ we have $x^{q^{n}}=x$. Most likely this is what is intended. Nevertheless, the distinction between a polynomial and a polynomial function is good to keep in mind. – Jyrki Lahtonen Feb 14 '14 at 06:08
  • @JyrkiLahtonen thanks! – juaninf Feb 14 '14 at 12:16

1 Answers1

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This is just a matter of chopping the sum into two different pieces and reassembling. Given that, for instance, $f(j,k)$ only depends on the value of $j\bmod n$ (i.e., that $f(j+n, k) = f(j,k)$), then we can write: $$\begin{align} \sum_{k=0}^{n-1}\sum_{j=0}^{n-1}f(j+k, k) &= \sum_{k=0}^{n-1}\sum_{m=k}^{n+k-1}f(m,k) &\text{(substituting $m=j+k$)}\\ &= \sum_{k=0}^{n-1}\left(\sum_{m=k}^{n-1}f(m,k)+\sum_{m=n}^{n+k-1}f(m,k)\right)&\text{(splitting at $m=n$)}\\ &=\sum_{k=0}^{n-1}\left(\sum_{m=k}^{n-1}f(m,k)+\sum_{m=0}^{k-1}f(m,k)\right)&\text{(shifting the second sum)}\\ &=\sum_{k=0}^{n-1}\left(\sum_{m=0}^{k-1}f(m,k)+\sum_{m=k}^{n-1}f(m,k)\right)&\text{(swapping the two terms)}\\ &=\sum_{k=0}^{n-1}\sum_{m=0}^{n-1}f(m,k)&\text{(recombining)}\\ \end{align} $$

You can apply this technique to both the sums over $i$ and over $j$ in your equation.

Steven Stadnicki
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