# Simplification of a double summation in a polynomial ring over a finite field

I am looking out to simplify the following double summation in $\mathbb{F}_q[x_1,x_2]$, where $p$ is a prime and $q=p^k$ for some positive integer $k$ and a positive integer $r$ such that $0 \leq r \leq q-2$:

$\sum\limits_{i=0}^{q-2} \sum\limits_{j=0}^r {(q-i)(q-1) \choose j}x_1^{(q-i+j)(q-1)} {(i+2)(q-1) \choose r-j} x_2^{(i+2+r-j)(q-1)}$

(Assume that the sum is taken over only those terms where the binomial coefficients make sense) This sum arose during my study of invariant theory, on the application of Steenrod operations to certain polynomials in $\mathbb{F}_q[x_1,x_2]$ (you can comment if you want more details). I expect the sum to be a non-zero scalar multiple of the following sum:

$\sum\limits_{i=0}^{q-r-2}x_1^{(q-i)(q-1)}x_2^{(i+r+2)(q-1)}$

I tried changing the order of summation and some changes of variables which didn't turn out to be very helpful. Any help regarding any techniques that could be used here or references to books/results that could be used for the simplification will be appreciated. Thanks in advance!

• so you mean you are in $\mathbb{F}_p[x_1,x_2]$ ? (the ring of polynomials in $x_1,x_2$ with coefficients in $\mathbb{F}_p$) – reuns Jun 9 '16 at 2:23
• @user1952009 Yes. I realize it's my mistake. I'll edit my question. – MathManiac Jun 9 '16 at 2:24
• my first comment was also that many of the binomial coefs contain $p=0$ as a factor (and write $f(x_1,x_2) = \ldots, g(x_1,x_2) = \ldots$ and I suspect $f = g$) – reuns Jun 9 '16 at 2:26

In order to prove this identity, we first transform it to a simpler form. Firstly, cancel $x_1^{q(q-1)}x_2^{(r+2)(q-1)}$ from both sides. Next, replace $x_1^{q-1}$ and $x_2^{q-1}$ by $z_1$ and $z_2$ respectively, and next, replace $z_2/z_1$ by $z$. Thus, the identity to be proven transforms into:

$$\sum\limits_{i=0}^{q-2}\sum\limits_{j=0}^r {(q-i)(q-1) \choose j}{(i+2)(q-1) \choose r-j} z^{i-j} = {q-1 \choose q-r-1}\sum \limits_{i=0}^{q-r-2} z^{i}$$

Now, as we are in $\mathbb{F}_q$, we have ${(q-i)(q-1) \choose j} = {i \choose j}$ and ${(i+2)(q-1) \choose r-j} = {q-i-2 \choose r-j}$ by Lucas' Theorem. Finally, putting $k=i-j$ and suitably modifying the limits for summation, we get:

$$\sum\limits_{i=0}^{q-2}\sum\limits_{k=i-r}^i {i \choose k}{q-i-2 \choose q-2-r-k} z^k = {q-1 \choose q-r-1}\sum \limits_{k=0}^{q-r-2} z^k$$

Next, we notice that for $k>i$ and for $k<i-r$, the summand on the LHS becomes 0. So, we can sum over $k$ from $0$ to $q-2$. Then, the coefficient of $z^k$ on the LHS becomes equal to $\sum\limits_{i=0}^{q-2} {i \choose k}{q-i-2 \choose q-2-r-k}$. (For $k>q-2-r$, this is zero exactly as in the RHS). This is the coefficient of $a^kb^{q-2-r-k}$ in:

\begin{align*} \sum\limits_{i=0}^{q-2} (1+a)^i(1+b)^{q-i-2} &= (1+a)^{q-2} \sum\limits_{i=0}^{q-2} \bigg(\frac{1+b}{1+a} \bigg)^{q-i-2} \\&=(1+a)^{q-2} \frac{1-\bigg(\frac{1+b}{1+a}\bigg)^{q-1}}{1-\frac{1+b}{1+a}} \\&=\frac{(1+a)^{q-1} - (1+b)^{q-1}}{a-b} \\&=\frac{\sum\limits_{i=0}^{q-1}{q-1 \choose i}(a^i-b^i)}{a-b} \\&=\sum\limits_{i=0}^{q-1} {q-1 \choose i} \sum\limits_{j=0}^{i-1}a^jb^{i-j-1} \end{align*}

The coefficient of $a^kb^{q-2-r-k}$ in the above expression is exactly ${q-1 \choose q-r-1}$ which is exactly the coefficient of $z^k$ in the RHS of the identity we are trying to prove. Hence the two expressions are equal in $\mathbb{F}_q[x_1,x_2]$.