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Is a collection of reciprocals of monic reducible quadratic polynomials, that is functions of the form

$$ \{ \left( (x-a_i)(x-b_i) \right)^{-1} \}_{i=1}^{k}, $$

linearly independent over a finite field? This can be seen for the reciprocals of linear functions from the invertibility of Cauchy matrices.

Edit: To address ABX's very nice observation/obstruction, assume that $k$ is very small compared to the characteristic of the field.

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  • $\begingroup$ What if I assume I have far fewer than $p$ such functions? $\endgroup$
    – user119164
    Commented Jan 2, 2018 at 5:45
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    $\begingroup$ Not always: since $\dfrac{1}{(x+a)(x+a+1)} =\dfrac{1}{(x+a)}-\dfrac{1}{(x+a+1)}$, we have $\sum\limits_{a=0}^{p-1}\dfrac{1}{(x+a)(x+a+1)}=0\ $ over any field of characteristic $p$. $\endgroup$
    – abx
    Commented Jan 2, 2018 at 5:49
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    $\begingroup$ For $a_i\neq b_i$ These are $q(q-1)/2$ functions inside a subspace of dimension $q$... if $a,b,c$ are distinct, $1/((x-a)(x-b))$, $1/((x-b)(x-c))$, $1/((x-c)(x-a))$ are proportional to $1/(x-a)-1/(x-b)$, $1/(x-b)-1/(x-c)$, $1/(x-c)-1/(x-a)$ which have sum zero. So the answer is no as soon as $k,q\ge 3$. Did I miss something? $\endgroup$
    – YCor
    Commented Jan 2, 2018 at 6:07
  • $\begingroup$ Just to be specific, $1/(x(x+1))+1/((x+1)(x+2))+(p-2)/(x(x+2))=p/(x(x+2)).$ $\endgroup$
    – Aaron Meyerowitz
    Commented Jan 2, 2018 at 20:15
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    $\begingroup$ You can't edit a comment once 5 minutes have passed, but you can still delete and replace your own comment. $\endgroup$
    – Noam D. Elkies
    Commented Jan 2, 2018 at 22:50

1 Answer 1

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$$\frac1{x(x+1)}+\frac1{(x+1)(x+2)}+\frac{p-2}{x(x+2)}=\frac{p}{x(x+2)}.$$

Also, $$\frac1{(x+1)(x+2)}-\frac1{(x+3)(x+4)}={\frac {4x+10}{ \left( x+2 \right) \left( x+1 \right) \left( x+3 \right) \left( x+4 \right) }}.$$ So, over $\mathbb{Z}_5,$ the only $x$ for which this is defined is $x=0$ and there the numerator is $0.$ However the two are independent in $\mathbb{Z}_5(x).$

I think that if the $a_i$ and $b_i$ are all unequal then the expressions are linearly independent. In fact

If $\{\frac1{(x-a_i)(x-b_i)}\}_{i=1}^k$ is dependent in the over some field, but all subsets are independent, then in the list $a_1,b_1,a_2,b_2,\cdots , a_k,b_k$ every entry appears at least twice.

Consider $$\sum_1^k\frac{c_i}{(x-a_i)(x-b_i)}$$ where the $a_i,b_i,c_i$ are $3k$ parameters. This is a rational function with numerator a polynomial of degree $2k-2.$ The coefficient of $x^{k-2}$ is $\sum c_i$ so one could arrange for the leading coefficient, and others, to be zero. The question, as I will answer it, is if all the coefficients could be zero, i.e. if the numerator could be identically zero as a formal polynomial. Alternately, one might wonder if the numerator evaluates to zero for all $x$ in the field (not of the form $x=a_i$ or $x=b_i.$) The example above shows an instance of this. However I am not considering that, and for $k$ small with respect to the order of the field this can't happen.

I will assume that the parameters are chosen so that $a_1$ is not repeated among the other $a_i$ and $b_i$ and that $c_1 \neq 0.$ Without loss of generality we can take $a_1=0$ and $c_1=1.$ Then the numerator has constant term $\prod_{i=2}^k(a_ib_i)\neq 0.$

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    $\begingroup$ How did $x+4$ appear in the denominator of the second displayed formula?! $\endgroup$
    – Vladimir Dotsenko
    Commented Jan 2, 2018 at 22:54
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    $\begingroup$ Good point! The numerator didn't work either. The denominator of the second fraction was intended to be (and now is) $(x+3)(x+4).$ Thanks. $\endgroup$
    – Aaron Meyerowitz
    Commented Jan 3, 2018 at 3:56

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