0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What if I assume I have far fewer than $p$ such functions? $\endgroup$ – user119164 Jan 2 '18 at 5:45
  • 2
    $\begingroup$ 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 Jan 2 '18 at 5:49
  • 4
    $\begingroup$ 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 Jan 2 '18 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 Jan 2 '18 at 20:15
  • 1
    $\begingroup$ 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 Jan 2 '18 at 22:50
1
$\begingroup$

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

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 $\prod_{i=2}^k(a_ib_i)\neq 0.$

$\endgroup$
  • 2
    $\begingroup$ How did $x+4$ appear in the denominator of the second displayed formula?! $\endgroup$ – Vladimir Dotsenko Jan 2 '18 at 22:54
  • 1
    $\begingroup$ 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 Jan 3 '18 at 3:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.