Mathematics Asked on November 26, 2021
Let $p$ be a prime number and $ninmathbb N$. Consider the determinant
$$M_n = begin{vmatrix}frac1{x^{p^{n+1}}-x}&frac1{x^{p^{n+1}}-x^p}\ frac1{x^{p^{n+2}}-x}&frac1{x^{p^{n+2}}-x^p}end{vmatrix} in mathbb F_p(x)$$
Numerical computations suggest that
$$deg(M_n)=p-(p+2)p^{n+1}$$
Is it true? Is yes, does anyone have an idea to prove it?
Assuming that by degree you mean degree of the numerator minus the degree of the denominator, this is a trivial calculation. The determinant is $$frac{1}{(x^{p^{n+1}}-x)(x^{p^{n+2}}-x^p)}-frac{1}{(x^{p^{n+2}}-x)(x^{p^{n+1}}-x^p)}.$$ Multiplying the denominators of these fractions to combine them gives a denominator of degree $2(p^{n+1}+p^{n+2})$. In the combined numerator, the $x^{p^{n+1}+p^{n+2}}$ terms will cancel, leaving the highest degree term in the numerator as $x^{p^{n+2}+p}$. So the degree of the numerator minus the degree of the denominator is $$(p^{n+2}+p)-2(p^{n+1}+p^{n+2})=p-(p+2)p^{n+1}.$$
Answered by Eric Wofsey on November 26, 2021
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