TransWikia.com

Degree of a determinant

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?

One Answer

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

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP