1
$\begingroup$

Let $\{f_n\}_{n=1}^\infty\in \mathbb{C}[x,y]$ be a sequence of polynomials given by the following expressions $$ f_n(x,y)=\sum_{i=0}^{[\dfrac{n}{2}]}(-1)^{n-i}{{n-i}\choose i}x^{n-2i}y^i. $$

Let $(f_n,f_{n+1})$ denote the ideal generated by $f_n,f_{n+1}$.

Let $h(n)$ be the smallest positive integer such that $x^{h(n)}\in (f_n,f_{n+1})$.

I want to prove $h(n)=2n-1$.

The result is tested by computer "sagemath". I do not know how to prove it?

My attempt:

Step~1. By Euclid Algorithm, I have proved $x^{2n-1} \in (f_n,f_{n+1})$. Hence $h(n)\leq 2n-1$.

Step~2. Let $k$ be an arbitrary positive integer such that $x^k\in (f_n,f_{n+1})$. I want to prove $k\geq 2n-1$.

enter image description here

$\endgroup$
2
$\begingroup$

Perhaps you could try using Groebner bases. The two examples that I computed using Macaulay2 (displayed below) suggest that there is a Groebner basis for $(f_n, f_{n+1})$ consisting of polynomials with leading terms of degree $n$. (These are $f_n, yf_{n-1}, y^2f_{n-2}, \cdots, y^n$, up to signs.) The examples also suggest that when we start reducing x^{2n-2} with respect to this Groebner basis, $y^{n-1}$, which cannot be reduced to zero, shows up at some stage.

Macaulay2, version 1.5
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases,
               PrimaryDecomposition, ReesAlgebra, TangentCone

i1 : R = kk[x,y];

i2 : f = n -> sum apply(floor(n/2)+1, i -> (-1)^(n-i)*binomial (n-i,i)*x^(n-2*i)*y^i)

o2 = f

o2 : FunctionClosure

i3 : gens gb ideal (f 5, f 6)

o3 = | y5 xy4 x2y3-y4 x3y2-2xy3 x4y-3x2y2+y3 x5-4x3y+3xy2 |

             1       6
o3 : Matrix R  <--- R

i4 : x^8 % ideal (f 5, f 6)

        4
o4 = 14y

o4 : R

i5 : x^8 + x^3*(f 5)

       6      4 2
o5 = 4x y - 3x y

o5 : R

i6 : x^8 + (x^3+4*x*y)*(f 5)

        4 2      2 3
o6 = 13x y  - 12x y

o6 : R
i7 : x^8 + (x^3+4*x*y)*(f 5) - 4*y^2*(f 4)

       4 2     4
o7 = 9x y  - 4y

o7 : R

i8 : x^8 + (x^3+4*x*y)*(f 5) - 13*y^2*(f 4)

        2 3      4
o8 = 27x y  - 13y

o8 : R

i9 : x^8 + (x^3+4*x*y)*(f 5) - 13*y^2*(f 4) - 27*y^3*(f 2)

        4
o9 = 14y

o9 : R

i10 : gens gb ideal (f 6, f 7)

o10 = | y6 xy5 x2y4-y5 x3y3-2xy4 x4y2-3x2y3+y4 x5y-4x3y2+3xy3 x6-5x4y+6x2y2-y3 |

              1       7
o10 : Matrix R  <--- R

i11 : x^10 % ideal (f 6, f 7)

         5
o11 = 42y

o11 : R

Perhaps there is a pattern above which can be exploited to prove (if it indeed is true!) that $x^{2n-2}$ reduces to $y^{n-1}$.

$\endgroup$

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.