# a problem about ideals of polynomial rings

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 $k\geq 2n-1$.

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 $y^{n-1}$.