0
$\begingroup$

Consider the $\mathbb Z$-module that consists of the polynomials in $\mathbb Z[x,y]$ that are homogeneous polynomials of degree $d$ in the indeterminates $x$ and $y$ (homogeneous meaning that all terms of the form $c_{ij} x^i y^j$ are such that $i+j = d$, where $d$ is the degree of the polynomial). One can see that there is a unique way of writing an homogeneous polynomial of degree $d$ in the form $ \sum_{j=0}^d c_k y^{d-k} \prod_{i=0}^{k-1} (x-iy) $ because there is precisely one term where $x$ is at the $k^{th}$ power for any $k$ in the range $[0,d]$, hence we can compute coefficients. Therefore, the polynomials $y^{d-k} \prod (x-iy)$ form a basis of the $\mathbb Z$-module.

Question.

Is it possible to write any homogenous polynomial of degree $d$ in the form $ \sum_{k=0}^d c_k \left( \prod_{i=0}^{d-k-1} (y-ipx) \right) \left( \prod_{i=0}^{k-1} (x-iy) \right) $ where $p$ is a prime number? (The larger context is a number theory context, thus the prime is the thing I need. The fact that $p$ is prime might not be needed to prove this though!) Note that the polynomials formed by the 2 products are actually homogenous polynomials of degree $d$ in $x$ and $y$, so this would be a basis of the $\mathbb Z$-module of homogenous polynomials of degree $d$.

The reason for this question is because I am looking for a characterization for a certain class of polynomials that are always 0 with respect to a prime power modulus and the existence of this decomposition would help me very much. =)

If you have any suggestions please feel free to comment.

  • 0
    The result I had in mind goes as follows. Write $p_k(x)=\prod_{i=0}^{kp-1}(x-i)$. Then the values $p_k(n), n\in\mathbf{Z}$ are always divisible by $(kp)!$, so if you multiply these polynomials with the appropriate power of $p$ you get polynomials with values vanishing $\pmod{p^\ell}$. If you take such polynomials up to the first monic one, you have a generating set for the ideal of polynomials in $\mathbf{Z}[x]$ that vanish module $p^\ell$.2011-08-29

1 Answers 1

1

I don't think that you have a $\mathbf{Z}$-basis. I claim that if $d>1$, then all the polynomials $q(x,y)$ in the $\mathbf{Z}$-span of your generators have the property $p-1\mid q(1,1)$. Thus, if $p>2$ the span cannot consist of all the homogeneous polynomials of degree $d$.

Proof: If you evaluate the polynomial $p_k(x,y)=\prod_{i=0}^{d-k-1}(y-ipx) \prod_{i=0}^k(x-iy)$ at $x=y=1$, the second product shows that $p_k(1,1)=0$ unless $k=0$ (if $k>0$, then the factor $x-y$ coming from $i=1$ makes this happen). But if $k=0$ and $d>1$, then the first product has a factor $y-px$ (again coming from $i=1$), and this forces $p_0(1,1)$ to be divisible by $y-px\vert_{(x,y)=(1,1)}=1-p$. For all $k$ we have $p-1\mid p_k(1,1)$, so the same holds for all the $\mathbf{Z}$-linear combinations of the polynomials $p_k(x,y)$.

  • 0
    I know, but that is not interesting me because it would get too complicated and I am looking for something readable for humans. =P2011-08-29