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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.


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.

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One remark for those who might consider inverting some kind of matrices : the fact that it is easy to generate polynomials when there is no "$ipx$" term in the linear factors for $y$ is that you can rewrite this problem in the form of solving a linear equation of the form $AX = Y$, where $X$ and $Y$ are the coefficient vectors and $A$ is the coordinate switch matrix. The "easiness" comes from the fact that $A$ is either upper or lower triangular (depending on which way you place coefficients...), hence it is easy to see that $A$ needs to be invertible with integer non-zero determinant. –  Patrick Da Silva Aug 29 '11 at 0:51
Re: Your background question. You do know the classification of single variable polynomials with values (at integer points) vanishing modulo a prime power, don't you? That not-too-well-known result has been rediscovered periodically (Singmaster in '69, Niven & Warren in '57, Carlitz says that it was in Dickson's old book,...) –  Jyrki Lahtonen Aug 29 '11 at 5:56
Yes, I know it, and it is quite simple to read (but took me weeks to prove...). Write $$ f(x) = \sum_{k=0}^d c_k(x)_k. $$ It is always possible to do this in $\mathbb Z[x]$. Now note that $f(\ell) = \sum_{k=0}^{\ell} c_k k! \begin{pmatrix} \ell \\ k \end{pmatrix}$, hence you can show by induction that $c_k k! \equiv 0$ $\mathrm{mod}$ $p^m$ by assuming that $f$ vanishes $\mathrm{mod}$ $p^m$. Conversely, a polynomial with this property vanishes because $f(\ell) = \sum_{k=0}^{\ell} c_k k! \begin{pmatrix} \ell \\ k \end{pmatrix} \equiv \sum_{k=0}^{\ell} 0 = 0.$ –  Patrick Da Silva Aug 29 '11 at 7:32
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 $p^\ell$. –  Jyrki Lahtonen Aug 29 '11 at 7:55
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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 $p_k(x,y)$.

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Given that $p-1$ is a unit in your eventual ring this may not ruin your approach, but it does seem to settle the question over $\mathbf{Z}$. –  Jyrki Lahtonen Aug 29 '11 at 6:10
I am a little confused by your argument because there are many assumptions along the way... but maybe it's clear and I should read this tomorrow XD But in my ring, $p-1$ is indeed a unit, so I think that if you're right there is still work to do, and I would be really pissed to find out that it's not possible to do this... –  Patrick Da Silva Aug 29 '11 at 7:36
For your curiosity : I have shown a few hours ago (heehee) that for $f(x,y)$ homogeneous of degree $d$ and $m \le p$, vanishing mod $p^m$ is equivalent to be able to write it in the following form : $$ f(x,y) = \sum_{j=0}^{m} \left( \left( p^{m-j} y^j \prod_{i=0}^{jp-1} (x-iy) \right) \gamma_j (x,y) \right) $$ where the $\gamma_j(x,y)$'s are homogeneous polynomials of degree $d-j(p+1)$ that can be arbitrary. When working on these kind of vanishing question, understanding what happens up to $p^p$ and then moving to $p^{p+1}$ and $p^{p+2}$ in a clever manner is the key point. –  Patrick Da Silva Aug 29 '11 at 7:38
@Patrick: I made an attempt to make the structure of my argument clearer. –  Jyrki Lahtonen Aug 29 '11 at 10:18
Okay, so $p-1 | q(1,1)$. So since for instance $x^2$ (or $x^d$ for generality) is such that $p-1 \nmid q(1,1) = 1$, we're done. =) Thank you! –  Patrick Da Silva Aug 29 '11 at 14:38
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