Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 1 down vote accepted

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 $p_k(x,y)$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.