Question

Question

Does there exist a computer package that will solve families of systems of linear equations over a field of prime characteristic?

An Example

Suppose I wanted to know when the following system of equations had a non-zero solution x over a field of characteristic p:

$\binom{N+i}{i} x =0 \quad \forall \quad 1 \leq i \leq M$

where $N,M \geq 1$. Obviously if I choose $N,M,p$ there are any number of packages that will work. But is there anything that will tell me in general when there is a solution, that is: There is a solution if and only if $p^{\ell(M)} \mid N+1$, where $\ell(M)$ is minimal such that $p^{\ell(M)} >M$.

Motivation

Over a field of characteristic $p$, it is easy to show that the Specht module $S^{(n-a,a)}$ has a submodule $D^{(n)}$ if and only if $\binom{n-a+1}{1},\ldots,\binom{n}{a} \equiv 0 \mod p$; this reduces to the problem above. One can also look at $Hom(S^\lambda,S^\mu)$. In some cases, this can also be reduced to solving systems of linear equations which are easy (for a computer) to solve for individual problems but it seems hard to find a general condition on $\lambda, \mu$. (Subject to constraints on the partitions - the absolutely general case is very far away.) Even where explicit solutions for the family of equations can be found, the proof is usually by case-by-case analysis, hence my desire for a computer program to do the work.

Please...

If you think a package will work, could you give me some indication of how to implement it, for example for the family of systems above. Hopefully I can then adapt that to solve the problem I actually want!

Answer 1

I am not sure if I am interpreting the question correctly, but if all you want is to decide whether some binomial coefficients are zero modulo $p$, then you shouldn't need a software package. You can use the following old elementary result, originally due to Lucas. If $n = \sum a_i p^i, m = \sum b_i p^i, 0 \le a_i, b_i < p$ are integers given by their base $p$-expansion, then ${n \choose m} \equiv \prod {a_i \choose b_i} \mod p$. In particular, ${n \choose m} \equiv 0 \mod p$ if and only if $a_i < b_i$ for some $i$.

Answer 2

"LinBox provides tools for linear algebra computations over the integers, the rational numbers, and finite fields and rings."

Answer 3

If I interpret the question correctly, the determinant you are looking for is computed exactly in http://arxiv.org/abs/math/9902004, Section 2.