Importance: Medium ✭✭
Subject: Combinatorics
» Matrices
Recomm. for undergrads: no
Prize: none
Posted by: mdevos
on: March 8th, 2007

\begin{conjecture} If $B_1,B_2,\ldots B_p$ are invertible $n \times n$ matrices with entries in ${\mathbb Z}_p$ for a prime $p$, then there is a $n \times (p-1)n$ submatrix $A$ of $[B_1 B_2 \ldots B_p]$ so that $A$ is an AT-base. \end{conjecture}

Definition: If $A$ is an $n \times (p-1)n$ matrix over a field of characteristic $p$, then we say that $A$ is an Alon-Tarsi basis (or AT-basis) if the permanent of the $(p-1)n \times (p-1)n$ matrix obtained by stacking $p-1$ copies of $A$ is nonzero.

It follows from the Alon-Tarsi polynomial technique that if $A$ is an AT-base then for every $X_1,X_2,\ldots,X_{(p-1)n} \subseteq {\mathbb Z}_p$ of size 2 and for every $y \in {\mathbb Z}_p^n$, there exists a vector $x \in X_1 \times X_2 \ldots \times X_{(p-1)n}$ so that $Ax=y$ (using the notation from \OPref[A nowhere-zero point in a linear mapping]{a_nowhere_zero_point_in_a_linear_mapping}, $A$ is (2,1)-choosable). It follows from this that every Alon-Tarsi base over ${\mathbb Z}_p$ is also an additive basis. Thus, the above conjecture, if true, would imply \OPref[The additive basis conjecture]{the_additive_basis_conjecture}. The following strengthening of this conjecture was suggested in [D]

\begin{conjecture}[The strong Alon-Tarsi basis conjecture (DeVos)] If $B_1,B_2,\ldots,B_p$ are invertible $n \times n$ matrices with entries in a field of characteristic $p$, then we may partition the columns of $[B_1 B_2 \ldots B_p]$ into an $n \times (p-1)n$ matrix $A$ and an $n \times n$ matrix $C$ so that $A$ is an AT-base and $C$ is invertible. \end{conjecture}

In addition to implying the conjecture, above, if true, this conjecture would imply both \OPref[The permanent conjecture]{the_permanent_conjecture} and \OPref[The choosability in ${\mathbb Z}_p$ conjecture]{a_nowhere_zero_point_in_a_linear_mapping}.


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