In this case $ n $ represents the number of points and $ m $ represents the dimension of the vector space the points lie in. So for $ n $=4 the matrix $ \bold{D} $ constructed below is 3 x 3. Thus there are $ \binom{3}{2} \times \binom{3}{2} = 9 $ ($ 2 \times 2 $)-minors of $ \bold{D} $. By the construction of $ \bold{D} $, there is a certain symmetry to the matrix(this symmetry occurs for all $ n\ge{4} $). For this reason, with $ n $=4, the polynomials given by the ($ 2 \times 2 $)-minors are not all unique.$ \bold{(*)} $ In fact, there are only 6 distinct polynomials. Below is what I'm thinking may be the case in general.
Idea: For $ 2\le{m}\le{n-2} $, let $ D_{i,j} $ be indeterminates $ (1\le{i}<j\le{n}) $ and let
$ \bold{D} $ = ($ D_{i,j} - D_{i,n} - D_{j,n} $)$ _{i,j=1,...,n-1} $
be the matrix where we set $ D_{i,i} $ := 0 and $ D_{i,j} $ := $ D_{j,i} $ for $ i>j $. Then there are $ \binom{\binom{n-1}{m}+1}{2} $ distinct ($ m \times m $)-minors of $ \bold{D} $.
I'm thinking that the number of distinct minors of $ \bold{D} $ is actually the number of polynomials of $ n\times m $ variables that we are looking for to determine constructibility of the $ n $-point configurations. My reasoning for this is that in the algorithm presented in the proof of Theorem 1.6, we set $ F:=F_1F_2 $ where $ F(d_{1,2},...,d_{n-1,n})=f(P_1,...,P_n)\ne 0 $. Here, I'm thinking that $ F_1 $ can only be one polynomial. $ F_2 $ on the other hand can be any ($ m \times m $)-minor of $ \bold{D} $. If I'm right, then the number of distinct ($ m \times m $) minors of $ \bold{D} $ is the same number of distinct polynomials $ F $, and thus the same for $ f $.
$ \bold{(*)} $ Let $ M_{a,b} $ be the minor given by removing the $ a^{th} $ row and the $ b^{th} $ column from $ \bold{D} $. Then with $ n $=4 and $ m $=2, we get 6 distinct minors which are listed below.
$ M_{1,1} $
$ M_{2,2} $
$ M_{3,3} $
$ M_{1,2} = M_{2,1} $
$ M_{1,3} = M_{3,1} $
$ M_{2,3} = M_{3,2} $
--Davis29 16:36, 12 June 2009 (UTC)
