In this case <math>n</math> represents the number of points and <math>m</math> represents the dimension of the vector space the points lie in. So for <math>n</math>=4 the matrix <math>\bold{D}</math> constructed below is 3 x 3. Thus there are <math>\binom{3}{2} \times \binom{3}{2} = 9</math> (<math>2 \times 2</math>)-minors of <math>\bold{D}</math>. By the construction of <math>\bold{D}</math>, there is a certain symmetry to the matrix(this symmetry occurs for all <math>n\ge{4}</math>). For this reason, with <math>n</math>=4, the polynomials given by the (<math>2 \times 2</math>)-minors are not all unique.<math>\bold{(*)}</math> In fact, there are only 6 distinct polynomials. Below is what I'm thinking may be the case in general.
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Idea: For <math>2\le{m}\le{n-2}</math>, let <math>D_{i,j}</math> be indeterminates <math>(1\le{i}<j\le{n})</math> and let
be the matrix where we set <math>D_{i,i}</math> := 0 and <math>D_{i,j}</math> := <math>D_{j,i}</math> for <math>i>j</math>. Then there are <math>\binom{\binom{n-1}{m}+1}{2} = \frac{\binom{n-1}{m}^2 + \binom{n-1}{m}}{2}</math> distinct (<math>m \times m</math>)-minors of <math>\bold{D}</math>.
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I'm thinking that the number of distinct minors of <math>\bold{D}</math> is actually the number of polynomials of <math>n\times m</math> variables that we are looking for to determine constructibility of the <math>n</math>-point configurations. My reasoning for this is that in the algorithm presented in the proof of Theorem 1.6, we set <math>F:=F_1F_2</math> where <math>F(d_{1,2},...,d_{n-1,n})=f(P_1,...,P_n)\ne 0</math>. Here, I'm thinking that <math>F_1</math> can only be one polynomial. <math>F_2</math> on the other hand can be any (<math>m \times m</math>)-minor of <math>\bold{D}</math>. If I'm right, then the number of distinct (<math>m \times m</math>) minors of <math>\bold{D}</math> is the same number of distinct polynomials <math>F</math>, and thus the same for <math>f</math>.
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<math>\bold{(*)}</math> Let <math>M_{a,b}</math> be the minor given by removing the <math>a^{th}</math> row and the <math>b^{th}</math> column from <math>\bold{D}</math>. Then with <math>n</math>=4 and <math>m</math>=2, we get 6 distinct minors which are listed below.
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<math>M_{1,1}</math>
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<math>M_{2,2}</math>
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<math>M_{3,3}</math>
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<math>M_{1,2} = M_{2,1}</math>
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<math>M_{1,3} = M_{3,1}</math>
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<math>M_{2,3} = M_{3,2}</math>
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--[[User:Davis29|Davis29]] 18:53, 12 June 2009 (UTC)
