The elusive role of elimination


•The above procedure is simply sintactical manipulation...

...how could we expect to
learn something from it?


• Eliminate variables in the system

H=0 &¬T=0 (
non-degeneracy conditions )

because, obviously

H=0 & ¬(H=0 &¬T=0) ===> T=0

else in

H=0 & T=0 (
discovery conditions )

because, obviously

H=0 & (H=0 & T=0) ===>T=0


•Elimination lead us to rewrite our given data in terms of

some
privileged variables

•Privileged in some
human determined sense...

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The construction yields (sometimes) the following ideal

H:=Ideal (ly-l^2+2lx,
2lw-2l^2+lz,
x-y,
z-w)

The expected dimension should be one, with "l" as the only privileged variable, but

Use R::=Q[l,x,y,z,w]

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Dim(R/Ideal(-l^2 + 2lx + ly, -2l^2 + lz + 2lw, x - y, z - w))

2
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•Deciding which are the privileged variables is crucial.

Example :
For any parallelogram, the center lies on one side (ag=0).

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We should have as privileged variables just a triple such as {a, b, c},

but

Use R::=Q[fgedabc]

Dim(R/Ideal(ad-ac,c(e-a)-bd, (f-a)c-(b-a)g,fd-ge))

4
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So we look for four independent variables:


Elim([g,e,a], Ideal(ad-ac,c(e-a)-bd, (f-a)c-(b-a)g,fd-ge))

Ideal(0)
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So we might get confused and use {f,d,b,c} as privileged variables,
thus we eliminate the remaining variables in our procedure.

Use R::=Q[tfgabcde]

Elim([g,e,a,t],Ideal(ad-ac,c(e-a)-bd, (f-a)c-(b-a)g,fd-ge,
agt-1))


Ideal(c - d)
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So the theorem holds, for instance, when c-d≠0. And this condition is not incompatible with the construction:

Use R::=Q[htfgabcde]

NF(1, Ideal(
ad-ac,c(e-a)-bd, (f-a)c-(b-a)g,fd-ge,(c-d)h-1))

1
-------------------------------

In fact, we can check the thesis holds under this new set of hypotheses:

NF(1,Ideal(ad-ac,c(e-a)-bd, (f-a)c-(b-a)g,fd-ge,
agt-1,(c-d)h-1))

0
-------------------------------