**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 &ÂT=0 (
__non-degeneracy conditions__
)

because, obviously

H=0 & Â(H=0 &ÂT=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

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]

-------------------------------

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

2

-------------------------------

¥Deciding which are the privileged variables is crucial.

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

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

-------------------------------

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)

-------------------------------

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)

-------------------------------

So the theorem holds, for instance, when c-d0. 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

-------------------------------