We assume that our theorem is not algebraically true and that Sat( I ,t,x ) is zero (ie. theorem not generally true). This is normal....
Then we compute I_d=( I ,t) (or I_d,x = Elim( y , I_d )).
Since t does not vanish over all I , we have that I_d gives a strictly smaller variety, where the theorem surely will be true...but trivially true!!
More interesting is to consider I_d,x .
I_d,x is zero ==> t vanishes over some privileged components of the hypotheses variety.
I_d,x is zero <== t vanishes identically over some privileged components of the hypotheses variety.
If for every privileged component t does not vanish, we take a
x -polynomial g that vanishes over all remaining components. Then g vanishes
on the intersection of V ( I ) and t =0, ie. g is in the radical of I_d , so I_d,x is not zero.
Conversely, if I_d,x is not zero, take g in I_d,x . We have that g is a combination of I , t , so it vanishes over any component where t vanishes. It follows that t can not vanish identically over privileged components.
Notice : this result can be improved with further conditions on the privileged variables.
If I_d,x =(h'_1,...,h'_r) is not zero, we are in the generally false case. The thesis does not hold globally over any of the privileged components.
Then we may consider as new hypotheses variety ( I , I_d,x ).
It is a proper subvariety, since the set of privileged variables x is now dependent.
We start all over again....considering a subset x' of independent variables. This time ( I , I_d,x, t ) does not contain non zero x' polynomials, so we can hope :-) it is generally true ...
But see the next two examples: