Non-degeneracy conditions II

•If we care for a set of independent privileged variables, say [ x (privileged), y ], it makes sense to eliminate y in Sat( I , t ).

Let Sat(
I ,t,x ) denote such ideal and let be (g_1, ...,g_s) a set of generators.

Thus non-degeneracy conditions are now expressed in terms of privileged variables.

Of course, the zero set of (g_1, ...,g_s) in
x -space is the closure of the projection of those components where the theorem fails.

•There will be components of the hypothesis variety where the privileged variables remain independent, ie.

no polynomial in these variables vanishes over the component <===> its projection onto
x -space is the whole space.

• It can happen that theorem holds over all components where the privileged variables remain independent.

The we say that we have a
generally true theorem.

(g_1, ...,g_s)≠(0) iff
t =0 over all components of V (I )where the privileged variables remain independent.


t ≠0 over some component where x is independent, since we have that g_i* t =0 over the whole hypotheses variety, it follows that g_i must vanish over such component, contradicting the independence of x over such component.

Conversely, assume
t =0 over all these components. The we compute a non-zero polynomial g in x such that it vanishes over the remaining components (by definition of non-independence). So g* t vanishes over the whole variety, and thus some power of g is in Sat( I ,t ). It follows that Sat( I ,t,x ) is non zero.

In the example Bisecting Diagonals, privileging x_1 and x_2, we have

Elim([x[3],x[4]], P1)
Elim([x[3],x[4]], P2)
Elim([x[3],x[4]], P3)
Elim([x[3],x[4]], P4)
Ideal(x[2], x[1])

So only P_1 is a prime component where the two variables remain independent.

Then we compute Sat(
I ,t,x ):

Elim([x[3],x[4]],Ideal(x[3]x[4], x[1]x[4], x[2]x[3], x[1]x[2]))


It can happen that Sat(
I ,t,x )=(g_1, ...,g_s)=0, ie. that theorem fails over some privileged component, ie. a NON generally true theorem. Then we face a problem....