# First order logic in polynomial equations

Have you ever wondered which points on a conic are the intersections of tangent lines of another surface through the origin? More generally, which points on a shape hold some specified relation to all the points on another shape. The theory of equations is supposed to describe things like this, and the modern (1940s at least) method of doing this is with ideals in commutative algebra.

I am having difficulty expressing some equations as ideals. In the following $K$ is an infinite field.

The first construction has two affine algebraic sets $V$ and $W$ in different variables and some polynomial equations between them, and produces the subset of $V$ satisfying those equations. $$\operatorname{Sub}(V,W,E) = \{ v \in V : \forall w \in W, \forall e \in E, e(v,w) \text{ is true } \}$$

The second construction is just a little more complicated: It takes three varieties $V,W,X$ in distinct variables, and equations involving the variables from all three varieties, and returns the subset of $V$ satisfying all the equations for at least one element of $X$.

$$\operatorname{Sub}(V,W,X,E) = \{ v \in V : \forall w \in W, \forall e \in E, \exists x \in X, e(v,w,x) \text{ is true } \}$$

Given the ideals for $V,W,X,E$, how do I find the ideal for $\operatorname{Sub}(V,W,X,E)$?

## Examples

If $V$ is the parabola $\{(x,y) \in K^2 ~\mid~ y=x^2\}$ and $W$ is the line $\{(z,w) \in K^2 ~\mid~ z=w+1 \}$ and the equations are $\{yz=x^2\}$, then the output should be $\{ (x,y) \in K^2 ~\mid~ x=0, y=0 \}$ because the only point on the parabola $y=x^2$ that always satisfies $yz=x^2$ is the point $(0,0)$.

In other words, given the ideal $(y-x^2,z-w-1,yz-x^2) \leq K[x,y,z,w]$ and the block of variables $\{x,y\}$ and $\{z,w\}$ it should give the ideal $(x,y) \leq K[x,y]$.

Sometimes the answer seems hard to describe: let $V$ be the algebraic set of $n \times n$ matrices, let $W$ be an irrelevant algebraic set, let $X$ be the algebraic set of scalars, and finally let $E$ be the single equation expressing that the determinant of the matrix from $V$ times the scalar from $X$ is $1$. So the ideal for the question is $((ad-bc)x-1) \leq k[a,b,c,d,x]$, but I believe this is also the answer.

Another example takes $V$ to be the special linear group, $W$ to be a closed subgroup, and the equations to be that every element of $V$ centralizes every element of $W$. When $W$ consists of a single (closed) point, this is easy, but in general this is causing my great difficulty.

Another example takes $V$ to be an algebraic group, $W$ to be a closed subgroup, $X$ to be another closed subgroup, and $E$ to be $\forall v \in V : \forall w \in W : \exists x \in X : vw=xv$, so that the answer is the closed subgroup of $V$ consisting of those elements that conjugate $W$ into $X$.

## False starts

Things I've tried that don't quite work:

• “elimination”, that is $I \leq K[x,y,z,w]$ is replaced by $I \cap k[x,y]$. For instance $(xz)$ is replaced by $(0)$ even though the correct answer is $(x)$.
• “localizing”, that is $I \leq k[x,y,z,w]$ is replaced by $I' \leq k(z,w)[x,y]$ and then by $I' \cap k[x,y]$. For instance $(xz,z)$ is replaced by $(1)$ even though the correct answer is $(0)$.

## Nice to haves

I'm also interested in an answer that handles the case that $W$ is not irreducible smoothly. I have a sequence of localizing over function fields that I cannot find counterexamples to, but I suspect this is due to my deficiencies rather than the sequence's adequacies.

I'm also interested in an answer that is amenable to algorithmic resolution: for instance, one that could be programmed in Maculay2, singular, sage, magma, etc.

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Perhaps no one but me finds this interesting, but I'm getting limitedly good results with "for all" meaning take the function (hopefully) field, and "exists" meaning elimination (if possible, but be careful, it is not always possible). It may be that magma is the only CAS out there that can handle the fields needed. – Jack Schmidt Jul 29 '12 at 22:44