V-equivalence classes

Previously we mentioned that the mapping $V : \wp(R[x_1,...,x_n]) \to \wp(\mathbb{A}^n)$ has partially ordered equivalence classes. As suborders of a power set $\wp(R[x_1,...,x_n])$ these V-equivalence classes also form set systems, and have set-theoretic features such as unions and intersections. Firstly, I need to show that $V$ is a complete semilattice homomorphism from union to intersection. This is a simple result of associativity and idempotence.

Proposition. $V(\cup f) = \cap V(f_i) $

Proof. $V$ can be expressed as the intersection of the roots of each of its polynomials. When the argument is given a union decomposition, then $V$ can be equivalently be expressed as the intersection of the roots of the polynomials of each its components. The nesting is cancelled by the associativity of intersection, and any overlap between sets in the union decomposition is cancelled by idempotence: \[ V(\cup f) = \] \[ \bigcap_{p \in \cup f} \{ a \in \mathbb{A}^n : p(a) = 0 \} = \] \[ \bigcap_{s \in f} (\bigcap_{p \in s} \{ a \in \mathbb{A}^n : p(a) = 0 \}) \] This produces two different representations of $V(\cup f)$ which are distinguished only by nesting and repetition. As nesting and repetition are cancelled out in a semilattice $V(\cup f) = \cap V(f_i)$ and $V$ is a complete semilattice homomorphism from union to intersection.

This construction works for any arbitrary family of polynomial systems, even if they are not ideals. There is no similarly general decomposition available for intersections. It is true that the restriction map of $V$ to ideals in an integral domain maps intersections to unions, but that only works for integral domains and ideals. With this out of the way, we get to the set-theoretic properties of V equivalence classes.

Corollary. $V$ equivalence classes are completely union closed.

Proof. Let $f$ be a subclass of a V-equivalence class, this means that there exists $S$ such that $\forall f_i \in f : V(f_i) = S$. Therefore, by idempotence of intersection the intersection $\cap V(f_i) = S$. By the previous theorem $V(\cup f) = \cap V(f_i) = S$, and now since $V(\cup f) = S$ the union of the subclass of the V-equivalence class is contained in it, which demonstates union closure.

Corollary. V equivalence classes are upper bounded by $\cup V^{-1}(S)$.

This is the maximal polynomial system that produces a given algebraic set, and by the invariance of ideal closure under $V$ this maximal polynomial system is an ideal. This is not necessarily a unique lower bound for a V-equivalence class. Consider two pairs of distinct lines that intersect in the same point, then their subsets have common roots, and so even for the equivalence class for a single point there doesn't need to be a corresponding lower bound. We will therefore focus on these upper bounds instead.

It is clear that the maximal element of any $V$ class is equal to $\mathcal{I}(S)$, because the maximal polynomial system must contain every polynomial that vanishes at $S$ which means $\mathcal{I}(S)$ is contained in it. To see the inverse inclusion, notice that every root of an element of the V class vanishes at $S$ so it is contained in $\mathcal{I}(S)$. Therefore, if we take $V$ as our central objects of study we can describe algebraic sets as emerging from its image and $\mathcal{I}$ as emerging from its inverse images. Let $M$ be the family of all maximal polynomial systems associated to an algebraic set and let $F$ be all algebraic sets, then the restriction mapping to $M$ is one to one. \[ V|M : M \to F \] This means that there is always a one to one mapping between a family of ideals $M$ and the family of all algebraic sets. Hilbert's nullstellensatz simply says that $M$ consists of all radical ideals for a given algebraically closed field. The beauty of Hilbert's nullstellensatz is that it is a topological correspondence, because the lattice of radical ideals has an order-dual set system presentation $Spec(R)$ that is order-isomorphic to the Zariski cotopology. As $Spec(R)$ is a sober topology, and determined entirely by its order this is a correspondence of topological properties between $Spec(R)$ and the Zariski topology.