Polynomial systems and algebraic varieties

This blog has mainly dealt with set systems, and so polynomial systems haven't been considered as much. I intend that to change. The way I think of it now, is that set systems are the nicest objects of study in order theory (for example every poset represented as a set system can be made into a lattice by adding certain missing sets) and polynomial systems are the nicest objects of study in commutative algebra. Lets get started.

Definition. let $R$ be a commutative ring, and let $R[x_1,...,x_n]$ be a polynomial ring with $n$ variables, then the complete family of polynomial systems is denoted $\wp(R[x_1,...x_n])$.

Anyone non-trivial study of set systems must take certain monotone maps as its central objects of study. How fitting then that the central object of study in polynomial systems is an antitone map. In this post, we will consider this antitone map from polynomial systems to point sets. \[ V : \wp(R[x_1,...x_n]) \to \wp(\mathbb{A}^n) \] This fundamental antitone map $V$ maps any polynomial system to its set of common roots. Its not hard to see that this is an antitone map, as the more polynomials you have the fewer common roots there are. \[ V(S) = \{ a \in A^n : \forall p \in S : p(a) = 0 \} \] The map $V$ is a morphism in the category of sets, but it need not be an epimorphism. We can consider the subset $F$ of $\wp(\mathbb{A}^n)$ which consistutes its image. We can say that the algebraic sets emerge here, as essentially elements of the image of the fundamental antitone mapping $V$. \[ F = image(V) \] The image of $V$ consisting of all algebraic sets is clearly a set system. We can use basic set theory, to get that this image is a Moore family (it has complete intersection closure). As an antitone function, it maps the empty set to the largest algebraic set $\mathbb{A}^n$ and the set of all polynomials to the smallest algebraic set $\emptyset$. As is common in these cases, the only thing remaining is to show that this family has finite union closure and then we will get a cotopology.

In order to get that this forms a cotopology, note that in integral domains the roots of two polynomials $p$ and $q$ can both be combined in their product polynomial $pq$ and since integral domains have no non-trivial zero divisors $p(a)q(a) = 0$ is logically equivalent to $p(a)=0$ or $q(a)=0$. This can be extended to get finite union closure. It is not hard to see then, that in the special case of an integral domain the algebraic sets form a cotopology (the zariski cotopology). Now that we have considered the image of $V$ now we can consider the inverse image of an affine set $S$. \[ V^{-1}(S) \] Given an algebraic set, then we can get a family of polynomial systems which produce it as an output. This family of polynomial systems is clearly partially ordered by inclusion, so we can get smaller or larger polynomial systems that have the same common roots. One way that we can get a larger polynomial system is taking the ideal closure, because given any two polynomials, if they have a common root then their sum is a root as well and ideals are uneffected by scaling. This is the natural manner in which ideals emerge from polynomial systems.

Further, this partially ordered family of polynomial systems has a maximal element defined by $\mathcal{I}(S)$ which is the family of all polynomials that have $S$ as roots. This is maximal because, suppose there is some other polynomial not in $\mathcal{I}(S)$ that has $S$ as a root, then by definition it must be contained in $\mathcal{I}(S)$ as it contains everything with $S$ as roots. In the special case of algebraically closed fields, Hilbert's Nullstellensatz states that these maximal sets of the inverse images are radical ideals, which means that there is a one-to-one restriction mapping of $V$ from radical ideals to algebraic sets.