Applications of commutative semigroup rings

Let $R$ be a commutative ring. Then every commutative semigroup $S$ is naturally associated to a commutative ring extension of $R$, the commutative semigroup ring of $R$ by $S$. It is not hard to see that this construction is full of applications in commutative algebra and algebraic geometry. We will utilize commutative semigroup rings as an organizing principle in the theory of polynomial rings, which is an important part of algebraic geometry.

Polynomial rings:

The free $\mathbb{N}$-semimodule $F(X)$ is a very familiar object of commutative semigroup theory. It is not hard to see that the polynomial ring $R[x_1,x_2,...]$ is merely the commutative semigroup ring of $R$ by $F(x_1,x_2,...)$ : $RF(x_1,x_2,..)$. As a consequence, the polynomial rings that are so fundamental in algebraic geometry, can be considered to be a special case of a commutative semigroup ring.

Subalgebras of polynomial rings

Let $S$ be a finitely generated torsion-free cancellative J-trivial commutative semigroup. Then $S$ embeds into the free commutative semigroup $F(X)$ on a finite set of generators $X$. As a consequence, we can embed the commutative semigroup ring $RS$ into the polynomial ring $R[x_1,x_2,...]$.

As an example, any numerical semigroup can be embedded in the polynomial ring on a single generator. The polynomial subring $R[x^2,x^3]$ for example is merely the commutative semigroup ring of the numerical semigroup $\{2,3\}$. If we had $R[x^2y,yz^3]$ for example it would be generated by the commutative semigroup $(x^2y,yz^3) \in F(x,y)$, and so on.

Extensions of polynomial rings:

It is a basic fact of commutative algebra that $F(X)$ is a cancellative semigroup. Therefore, the free $\mathbb{N}$ semimodule $F(X)$ can be embedded in the free $\mathbb{Z}$-module $F^{\circ}(X)$. As a consequence, the commutative semigroup ring of multivariable polynomials $RF(X)$ can be embedded in the ring of multivariable Laurent polynomials $RF^{\circ}(X)$.

This can be further extended by considering rings of Puiseux polynomials $R\mathbb{Q}^n$ consisting of polynomials that have rational exponents, or this could even be embedded in $R\mathbb{R}^n$ to have arbitrary real exponents, so that we can have a complete extension of the ordinary polynomial ring $R[x_1,x_2,...]$.

Coordinate rings of varieties

Let $Y$ be an algebraic variety defined by a system of polynomial equations in $R[x_1,x_2,...]$. Then by now means is it the case that the coordinate ring $A(Y)$ can always be defined by a commutative semigroup ring. However, there is an important case in which they can be: algebraic varieties defined by differences of monomials. These correspond to relations in the presentation of a commutative semigroup.

Therefore, we can use commutative semigroup rings in algebraic geometry in order to deal with the important special case of varieties determined by differences of monomials. For example, consider the hyperbola $\frac{R[x,y]}{xy=1}$. Then this clearly produces a presentation of the commutative group $\mathbb{Z}$ so this is a ring of Laurent polynomials. As you can see, this is a very useful concept of commutative algebra.

References:
[1] Commutative semigroup rings by Gilmer

The affine scheme of Spec(R)

The construction of the affine scheme on Spec(R) hinges on the fact that you can you localize at the points of the spectrum: the irreducible radical ideals. This is possible because irreducible radical ideals are prime, which means that their complement is multiplicatively closed. This provides a link between the lattice of radical ideals and the lattice of multiplicative sets, which isn't entirely obvious. As a preliminary, I will therefore explain why irreducible radical ideals are prime ideals, and why you can indeed localize over them. This doesn't require much more then the proof that the nilradical is the intersection of all prime ideals and the lattice theorem.

Radical ideals are the complement of multiplicative-iteration closed sets and prime ideals are the complements of multiplicatively closed sets, so prime ideals are quite obviously radical. Prime ideals are also clearly irreducible (not the intersection of any other ideals) because if they are reducible then we can form elements from each set not in P whose product is in P. It remains only to show that irreducible radical ideals are prime. The proof is based upon Zorn's lemma, which is applicable because the lattice of ideals is a complete lattice.

Lemma. the nilradical $N$ is the intersection of all prime ideals

The nilradical is the smallest radical ideal, so it is contained in all prime ideals. We will show the converse. Let $x \not\in N$ be a non-nilpotent. Let $F$ be the family of all ideals not containing $x^n$ for $n \ge 0$. We can apply Zorn's lemma to get a maximal element $I_{max} \in F$. If $g,h \not\in I_max$ and $gh \in I_{max}$ we have $I_{max} + (g), I_{max} + (h) \not\in F$. Which implies that there exists $n,m \in \mathbb{N}$ such that $x^n \in I_{max} + (g)$ and $x^m \in I_{max} + (h)$ so $x^{n+m} \in I_{max}$ which contradicts the supposition that $I_{max}$ contains no powers of $x$. Non-nilpotents are all contained in prime ideals, so elements contained in all prime ideals are nilpotent. $\square$

The proof that the nilradical is the intersection of all prime ideals pays off, because everything else to do with the intersection between prime and radical ideals follows from it. The rest of the proofs are suprisingly easy.

Lemma. let $R$ be a commutative ring then if $(0)$ is an irreducible radical ideal it is prime.

Proof. If $(0)$ is a radical ideal it is the nilradical and the intersection of all prime ideals. By the fact that $(0)$ is irreducible it is contained in this set of all prime ideals which means it is a prime ideal. $\square$

Theorem. irreducible radical ideals are prime

Proof. let $I$ be an irreducible radical ideal of a commutative ring $R$. By the lattice theorem $(0)$ is an irreducible radical ideal in $\frac{R}{I}$ which means it is prime. The inverse image of prime ideal is prime, so this $(0)$ ideal in $\frac{R}{I}$ can be reflected back to get a prime ideal $I$. $\square$

With these preliminaries out of the way, we can now construct the affine scheme associated with a commutative ring $R$. In summary, the lattice of radical ideals is dual to a locale whose points are prime ideals. That the cotopology $Spec(R)$ is order-dual to the lattice of radical ideals is relevant, as for example coatomic ideals (which are called maximal ideals for some reason, although I have already gotten used to it) are translated into atomic sets in the cotopology which are of course singletons $\{x\}$.

The stalk of the affine scheme is defined by the localisation, which is a morphism from the lattice of multiplicative sets $Sub(R,*,1)$ to the objects of the category of rings $Ob(Rings)$. In this case of integral domains, the larger the zero free multiplicative set the larger the resulting localisation, as the localisation approaches the field of fractions, which means in certain cases localisation is a functor on a thin category. \[ \ell : Sub(R,*,1) \to Ob(Rings) \] By their very definition, the complement of a prime ideal is a multiplicative. This is formalized by the map $f : PrimeIdeals(R) \to Sub(R,*,1)$. The input action of this map on the localisation map $\ell$ produces a new morphism in the category of sets: \[ \ell : PrimeIdeals(R) \to Ob(Rings) \] We can now apply the image functor, to turn this into a map from sets of prime ideals (including open sets of our topology $Spec(R)$) to sets of rings. \[ \ell : \wp(PrimeIdeals(R)) \to \wp(Ob(Rings)) \] We now have a set of rings associated to any open set in the spectrum $Spec(R)$, but we want to get a single ring. There is a very obvious way to get a single object from a set of objects in a category: the product. This is precisely what we are going to use to construct the presheaf of rings. \[ \times : \wp(Ob(Rings)) \to Ob(Rings) \] We now have an adjusted localisation map $\ell_*$ which assigns a ring to any set of prime ideals by the product of localisations. \[ \ell_* : \wp(PrimeIdeals(R)) \to Ob(Rings) \] This is functorial on open sets because given any product of a set of objects we can form restriction morphisms to any product of a subset of objects, so the restriction maps merely restrict products of localisations to their subsets, which forms a ring homomorphism as it would in any category. So we have a presheaf of rings $\ell_*$ which we constructed from the localisation. This isn't necessarily a sheaf, so all that remains is to use sheafification to get a sheaf $\ell_*^{\#}$.

The sheafification, which is adjoint to the inclusion functor from the category of presheaves to the category of sheaves, is very convenient because it is much easier to construct presheafs then it is sheafs. Presheafs are abound in mathematics and in category theory, and all we need to do turn them into sheafs is to apply a single functor. This sheaf is a scheme because it is defined on the spectrum $Spec(R)$ of a topological space.

It is amazing that this scheme construction is even possible, which is why I spent so much time at the beginning describing why it is. There are obviously limitations of the traditional set-theoretic approach to algebraic geometry, such as that it doesn't take care of multiplicities which is responsible for the utility of this scheme construction. I will examine some of the details of that later.

Source:
Algebraic geometry by Robin Hatshorne

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.

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.