Ring subsets

Let $R$ be a ring. Then the subsets of $R$ can be classified by which operations and relations they are closed under. This produces an ontology of Moore families.

Every single one of these Moore families produces a type of lattice: lattices of additive subgroups, lattices of subrings,lattices of ideals, etc. Subfamilies in the inclusion ordering form lattice-ordered suborders of one another as lattices. These may or may not be sublattices depending upon the families in consideration.

Lattice of ideals:
Theorem. the lattice of ideals forms a sublattice of the lattice of additive subgroups

Proof.
Meet closure:
Ideals and additive subgroups are both Moore families, so they both preserve intersection as their meet operation. Therefore, the lattice of ideals is a meet subsemilattice of the lattice of additive subgroups.

Join closure:
Let $R$ be a ring, then the additive group of the ring $(R,+)$ is an commutative group. The join of two groups is equal to the set of all sums of sequences formed by elements of the two groups, but with commutativity this reduces to all sums of two elements with one chosen from each group. Let $A,B$ be subgroups of $(R,+)$ then $A \vee B$ is equal to the set {a+b} of sums of elements from each group.This forms a subgroup, so the only thing to remain is to show that this set {a+b} is also an ideal.

Let $x$ be an element of the ring $R$. Then let $a+b$ be an element of $A \vee B$. The product of $x$ and $a+b$ is equal to $x(a+b)$. By the distributive law, this is equal to $xa+xb$. The set $A$ is an ideal so there exists some element of $A$ denoted $a_0$ which is equal to the product $xa$ and since $B$ is an ideal there is some element $b_0$ equal to the product $xb$. By substitution $xa+xb = a_0+b_0$, but now $a_0+b_0 \in {a+b} = A \vee B$ and therefore the product is in the original set. Ideals therefore form a join subsemilattice of the lattice of additive subgroups.

Remarks. the join of two ideals in the lattice of ideals is typically referred to as $A+B$ rather then $A \vee B$. The fact that ideals are a sublattice of additive subgroups puts this in context. Therefore, from one now one the join of ideals can be referred to as $A+B$.

Corollary. the lattice of ideals is a sublattice of $Sub(R)$.

Proof. Additive subgroups, subrings, and ideals form a chain of lattices. Ideals are a sublattice of additive subgroups, and therefore they must be a sublattice of the intermediate lattice of subrings. To see this, consider the join of two ideals $A,B$ in the lattice of subrings. This is the smallest set that is both an additive subgroup and multiplicatively closed which contains both of them, but the smallest additively closed set is also multiplicatively closed. Therefore, the subring join of ideals is the additive subgroup join which is an ideal. Therefore, the join of ideals is an ideal. The meet of ideals is an ideal because both subrings and ideals are Moore families.

Overview. ideals form a union-free modular Moore family

Lattice of radical ideals
Another lattice of ring subsets is the lattice of radical ideals. This does not necessarily form a sublattice of the lattice of ideals. The closure operation associated with this family is the radical of an ideal $\sqrt{I}$. Prime ideals are the intersection-irreducible members of this family, therefore by considering the lattice of radical ideals we can give the prime ideals a lattice-theoretic perspective.

Subalgebra lattices of finite fields

When proving a theorem using Galois theory it is useful to consider groups and fields separately and then finally use Galois theory to connect the results about the two of them together. Therefore, groups will be considered first before connecting them to fields.

Finite cyclic groups:
Oystein Ore proved that a group is locally-cyclic if and only if it is subalgebra-distributive. For a finite cyclic group $G$ we can go one step further and prove that the lattice of subgroups $Sub(G)$ is the most fundamental type of distributive lattice: a multiset inclusion lattice / divisibility lattice.

Theorem. let $G$ be a cylic group of order $n$. Then $Sub(G)$ is the divisibilty lattice of the factors of $n$. [1]

Cyclic extensions:
In order to relate cyclic groups and their subalgebra lattices to finite fields, we need to show that finite extensions of finite fields are cyclic. This can be shown using the Frobenius automorphism [2] which generates the Galois group.

Theorem. let $F$ be a finite field of order $p^n$, then $Gal(F/GF(p))$ is a cyclic group of order $n$. [2]

Galois theory:
With the group-theoretic prelimanaries out of the way, we can now prove that the subalgebra lattice of a finite field $Sub(F)$ is a multiset inclusion lattice.

Theorem. let $F$ be a finite field, then $Sub(F)$ is a multiset inclusion lattice.

Let $F$ be a finite field of order $p^n$, then the Galois group of $F$ over its prime field is a cyclic group of order $n$. By the fundamental theorem of Galois theory, $Sub(F)$ is therefore order-dual to the subalgebra lattice of a cyclic group of order $n$. This subgroup lattice is a multiset inclusion lattice, but since finite total orders are order-dual so is its order dual. Therefore, the lattice of subfields of the finite field $Sub(F)$ is a multiset inclusion lattice as well.

References:

[1] Subgroup of Finite Cyclic Group is Determined by Order
https://proofwiki.org/wiki/Subgroup_of_Finite_Cyclic_Group_is_Determined_by_Order

[2] The Frobenius Automorphism for a Finite Field
http://mathonline.wikidot.com/the-frobenius-automorphism-for-a-finite-field

Normal subgroups form a sublattice

Theorem. let $G$ be a group, then the subset of $Sub(G)$ consisting of all normal subgroups forms a sublattice.

Proof. Let $N$ be the subset of $Sub(G)$ consisting of all normal subgroups. In order to prove that normal subgroups form a sublattice, we will prove that it is a meet subsemilattice and a join subsemilattice separately. Combined these make the normal subgroups a sublattice.

Meet closure:
Let $A,B$ be normal subgroups. Then the meet of $A$ and $B$ is the intersection $A \cap B$. Let $c$ be an element of $A \cap B$ and let $g$ be an element of $G$ then since $A$ is normal $gcg^{-1}$ is in $A$ and since $B$ is normal $gcg^{-1}$ is in $B$. Therefore, $gcg^{-1}$ is in $a \cap b$, which implies it is normal.

Join closure:
Let $A,B$ be normal subgroups, and let $A \vee B$ by their join in the lattice of subgroups. Let $c_1 ... c_n$ be the product of elements in $A$ and $B$, by closure this is in $A \vee B$. Now, the conjugate $gc_1 ... c_ng^{-1}$ is equal to $(gc_1g^{-1})...(gcg^{-1})$ which is the product of conjugates of elements in the normal subgroups, so it is in each of them. So by subgroup closure, this conjugate is in the subgroup $A \vee B$. By conjugate closure, the subgroup $A \vee B$ is normal.

Remarks: if $Sub(G)$ is the lattice of subgroups, then normal subgroups are a member of $Sub(Sub(G))$, the lattice of sublattices of the lattice of subgroups. This shows how the subalgebra lattice construction can be nested.

Subgroup lattices are union-free

Theorem. let $G$ be a group, then $Sub(G)$ is union-free. In particular, if $A,B \subseteq G$ such that $A \not\subseteq B$ and $B \not\subseteq B$ then $A \cup B$ is not a group.

Proof. since $A$ and $B$ are incomparable there must be some $a$ in $A-B$ and some $b$ in $B-A$. Then we can consider their product $ab$, since it is in $A \cup B$ it must be in either $A$ or $B$. If it is in $A$ then $a^{-1}ab = b$ is in $A$ which contradicts the supposition that $b$ is not in $A$, likewise if it is in $B$ then $abb^{-1}=a$ is in $B$ which contradicts the supposition that $a$ is not in $B$. So by proof by contradiction, the union of subgroups is not a subgroup.

Corollaries:
Modules, ideals, rings, or any structure more algebraically complicated then a group is also union-free. The union of any of them cannot even be an additive subgroup, much less satisfy any other conditions. This theorem lets us show that the union of two ideals is not an ideal using only group theory.

Lattice-ordered suborders and semigroups

Let $L$ be a lattice, and $S$ be a suborder of $L$ that forms a lattice. Then the two lattices can be compared to determine when they are similar and when they are different. In particular, for any two elements $a$ and $b$ they can be either join-preserving or meet-preserving based upon their relation to the parent order:

• Join-preserving: two elements $a,b$ are join preserving if $a \vee_S b = a \vee_L b$
• Meet-preserving: two elements $a,b$ are meet preserving if $a \vee_S b = a \vee_L b$

The join-preservation and meet-preservation both form simple graphs on the set of elements of the suborder. The extrema-preservation graph is the intersection of these two graphs. Clearly, if $a \subseteq b$ or if $b \subseteq a$ then the two elements are extrema-preserving.

Proposition. the comparability graph is a subgraph of the extrema-preserving graph

Proof. let $P$ be a pair of two elements in the comparability graph, as comparable elements we know that there must be some $a,b$ such that $a \subseteq b$ in P, then the join in the parent lattice is still $b$ and the meet is still $a$, because the suborder is always order-preserving (even though it isn't always meet preserving and join preserving).

As the comparability graph is a subgraph of the extrema-preserving graph, a suborder is either join-free and meet-free if its incomparable elements are not join-preserving or meet-preserving respectively. It is in this sense that we say that preorder containment families are union-free: they have no non-trivial unions. A suborder is a meet subsemilattice if it is totally meet preserving, a join subsemilattice if it is totally join preserving, and a sublattice if it is both.

In those cases where the suborder is not meet or join preserving, it is interesting to consider the subset of cases in which it is. In particular, a Moore family is a meet-preserving lattice-ordered suborder in the lattice of sets. In this cases, we can consider the subset of pairs for which the union operation is preserved.

Chains of lattices:
Given a chain of three lattices $A$, $B$, $C$ then it is possible for $A$ to be a sublattice of $C$ even though its intermediate lattice $B$ is not a sublattice. Consider a finite group $G$ which is not a prime power cyclic group and let $Sub(G)$ be its lattice of subgroups, then $Sub(G)$ is not a sublattice of $\mathscr{P}(C)$. Then let $A$ be a chain of groups, such as a composition series, then $A$ is a sublattice of $C$ because it is a chain. Sublattices of the parent lattice come from extrema-preserving cliques that are sublattices.

Lattices of subsemigroups
Let $S$ be a semigroup, and let $Sub(S)$ be its lattice of subsemigroups, then generally the lattice of subsemigroups is not a sublattice. One example where it is, is the lattice of subsemigroups of a total order semilattice, a pure rectangular band, or a totally ordered sum of pure rectangular bands. In fact, in those cases every subset of the semigroup is a subsemigroup. As $Sub(S) = \mathscr{P}(S)$ it forms a sublattice. Another case is the lattice of subsemigroups of a prime power cyclic group in which case $Sub(S)$ forms a chain, which is always a sublattice.

In those cases when the lattice of subsemigroups $Sub(S)$ does not necessarily form a sublattice, then it is potentially interesting to consider the cases when the join is simply the union. An obvious case is a group with zero, in that case the singleton set containing the zero element and the rest of the set form a join-preserving pair of subsemigroups, with the union being the entire semigroup. Another example, is the set of even integers $2\mathbb{Z}$ and the set of odd integers $2\mathbb{Z}+1$ both of which form multiplicative subsemigroups and their union is the entire semigroup, so they form a join-preserving pair. In general, the multiplicative semigroup of a prime ideal and its complement form an example of a join-preserving pair.

Implications of Lagrange's theorem

In a general semigroup, the order of any subsemigroup doesn't need to divide the order of the semigroup. As a consequence, there are little to no limits placed on the types of graphs that can be commuting graphs of semigroups. On the other hand, by Lagrange's theorem the order of any subgroup of a group divides the order of the group. This implies that the size of any commutatively necessary subsemigroup must divide the order of the group. Therefore, from Lagrange's theorem alone we can get these restrictions placed on the class of commuting graphs of groups:

• The degree of any vertex must divide the order of the graph
• The total number of dominant vertices must divide the order of the graph
• The clique number must divide the order of the graph
• In general, any commutatively-necessary subalgebra must divide the order of the graph

It is clear from this that there is a significant difference between the commuting graphs of groups and semigroups. All sorts of semigroups with commuting graphs unlike anything that could emerge from a group exist (like rectangular bands for example). The starting point of the theory of commuting graphs of groups should be the implications of Lagrange's theorem on the possible commuting graphs of a group, but there is more that can be infered beyond that.

Beyond Lagrange's theorem:
Groups are strongly divisibility commutative, and this is characterized by the conjugacy classes of elements, which are always non-empty. The first thing we can say beyond Lagrange's theorem is that the size of the conjugacy classes of the group are equal to the index of the centralizer. The size of each of these conjugacy classes must then sum up to equal the order of the group. We can from this, for example that groups of prime-squared order are necessarily commutative.

Modular lattices and subalgebras

Modular lattices can be considered from two perspectives: (1) the general order-theoretic perspective which examines the class of modular lattices and (2) the properties of its members and the specific modular lattices that arise in various other branches of mathematics. We will now take the later approach. Here are some of the types of modular lattices that arise in most of the branches of abstract algebra:

• Lattices of normal subgroups
• Lattices of submodules
• Lattices of ideals of a ring
• Lattices of normal intermediate field extensions of galois extensions

In general, if we have an algebraic structure that has a kernel associated with it, and associated special subalgebras that define all quotients then in all the standard cases (groups, rings, modules, etc) these subalgebras form a modular lattice as a subalgebra system. That modules form a modular lattice was proven by Dedekind, and it is not hard to see that ideals are a special case of submodules. The last case is simply a corollary of the fundamental theorem of galois theory.

Proposition. the lattice of normal intermediate field extensions of a galois extension form a modular lattice

Proof. the lattice of normal subgroups of the galois group of the field extension is modular. By the fundamental theorem of galois theory (FTGT), we know that the lattice of normal intermediate field extensions of a galois extension is order-dual to the lattice of normal subgroups. But since modularity is a self-dual condition, the lattice of normal intermediate field extensions is modular.

Subalgebra-modular structures:
Supposing that we have an algebraic structure like a ring then we can form from these structures subalgebra systems (suborders of Sub(A)) that are modular. A separate question is under what conditions is Sub(A) modular? We can now answer this to some extent using the concepts already covered. Here are algebraic structures for which Sub(A) is modular:

• Dedekind groups
• Commutative groups (a special case of dedekind groups)
• Modules
• Fields which are galois-dedekind over their prime subfield
• The ring of integers Z (all subrings are ideals)

Commutative groups can be seen to be subalgebra-modular in two different ways: (1) by the fact that they are dedekind or (2) by the fact that commutative groups can be represented as (not necessarily faithful) Z-modules. That commutative groups are subalgebra-modular has implications in group theory, because it means from the commuting graph we can infer not only subalgebras but also properties of the lattice structure. Dedekind groups are not the only subalgebra-modular groups and in general subalgebra-modular groups are refered to as iwasawa groups.

Complements:
Most of these subalgebras are related normal subgroups in some way, but normal subgroups have another important property. In the lattice of normal subgroups, complementary members characterize direct product representations of the group. It is not hard to see then, how complements are unordered, as a given subgroup which characterizes the group by direct product with some other complementary subgroup is not going to produce the same direct product with a subalgebra of that other complementary group. Therefore, the property that complements in a modular lattice are unordered (which was proven entirely using order theory) is a pre-requisite for the complements to determine direct products.

Modular lattices

Modular lattices are a generalization of distributive lattices, especially useful in the theory of lattices of subalgebras. They can be defined either by the modular law: $a \le b$ implies that $\forall x : a \vee (x \wedge b) = (a \wedge b) \vee b$ or by the forbidden lattice characterization. The following lattice is the simplest non-modular lattice:

It can easily be seen from this that every sublattice of a modular lattice is modular (this is the case for all classes of lattices defined by forbidden sublattices).

Proposition. every sublattice of a modular lattice is modular.

Additionally, complements play a special role in the theory of modular lattices. Suppose that we have an element of a modular lattice, then the set of complements of that element forms an antichain. In other words, complements are unordered. Distributive lattices are a special case in which complements are unique.

While distributive lattices are important in the foundations of mathematics and set theory, modular lattices are especially important in the study of subalgebras of algebraic structures.