Commutativity necessary subrings

The commuting graphs of semigroups induce a number of different commutativity necessary subsemigroups: centralizers, maximal commuting cliques, commutative principal filters, the center, etc. As it it happens all of these concepts can be transferred to non-commutative rings, so that we can understand non-commutative algebra using graph-theoretic techniques.

Definition. let $R$ be a ring then the commuting graph $Com(R)$ of $R$ is the commuting graph of its multiplicative semigroup. \[ Com(R) = \{(x,y) : xy = yx \} \] The first step towards proving the existence of commutativity necessary subrings for a ring $R$ is to demonstrate that centralizers are subrings. The existence of the other types of commutativity necessary subrings immediately follows.

Theorem. let $R$ be a ring, $x$ an element of $R$, then the centralizer $S$ of $x$ is a subring. \[ S = \{ c : cx = xc \}\] Proof. (1) by basic semigroup theory, then centralizer is a multiplicative subsemigroup.

(2) by the nullary distributive law we have \[ \forall x*0 = 0 = 0*x \] It follows that $0$ is a central element which implies $0 \in S$.

(3) by the binary distributive law we have \[ x(a+b) = xa + xb = ax + bx = (a+b)x \] So the centralizer $S$ is additively closed.

(4) let $c \in R$ then $(-c)x$ is equal to $-(cx)$ by the distributive law and the definition of additive inverses. \[ cx + (-c)x = (c-c)x = 0*x = 0 \] In the other direction, $x(-c)$ is equal to $-(xc)$ for the same reasons. \[ xc + x(-c) = x(c-c) = x*0 = 0 \] Thus if $c$ is in the centralizer $S$ then so is $-c$ \[ (-c)*x = -(cx) = -(xc) = x*(-c)\] Thus $S$ is also negation closed, so that $S$ is a subring. $\square$

The first three conditions are equally applicable to semirings. Semirings axiomatically have both the binary and nullary distributive laws, which ensures that centralizers are additive submonoids. In the ring case, we see that centralizers also form additive subgroups.

All other commutativity necessary subrings can be formed by intersections of centralizers. The center is the intersection of all centralizers. Maximal commuting cliques are equal to the intersection of the centralizers of all their elements. Commutative principal filters are formed by the centralizer of the centralizer.

Centralizers:
Let $Mat_n(F)$ be the non-commutative ring of matrices over a field $F$. Then the centralizers of $Mat_n(F)$ can be computed by solving systems of linear polynomial equations. In general, the previous theorem shows that centralizers are always subrings.

Centers:
The center of any non-commutative ring is a commutative ring, over which the non-commutative ring is a ring extension. For example, the center of the quaternions $\mathbb{H}$ are the reals $\mathbb{R}$, so that $\mathbb{}$ is a four dimensional vector space over $\mathbb{R}$. In a non-commutative ring of matrices like $Mat_n(F)$ the center consists of the commutative ring of scalar matrices.

Commutative principal filters:
The commuting preorder of a ring is the adjacency preorder of its commuting graph. Then the multiplicative iteration preorder is a suborder of the commuting preorder. The principal filters of this preorder are all commutativity necessary subrings. The center is the minimal commutative principal filter.

Maximal commuting cliques:
Let $G$ be a finite graph, then every element of $G$ is contained in some maximal clique. In the infinite case we need Zorn's lemma, and the fact that the union of a chain of cliques is a clique. Then every element of an infinite graph is in a maximal clique as well. This implies that every element in a ring is contained in some commutative subring.

Theorem. monogenic rings are commutative

Proof. By Zorn's lemma every element is embedded in a maximal clique, which is a commutative subring. Then every element of a ring is embedded in a commutative subring. A monogenic subring is a minimal subring containing a given element, so it must be embedded in some greater then or equal commutative subring. The property of commutativity is hereditary, so monogenic rings are commutative. $\square$

This should work unless there is some issue with Zorn's lemma in the infinite case. This generalizes the well known fact that all monogenic semigroups are commutative, and it shows that rings can be built up from commutative building blocks.

Implications of Lagrange's theorem:
Let $R$ be a finite ring, then Lagrange's theorem implies that all the different commutativity necessary subrings have orders that divide the order of the ring $R$. In particular, the degrees of each element of the commuting graph are divisors of its order. This means that commuting graphs of rings are a lot like those for groups, but they aren't completely the same.

For example, the commuting graphs of rings don't necessarily have to have the same degree of commuting equality which is implied for in groups by Cauchy's theorem and the number of iteration equal elements $\phi(n)$ of prime cyclic groups of order $n$. Even so, Lagrange's theorem alone places a greater constraint on the set of possible multiplicative semigroups of rings.

In the case of finite division rings, we get that subalgebras must satisfy Lagrange's theorem in two different ways, corresponding to the two types of subgroups in a division ring: \[ (d)|(n) \] \[ (d-1)|(n-1) \] This places even greater constraints on the possible commuting graphs of division rings, but we can get around worrying about that altogether by Wedderburn's little theorem [1] which shows that all finite division rings are commutative. The double Lagrange's theorem still applies to finite fields.

See also:
Commutativity necessary subsemigroups

Subalgebra lattices of finite fields

References:
Weddernburn's theorem