Centralizers of division rings

We have that centralizers of commuting graphs determine subsemigroups and subrings, and corresponding to this there is the fact that centralizers of groups are subgroups and centralizers of division algebras are not simply subrings but division subalgebras. This can be used for example to determine the maximal subfields of a division algebra.

Subalgebras of division rings:
Let $R$ be a division ring, then $R$ as a ring is associated to a lattice of subrings. A meet subsemilattice of this lattice of subrings is the lattice of division subrings $Sub(R)$ of $R$, which consists of subrings that are also inverse closed with respect to non-zero elements.

The special case of centralizers:
By the theory of commutativity necessary subrings, we know that the centralizers of $R$ are subrings. The centralizers of $R$ together form a lattice: the centralizer lattice of the commuting graph of $R$. It remains to show that they are not only subrings but also division subrings.

Lemma 1. let $G$ be a group and $x$ an element of $G$ then the centralizer $C(x)$ is a subgroup of $G$.

Proof. the centralizer $C(x)$ is a submonoid, so let $y \in C(x)$ then $yx = xy$. Let $y^{-1}\in G$ then we want to show that $y^{-1}x = xy^{-1}$. By the fact that $G$ is a group, we can multiply both sides by $yx^{-1}$ to get $y^{-1}xyx^{-1} = 1$. Then by the fact that $x$ and $y$ commute this is equivalent to $y^{-1}yxx^{-1} = 1$ which is true, so that $y^{-1} \in C(x)$. $\square$

Lemma 2. let $S$ be a group with zero (respectively a Clifford 0-simple semigroup). Then centralizers of elements in $S$ are inverse closed.

Proof. let $x \in S$ and suppose that $x = 0$. Then the centralizer of $0$ is the entire semigroup, which is inverse closed. Suppose $x \not= 0$ then $x \in S-\{0\}$ which is a group, so the centralizer of $x$ in $S-\{0\}$ is a subgroup by lemma 1. It is therefore inverse closed. $\square$

Theorem. let $R$ be a division ring, then centralizers $C(x)$ are division subalgebras.

Proof. by commutativity necessary subrings, $C(x)$ is a subring and by lemma 2 it is inverse closed, so it is a division subalgebra. $\square$

The maximal cliques of the commuting graph of a division ring are precisely the maximal fields of the division ring. All the commutativity necessary subfields of a division ring are determined by maximal cliques and their intersections, which form a semilattice.

Centers of division rings:
The center of a division ring is precisely the intersection of all the maximal fields of the division ring. As it is the intersection of fields, it is a field. The division ring can then be seen to be a vector space over its central field. The measure of the commutativity of a division ring is the dimension of its extension over a central field: finite dimensional division rings (like the quaternions) are the most commutative.

Maximal subfields of a division ring:
An immediate corollary of this is that the maximal subfields of a division ring can be determined by the same general mechanism by which we determine maximal commutative subalgebras of semigroups, groups, rings, and semirings: they are maximal cliques of the commuting graph.

Proposition. let $D$ be a division ring then $F$ is a maximal subfield of $D$ iff $D$ is equal to its own centralizer: $C(D) = D$

It is not hard too see by the same reasoning that applies to any other algebraic structure like a group or ring, that every element of a division ring belongs to some maximal field.

Proposition. every element of a division ring belongs to some maximal field.

Maximal subfields of a division ring are a vital tool in the study of division rings. Their further properties can be studied using tensor products of division rings described as algebras over fields [1].

References:
[1] A first course in non-commutative rings
T.Y Lam