Set of Division Subrings forms Complete Lattice

Theorem

Let $\left({K, +, \circ}\right)$ be a division ring, and let $\mathbb K$ be the set of all division subrings of $K$.

Then $\left({\mathbb K, \subseteq}\right)$ is a complete lattice.

Proof

Let $\varnothing \subset \mathbb S \subseteq \mathbb K$.

By Intersection of Division Subrings:

• $\bigcap \mathbb S$ is the largest division subring of $K$ contained in each of the elements of $\mathbb S$.

• The intersection of the set of all division subrings of $K$ containing $\bigcup \mathbb S$ is the smallest division subring of $K$ containing $\bigcup \mathbb S$.

Thus:

• Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum.

• The supremum of $\mathbb S$ is the intersection of the set of all division subrings of $K$.

Therefore $\left({\mathbb K, \subseteq}\right)$ is a complete lattice.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 23$: Theorem $23.1$