Set of Division Subrings forms Complete Lattice

Theorem

Let $\struct {D, +, \circ}$ be a division ring.

Let $\mathbb K$ be the set of all division subrings of $K$.

Then $\struct {\mathbb K, \subseteq}$ is a complete lattice.

Proof

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

By Intersection of Division Subrings is Largest Division Subring Contained in all Division Subrings:

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

By Intersection of Division Subrings Containing Subset is Smallest:

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 $\struct {\mathbb K, \subseteq}$ is a complete lattice.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.1$