Set of Subfields forms Complete Lattice

Theorem

Let $\struct {F, +, \circ}$ be a field.

Let $\mathbb F$ be the set of all subfields of $F$.

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

Proof

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

By Intersection of Subfields is Largest Subfield Contained in all Subfields:

$\bigcap \mathbb S$ is the largest subfield of $F$ contained in each of the elements of $\mathbb S$.

By Intersection of Subfields Containing Subset is Smallest:

The intersection of the set of all subfields of $F$ containing $\bigcup \mathbb S$ is the smallest subfield of $F$ 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 subfields of $R$.

Therefore $\struct {\mathbb F, \subseteq}$ 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 $23.1$