Let $K$ be a field and $\overline{K}$ be its algebraic closure. The set $\operatorname{Latt}(K)$ of all intermediate fields $E$ (where $K\subseteq E\subseteq \overline{K}$ , ordered by set theoretic inclusion, is a poset. Furthermore, it is a complete lattice, where $K$ is the bottom and $\overline{K}$ is the top.

This is the direct result of the fact that any topped intersection structure is a complete lattice, and $\operatorname{Latt}(K)$ is such a structure. However, it can be easily proved directly: for any collection of intermediate fields $\lbrace E_i\mid i\in I\rbrace$ the intersection is clearly an intermediate field, and is the infimum of the collection. The compositum of these fields, which is the smallest intermediate field $E$ such that $E_i\subseteq E$ is the supremum of the collection.

It is not hard to see that $\operatorname{Latt}(K)$ is an algebraic lattice, since the union of any directed family of intermediate fields between $K$ and $\overline{K}$ is an intermediate field. The compact elements in $\operatorname{Latt}(K)$ are the finite algebraic extensions of $K$ The set of all compact elements in $\operatorname{Latt}(K)$ denoted by $\operatorname{Latt}_F(K)$ is a lattice ideal, for any subfield of a finite algebraic extension of $K$ is finite algebraic over $K$ However, $\operatorname{Latt}_F(K)$ as a sublattice, is usually not complete (take the compositum of all simple extensions $\mathbb{Q}(\sqrt{p})$ where $p\in \mathbb{Z}$ are rational primes).