Set of Ideals forms Complete Lattice

Theorem

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

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

Proof

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

By Intersection of Ideals:

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

• The intersection of the set of all ideals of $K$ containing $\bigcup \mathbb S$ is the smallest ideal 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 join of the set of all ideals of $\mathbb S$.

From Sum of Ideals is an Ideal, this supremum is:

$\displaystyle \sum_{S \mathop \in \mathbb S} \left({S, +, \circ}\right)$

i.e.:

$\left({S_1, +, \circ}\right) + \left({S_2, +, \circ}\right) + \cdots$

where addition of ideals is as defined in subset product.

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

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 22$: Theorem $22.4$: Corollary