Set of Ideals forms Complete Lattice

Theorem

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

Let $\mathbb K$ be the set of all ideals of $K$.

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

Proof

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

By Intersection of Ring Ideals is Largest Ideal Contained in all Ideals:

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

By Intersection of Ring Ideals Containing Subset is Smallest:

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 Ideal: General Result, this supremum is:

$\ds \sum_{S \mathop \in \mathbb S} \struct {S, +, \circ}$

That is:

$\struct {S_1, +, \circ} + \struct {S_2, +, \circ} + \dotsb$

where addition of ideals is as defined in subset product.

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

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.4$: Corollary